Bear - A Primer of Lebesgue Integration

PREFACE TO THE FIRST EDITION

This text provides an introduction to the Lebesgue integral for advanced undergraduates or beginning graduate students in mathematics. It is also designed to furnish a concise review of the fundamentals for more advanced students who may have forgotten one or two details from their real analysis course and find that more scholarly treatises tell them more than they want to know.

The Lebesgue integral has been around for almost a century, and the presentation of the subject has been slicked up considerably over the years. Most authors prefer to blast through the preliminaries and get quickly to the more interesting results. This very efficient approach puts a great burden on the reader; all the words are there, but none of the music. In this text we deliberately unslick the presentation and grub around in the fundamentals long enough for the reader to develop some intuition about the subject. For example, the Caratheodory definition of measurability is slick—even brilliant—but it is not intuitive. In contrast, we stress the importance of additivity for the measure function and so define a set £ e (0,1) to be measurable if it satisfies the absolutely minimal additivity condition: m(E) + m{E) = 1, where E^ = (0,1) — E and m is the outer measure in (0,1). We then show in easy steps that measurability of E is equivalent to the Caratheodory criterion, m(E n T) + m(E^ oT) = m(T) for all T. In this way we remove the magic from the Caratheodory condition, but retain its utility. After the measure function is defined in (0,1), it is extended to each interval (n, n + 1) in the obvious way and then to the whole line by countable additivity.

viii A PRIMER OF LEBESGUE INTEGRATION

We define the integral via the famihar upper and lower Darboux sums of the calculus. The only new wrinkle is that now a measurable set is partitioned into a finite number of measurable sets rather than partitioning an interval into a finite number of subintervals. The use of upper and lower sums to define the integral is not conceptually different from the usual process of approximating a function by simple functions. However, the customary approach to the integral tends to create the impression that the Lebesgue integral differs from the Riemann integral primarily in the fact that the range of the function is partitioned rather than the domain. What is true is that a partition of the range induces an efficient partition of the domain. The real difference between the Riemann and Lebesgue integrals is that the Lebesgue integral uses a more sophisticated concept of length on the line.

We take pains to show that both the Riemann-Darboux integral and the Lebesgue integral are limits of Riemann sums, for that is the way scientists and engineers tend to think of the integral. This requires that we introduce the concept of a convergent net. Net convergence also allows us to make sense out of unordered sums and is in any case something every young mathematician should know.

After measure and integration have been developed on the line, we define plane outer measure in terms of coverings by rectangles. This early treatment of plane measure serves three purposes. First, it provides a second example of the definition of outer measure, and then measure, starting with a natural geometric concept—here the area of a rectangle. Second, we show that the linear integral really is the area under the curve. Third, plane measure provides the natural concrete example of a product measure and is the prototype for the later development of general product measures.

The text is generously interlarded with problems. The problems are not intended as an intelligence test, but are calculated to be part of the exposition and to lure the reader away from a passive role. In many cases, the problems provide an essential step in the development. The step may be routine, but the reader is nevertheless encouraged thereby to pause and become actively

PREFACE TO THE FIRST EDITION Ix

involved in the process. There are also additional exercises at the end of each chapter, and the author earnestly hopes that these will add to the reader's education and enjoyment.

The author is pleased to acknowledge the help of Dick Bourgin, Bob Burckel, and Ken Ross, all of whom read the manuscript with great care and suggested many improvements in style and content.

PREFACE TO THE SECOND EDITION

The principal change from the first edition is the new one-shot definition of the Lebesgue integral. The integral is first defined for bounded functions on sets of finite measure, using upper and lower Darboux sums for finite partitions into measurable sets. This approach is designed to emphasize the similarity of the Lebesgue and Riemann integrals. By introducing countable partitions, we then extend the definition to arbitrary functions (bounded or not) and arbitrary sets (finite measure or not). This elegant touch, like many of my best ideas, was explained to me by A. M. Gleason.

Many of the errors and crudities of the first edition have been corrected, and the author is indebted to Robert Burckel, R. K. Getoor, K. R S. Bhaskara Rao, Joel Shapiro, and Nicholas Young for pointing out assorted mistakes. In addition, several anonymous reviewers of the second edition made many helpful suggestions. I feel confident, however, that there remain enough errors to challenge and reward the conscientious reader.

Finally, the author wishes to express his gratitude to Susan Hasegawa and Pat Goldstein for their superb work with the typing and proofreading.

H. S. Bear May 2001

1 THE RIEMANN-DARBOUX INTEGRAL

We start by recalling the definition of the familiar Riemann-Darboux integral of the calculus, which for brevity we will call the Riemann integral. Our later development of the Lebesgue integral will closely parallel this treatment of the Riemann integral. We consider a fixed bounded interval [a, b] and consider only real functions f which are bounded on [a,b].

A partition P of [ , ^] is a set P = {XQ, xi, Xi , . . . , Xn} of points of [a, b] with

a = XQ < Xi < X2 < ' " < Xn = b.

Let /" be a given function on [a, b] with

w < f(x) < M

for all X e [a, b]. For each / = 1, 2 , . . . , w let

mi = inf{ f(x) : X/_i < x < Xi},

Mi = sup{ f(x) : Xi-i < X < Xi).

In the usual calculus text treatment the infs and sups are taken over the closed intervals [x/_i,X/]. In our treatment of the Lebesgue integral we will partition [a, b] into disjoint sets, so we use the disjoint sets (x/_i, Xi) here. We are effectively ignoring a finite set of function values, f(xo), f(xi), fixi),..., f{Xn), and in so doing we anticipate the important result of Lebesgue that

2 A PRIMER OF LEBESGUE INTEGRATION

function values on a set of measure zero (here the set of partition points) are not relevant for either the Riemann or Lebesgue integral.

The lewder sum L{f, P) and the upper sum U(f, P) for the function f and the partition P are defined as follows:

n

L{f, P) = ^mi{Xi - x /_ i ) ,

i=l

Clearly m<mi< M/ < M for each /, so

m{b-a) < Uf,P) < U{f,P) < M(b-a).

For a positive function f on [a, b] the lower sum represents the sum of the areas of disjoint rectangular regions which lie within the region

S = {(x, y) :a <x<b,0 <y < f{x)].

Similarly, the upper sum U{f, P) is the area of a finite number of disjoint rectangular regions which cover the region 5 except for a finite number of line segments on the lines x = Xj. The function f is said to be integrable whenever

sup L(/-,P) = infU(/-,?). (1) p P

The integral of f over [a, b] is the common value in (1) and is denoted /^ f. The area of S is defined to be this integral whenever f is integrable and non-negative.

The integral is also sometimes written /^ f(x) dx^ particularly when a change of variable is involved. The "x" in this expression is a dummy variable and can be replaced by any variable except f or d. For example,

/ f{x)dx = [ f(t)dt = f f(a)da = f f{c)dc. J a J a J a J a

The last two versions are logically correct, but immoral, since they flout the traditional roles oi a,b,c as constants, and the

1 THE RIEAAANN-DARBOUX INTEGRAL 3

third integral discourteously uses the same letter for the limit and the dummy variable.

Now we make a few computations to derive the basic properties of the integral and show that the integral exists at least when f is continuous.

A partition Qis a refinement of the partition P provided each point of P is a point of Q. We will indicate this by writing P <Q without reference to the numbering of the points in P or Q. Clearly Q is a refinement of P provided each of the subintervals of [a, b] determined by Qis contained in one of the subintervals determined by P.

Proposition 1. IfP<Q, then L( f, P) < L(/*, Q) andU( f, Q)< U(f,P).

Proof. Suppose Q contains just one more point than P, and to be specific assume this additional point x* lies between the points xo and xi of P. If

m[ = inf{ f(x) : Xo < X < X*}

n/l = inf{ f(x) : X* < X < Xi}

mi = inf{ f(x) : Xo < X < Xi},

then 7f/^ > mi and m![ > mi so the sum of the first two terms in L( f, Q) exceeds the first term of L( f, P):

m[(x^ — Xo) + m^[(xi — x*) > mi(xi — Xo).

The remaining terms of L( /*, Q) and L( /*, P) are the same, so L(f, Q) > L(f, P). We can consider any refinement Q of P as obtained by adding one point at a time, with the lower sum increasing each time we add a point. The argument for the upper sums is similar, ill

Proposition 2. Every lower sum is less than or equal to every upper sum, as the geometry demands.

Proof. If P and Q are any partitions, and R = P U Q is the common refinement, then

L{f, P) < L(f, R) < U(f, R) < U{f, Q). ill


Proposition 3. f is integrable on [a, b] if and only if for each 8>0 there is a partition P of [a,b] such that U{fP) — L{fP)<8.

Proof This useful condition is equivalent to sup L{f P) = inf U( /*, P) in view of Proposition 2. ill

Proposition 4. If f is integrable on [a, b] and [a, ^] <z[a, b\ then f is integrable on[a, ^].

Proof Let s > 0 and let P be a partition of [a, b] such that U(f P) — L(f P) < 6. We can assume that a and ^ are points of P, since adding points increases L( /*, P) and decreases U(f P) and makes their difference smaller. If PQ consists of the points of P which are in [a, ^ ] , then PQ is a partition of [a, jS]. Note that

U{f P) - Uf P) = X](M, - m,)(x, ~ x,_i) (2)

and U( f PQ) — L( f PQ) is the sum of only those terms such that Xi-i and Xi e PQ. Since we omit some non-negative terms from (2) togetU(/ ' ,Po)-L(/* ,Po) ,

U(f Po) - L{f Po) < U(f P) - Uf P) < 8, ill

Problem 1. If f is integrable on [a, b]^ then —f and | f\ are integrable on [a, b], and Xf (-/") = - / . V , |xf/-| < Xf | f\. II

Problem 2. li a < c < b then / is integrable on [a, c] and on [c, /?] if and only if f is integrable on [a,b]. In this case

rc pb rb

f+ f= f. •" Ja Jc Ja

•Hill

Problem 3. If /"is integrable on [a, b] and g = / except at a finite number of points, then g is integrable and f^ g = J^ f. ""HI

Problem 4. li a = XQ < xi < - -- < Xn = b and f is defined on [a, b] with /"(x) = yi for x € (x/_i, X/), then f is integrable and

- x,_i). (Note that it is immaterial how f is defined on the %/, by the preceding problem.) ""HI


Problem S. We say that g is a step function on [a, b] if there is a partition a = XQ < Xi < xi < - - < Xn = b such that g is constant on each (x/_i, X/). (By the preceding problem, step functions are integrable with the obvious value for the integral.) Show that if f is integrable on [a, b] there are step functions gn and hn with {gn} increasing and {hn} decreasing and gn < f < hn for all n and all x, and \im J^gn = HmJ^hn = Xf f- (Note: We will be able to show later that gn alid hn approach f except possibly on a set of measure zero.) ""HI

Proposition 5. If f is continuous on [a, b], then f is integrable on [a, b\.

Proof. If f is continuous on [a,b\ then f is uniformly continuous. Hence iis > 0 there is 5 > 0 so that \f{x) — f{x')\ < s whenever \x — x^\ < S.li P is a partition with xi — Xi-\ < 8 for all /, then M/ — mi < s for each /, so

U(f P) - L(f P) = J2(M, - m,){x, - x,_i)

< 8 Y^{Xi - Xi-i)

= s(b-a), ill

Problem 6. (i) If f is continuous on [a, b] except at a (or b) and f is bounded on [a,b]^ then f is integrable on [a,b].

(ii) If f is bounded on [a, b] and continuous except at a finite number of points, then f is integrable on [a,b].

(iii) Suppose f is bounded on [a, b] and discontinuous on a (possibly infinite) set £ . Assume that for each ^ > 0 there are disjoint intervals (ai, b\),..., {a^, b^) contained in [a, b] such that E c (ai, bi) U ' " U (a^, ^N) and TJiLxih ^ ^i) < ^- Show that f is integrable. ""HI

So far we have the integral /^ f defined only when [a, b] is a bounded interval and f is bounded on [a,b]. Now we extend the definition to certain improper cases; i.e., situations where the interval is unbounded, or f is unbounded on a bounded interval. Typical examples of such improper integrals are

/ —z^dx and / -dx. Jo ^x Jo 1 + x^

A PRIMER OF LEBESGUE INTEGRATION

In both these examples the integrand is positive and the definition of the integral should give a reasonable value for the area under the curve. The definitions of the integrals above are

/ —zz: dx = lim / —7= dx, Jo y/x £-^0+ Js ^X

coo \ fb 1

/ :; : dx = lim / dx. Jo 1 + X^ b-^oQ Jo 1 -\- X^

Both these limits are finite, so both functions are said to be (improperly) Riemann integrable on the given interval. The Lebesgue definition of the integral will give the same values.

In general, if f is integrable on[a + s.b] for all £ > 0, but f is not bounded on [a, b\ (i.e., not bounded near a)^ we define

b

f= l imj f, J a s—^0+ Ja-\-s

provided this limit exists. Similarly, if f is integrable on every interval [a, b] ior b > a we define

POO rb

J a b—^00 J a

when the Hmit exists. Similar definitions are made for Xf f if f is unbounded near /?, and for /^^ /" if /* is integrable on [a, b] for all a < b.

These definitions lend themselves to the calculations of elementary calculus, but do not coincide with the Lebesgue definition if f is not always positive or always negative. For example, if f is {—ly/n on [n,n + 1), n = 1 ,2 , . . . , then f is improperly Riemann integrable on [1, oc). We will see later that f is Lebesgue integrable if and only if | /*| is Lebesgue integrable. Hence the above function is not Lebesgue integrable since ^ - = 00.

Problem 7. Show that /o~ '-^dx exists. "Hlll

Problem 8. (i) Exhibit a g on [0, oo) such that \g(x)\ = 1 and /o~ g exists.


(ii) Exhibit a function g on [0, oo) such that \g(x)\ —> oo as X —> oo and J^ g exists.

(iii) Exhibit a function g on [0, oo) so that \g(x)\ —> oo and J^ g z= 0. Hint: Don't think about formulas for continuous functions, think about step functions. ""HI

Problem9. Let f(x) = x^ on [0,1] andlet P^ = {0, i , ^ , . . . , 1}. Write formulas for L(f,Pn) and for U(f,Pn). Show that

Ji x^dx=\. < Problem 10. Write out the proof that U(f, Q) < U(f, P)

whenever P < Q. ""HI

Problem 11. Ii f is integrable on [a, b] and c is a number, then Xf cf = c Xf f. Show this for the case c < 0. ""HI

Problem 12. Ii f and g are integrable on [a, b] and f < g on [a,b],theni'f<i'g. <

Problem 13. Let xvy = max{x, y} and x A y = min{x, y}. For functions f and gy (f V g)(x) = f(x) V g(x)^ (/ A g)(x) = f(x) A g(x). Show that if f and g are integrable on [a,b]^ then fvgand f Ag are integrable, and fa f ^ g > la f ^ la S.

Iaf^g<Iaf^Iag' "

THE RIEMANN INTEGRAL AS A LIMIT OF SUMS

The engineers and scientists who work with integrals think of the integral as a hmit of sums:

/ Ja

b « f(x)dx = l i m ^ f(Xi)Axi (!)

where a = xo < Xi < xi < -- - < Xn = b is a partition of [a, b] and Axj = Xj — Xi-i. Historically the integral sign is a flattened S for sum, and the "Jx" suggests the typical (small) quantity Axi. The limit is the number that is approached as partitions are made finer and finer so that max Axi tends to zero.

Notice that the limit in (1) is a new animal. The sums in question depend on partitions. That is, we want the limit of a function of partitions, rather than a limit of a function of x or n. Assume for the moment that the limit in (1) makes sense for each of two functions / and g defined on [a, b]. Then

rb

/ (f + g) = limJ2(f(Xi) + g(Xi))Axi Ja

= lim ( ^ f(Xi)Axi +^g(Xi)Axi)

= l i m ^ f{Xi)Axi + X\mY^g{Xi)Axi nb rb

= 1 f^ Ja Ja

rb

g'

However we define the limit above, the limit of a sum is certainly the sum of the limits (i.e., the third equality), for that is a basic


property of addition: if a is close to A and b is close to B, then a + bis close to A + B.

In this chapter we define a kind of limit, generalizing the idea of the limit of a sequence, which makes precise the ideas above. The objects we use to generalize sequences are called nets, and convergence of nets is sufficiently general to describe any kind of limit which occurs in analysis or topology.

A directed set is a non-empty set D equipped with a partial ordering -< satisfying the following conditions:

(i) a -< a for all a e D\ (ii) if a ^ ^ and fi < y^ then a < y;

(iii) for any elements a and ^ in D there is y € D so that a < y and ^ < y.

We will write ^ : a to mean the same as a -< ^, and say that P is farther out than a when this relation holds.

A net is a function defined on a directed set. A sequence is a type of net, with D being the set of natural

numbers directed as usual: n < m means n < m. We will adopt the sequence notation for nets and write {Xa} for the net consisting of the real-valued function x defined on some directed set D of elements a. We use some index other than w, for instance or, to emphasize that D need not be the set N of natural numbers.

The net {Xa} converges to £, denoted Xa—>^^ or lim Xa=^^ provided that for every s > 0 there is ofo G D such that \Xa—t\<s whenever a >- ao. Of course the limits of nets are unique if they exist, justifying the notation lim x« = £. The uniqueness is a consequence of the fact that there is some y beyond any given a and )6. Hence if £i and £2 were different limits, with £2 — ^1 = ^ > 0, and \Xa—l\ < 6:/2whenof >- aoand|x^-£2l < 6:/2when^ > j8o? then if y >- ao and y : JSQ, the two contradictory conditions

\Xy — - ll < S/2, \Xy — III < ^/2

would both hold. The familiar limit of the calculus, lim f(x) = £, is another

X—>a

instance of a net limit. Here D consists of the points x near a and X > y means x is closer to a than y: 0 < \x — a\ < \y — a\.

2 THE RIEAAANN INTEGRAL AS A LIMIT OF SUMS 11

Problem 1. Describe D and < so these limits are limits of nets:

(i) lim fix) = i\ X ^00

(ii) lim fix) = i; X—>a+

(iii) Y. ai = i. ""Ill

Proposition 1. Let {Xa} and {);«} be real-valued nets on the same directed set D, with lim« x^ = t, limQ, ja = m. Then

(i) lim(x« + y^) = i + m; a

(ii) \im(Xa - ya) =i-m;

(iii) lim(Xaya) = ^ ;

(iv) lim(Xa/y(x) = ^ / ^ if^ 7 0, y 7 0 for all a.

Proof. The proofs of these statements, which are basically just familiar properties of addition and multiplication, are virtually the same as the corresponding statements for sequences or functions. We prove (iii) by way of illustration.

AssumeXa —> ândya —> m,orequivalently,Xa—i —> 0, ya — m —> 0. Let ra = Xa — i and Sa = ya — for all a. Then â —> 0 and Sa —> 0 and

Xaya = (^+ra){ni+Sa)

= im + rîm + Sai + TaSa-

Fix ^ > 0 and pick ofi, beyond which Ir ;! < 1 and in addition \ra\ is so small that \ram\ < s. We similarly pick 02 so that beyond 0 2, \Sa\ < s and |SQJ£| < s. There is QO SO that ao > OL\

and Qfo > 0 2. Hence if a :v ofo, we have

\roim\ < s, \rc\ < 1, \Sal\ < s, \Sa\ < s,

so

\Xaya - ^ni\ = \ram + Sai + raSal

< \ram\ + \Sal\ + KSal < ^S. ill


Problem 2. Prove parts (i) and (iv) of Proposition 1. ""HI

Problem 3. (i) If XQ. > 0 for all a in a directed set D and Xa —> I, then l>0.

(ii) If Xa < ya < Za for all a e D, and Xa —> I, Za —> ? then ja -^ L ""III

Problem 4. Let D be the set of all pairs (m, n) of positive integers. Partially order D as follows: (m, w) >- (m^ n^) if and only ii m + n > m^ + n\

(i) Describe geometrically v^hat (m, w) > (m^ n^) means. (ii) Show that if lim x^„ = £, then lim Xmn = ^ for all m,

and lim Xmn = ^ ior all w.

(iii) Let :x;„ „ = mn/{rr^ + n^). Show lim x^ „ = 0 for all n and lim :KV„ „ = 0 for all m but lim x^ „ fails to exist. ""HI

Problem 5. Let D be the set of all pairs (m, n) of natural numbers, with the partial ordering (m, n) > (mo, ^o) iff max{m, n} > max{mo, no).

(i) Describe geometrically the set of (m, n) such that (m, n) >-(mo, Wo) for a fixed (mo, wo).

(ii) Does lim x^n = i imply lim x^„ = ^ for all m. ""HI im,n) ' w—>oo

Problem 6. Let D be as above with the ordering (m, n) > {m\n') iff mn > m'n\ Give examples of nets {x , } which converge and nets which diverge. What is the connection, if any, between convergence in the ordering of D and the limits

lim Xnt^n. lim x^,„? ""HI m—>OQ n—^00

The nets we want to consider here, and later for the Lebesgue integral, are nets of Riemann sums. Again let f be any real function on [a, b] and let P = {XQ, Xi , . . . , X„} be any partition of [a,b].A choice function c for the partition P is a finite sequence Ci, ^ 2 , . . . , c:„ with Ci e (x^-i, Xi) for each /. The Riemann sum for /", P, and c is

P( / ' ,P ,c) = f^/*(c,)(x,-x,_i) .

2 THE RIEMANN INTEGRAL AS A LIMIT OF SUMS 13

The Riemann sum R(f, P,c)^ for fixed f, is a real-valued function which depends on the partition P and the choice function c. Thus {R(f, P,c)} becomes a net when we put an appropriate partial ordering -< on the pairs (P, c). We do this as follows:

(PuCt)>(P2,C2)iiiPi>P2.

Thus the pairs are ordered by the partitions themselves in the sense that (Pi, ci) is farther out than (P2, C2) if Pi is a refinement ofP2.

We say that a real-valued net {Xa] is increasing provided x^ > Xa whenever ^ > a. K decreasing net is defined similarly. The net {Xa] is bounded provided there are numbers b and B so that b <Xa < B ioY all a.

Problem 7. Show that an increasing bounded net [Xa] converges to £ = supjXoj '.oteD). ""Ill

The lower sums L{f, P) and the upper sums U{f, P) for a bounded function are nets, with the partitions ordered by refinement. If f is the function in question and m < f{x) < M, then the lower sums and upper sums are bounded. The lower sums form an increasing bounded net, and hence converge, and similarly for the upper sums. Clearly f is Riemann integrable if and only if

limL(/-,P) = lim U(f,P). (2)

If (F, c) is a partition of [a, b] with a choice function c, then

m < fid) < Mi

for each /, so

L{f,P)<R(f,P,c)<U(f,P). (3)

It follows immediately from (2), (3) that if f is integrable, then

limR(f,P,c)= tf. (4) ^ Ja


We want to show that the existence of the Umit Um R( /*, P, c) provides an alternative characterization of integrabiUty. Notice, however, that to talk of lower and upper sums we need to assume that f is bounded. The Riemann sums, on the other hand, are defined even if f is not bounded, as long as f is defined on all of [a, b]. Conceivably the limit in (4) could exist for an unbounded function f^ and we would have two distinct definitions of the integral. The next proposition resolves this question.

Proposition 2. If f is defined on [a, b] and lim R(f P, c) exists, then f is bounded.

Proof. Let lim R(fP,c) = I and assume that f is not bounded above. Let P be a partition of [a, b] such that for all choices c,

\R{fP,c)-I\<l,

In particular \R{ f P, c) — R(f P, c^)\ <2 for any choices c and c^ for P. Since f is unbounded, f is unbounded on some subin-terval, which we will assume is (XQ, Xi). Fix any choice c for P. Let c- = c/ for / = 2 , . . . , w, and choose c[ so f(c[) > N. Then

R{fP, d) -R(fP^c) = {f {c[) - f{cr)) Axi

> ( N - /'(ci))Axi.

We can choose N so large that the two Riemann sums differ by more than 2, which is a contradiction, ill

Proposition 3. A function f is integrable on [a, b] if and only if lim R(f P, c) exists, and of course the limit is the integral in

this case.

Proof. We have only to show that the existence of the limit implies that the function is integrable. To do this, fix ^ > 0 and choose a partition P so that \R(fP,c) — I\<6 for all choices c, which implies that for any two choices c, c^ for this P.,\R(fP,c)— R(f P, cOI < 2s. We will choose c and c^ so that

L(fP)<R(fP,c)<L(fP) + 8 (5)

U(fP)-8<R(fP,c')<U(fP). (6)


This implies that U( f, P) - L( /", P) < 3 s. To see how c and c^ can be chosen to satisfy (5) and (6), let

P = {XQ, Xi, . . . , Xn}'

For each /, choose Q so that

m < f(Ci) < nti + 8/(b - a).

Then

Û^ P) = Ylî^î <Y^f(Q)Ax, = R(f,P,c)

< ^ I m; + -; I AXi

= V miAxi + -J V Axi ^^ b - a

= L(f,P)+8.

We similarly choose c^ so that for each /,

M, > f{c'^>Mi-6/(b-a),

and we have (6) by the same argument as above, ill

Proposition 4. If f and g are integrable on [a, b] and k is a constant, then

(i) Ia(f + g) = Iaf + Iag; (ii)L'kf = ki'f; (Hi) i'f>Oiff>0.

Proof. For every partition-choice pair (P, c),

R(f + g, P, c) = R(f, P, c) + R{g, P, c)

R(kf,P,c) = kR(f,P,c).


Hence (i) and (ii) follow from the general results for nets in Proposition 1. Similarly, R(f, P,c) >0 for all (P, c) ii f > 0, solim RC/", P,c) > 0 . ill

Problem 8. If F is a continuous function on R and {Xa} is a net such that Xa —> x, then {F(Xa)} is a convergent net and F{^a)—>F(x). As a special case, if Xa—>x then \Xa\—>\x\. Apply this to show that if f and hence | f\ are integrable, so that Xf f= lim R{ /•, P, c) and Xf I f\= Hm R(| i |, P, c\ then |/,^ ^| <

liifi I Proposition 5. If f is integrable on [a, b] and F is a continuous

function on [a, b], and differentiable on {a, b) with F'{x) = f{x) on {a, b), then

I f=F(b)-F(a), Ja

Proof. Let £>0 and let PQ be a partition such that |P( f,P,c) — J^ f\ < 6 whenever P > PQ and c is any choice function for P. That is,

rb

Y^ f(Ct){Xi -Xi-i) - f Ja

Ja

< S

<e, (7)

whenever P > PQ and c is any choice for P. The hypotheses for the ordinary Mean Value Theorem for F hold on each interval [x,_i, Xi]. Therefore there is c, e (x,_i, x,) for each / so that

F{Xi) - F(xi-i) = F'(Ci)(Xi - Xi_i)

= f(Ci)(Xi - Xi^i). (8)

Let c be a choice function for P such that (8) holds. Then from (7) and (8) we get

J2(F(x,)-F(x,.,))- [ f < £.

2 THE RIEAAANN INTEGRAL AS A LIMIT OF SUMS 17

Since E(Fte) - FU_i)) ^F(b)-F(a).

•F(h)-F(a)- tf Ja

< S.

Since e is arbitrary, the left side is zero, and f^ f = F(b) — F(a). ill

Consider the problem of summing an arbitrary collection of numbers. Say a little boy hands you a basket of numbers and asks you to add them—what do you do? You empty the numbers out on the floor, kick them into a row, and start adding from left to right. If there is a finite number of numbers, then there is no problem. If there is an infinite number of numbers in the basket, then you keep adding from left to right until you determine a limit, and that is the sum. The difficulty with this process is that if you sweep up all the numbers, put them back in the basket, and repeat the process, you will likely get a different answer. Indeed, unless the positive numbers and negative numbers separately have finite sums you will almost surely get different answers on your second and subsequent trials. The point is that a conditionally convergent series is a very artificial thing, unless you have some real reason to want the numbers to appear in a given order. The unordered sum defined next gives a more convincing generalization of finite-sum addition.

Let A be any "index set" of elements a, and let Xa be a real number for each a e A. For any finite subset F c A define 5^ to be the finite sum J2aGF â- Partially order the finite subsets f of A by inclusion: Fi >- F2 if Fi ^ ^2- Then {Sp} is a net on this partially ordered set. If {5^} converges to L we write YâeAâ = L and say the x^ are summable.

Problem 9. (i) Show that if lim Sp = L exists, then at most countably many Xa are non-zero; i.e., there is a countable subset C C A so that Xa = 0 a a ^ C. (Hint: Suppose first that all Xa > 0. If uncountably many Xa > 0, then there is n such that Xa I for uncountably many a.) Observe that this shows that countable additivity is the most one can ever ask for. There is no such thing as (non-trivial) uncountable addition.


(ii) Show that if X a A a = ^? then the set of positive Xa is summable and the set of negative x« is summable and

aeA^ aeA~ aeA

w^here A+ = {a : Xa > 0}, A" = {a : Xa < 0}. (iii) If {Xa : a e A} is summable, then {\Xa\ : a G A} is sum

mable and

aeA aeA+ aeA^

(iv) If E~ ^ Xn is absolutely convergent, then {Xn : n e N} is summable, and conversely. In either case, E^l" = E ^ N - '"""

Let {Xa} be a net on the directed set D. We will say that {Xa} is a Cauchy net provided that for each s > 0 there is ao G D so that |x^ — x,,! < £ whenever ^ > a^ and y >- ofo-

Proposition 6. I/{Xc } /s a Cauchy net, then [Xa] converges.

Proof. For each n pick Qf„ so that |x^ — Xy| < ^ when jS, y >- a„. We can assume that ai < a2 < oti, < - - hy replacing 02 if necessary by 02 with 02 > a i , 02 ^ ofi, and 03 by 03 with 03 : 0 3 and 03 : ^2, etc. Then [XaJ is a Cauchy sequence, so there is a limit £, and given £ > 0 there is N with |XQ, — l\ < S. We can assume ^ < 6:. Then if ^ >- a^, l ys — Xa^\ < ^ and l ttN ~ -1 < 5 so Ix^ - €| < 26: if )S ^ ofN. •

Problem 10. Every convergent net is a Cauchy net. ""HI

Problem 11. If A is an index set and {Xa'.ae A} is summable, and A=UA„ where the A„ are disjoint subsets of A, then {Xai OL G A„} is summable for each n and

Y:^'^=Y.II^O.. •"I'll aeA ne^aeAn

Problem 12. (i) If {Xmn} is summable over N x N, then

(m,n) m=l w=l

00 00


(ii) Suppose the iterated (ordered) sums both exist. Does it follow that {x^n} is summable?

(iii) Suppose the iterated sums both exist and x^n ^ 0 for all m, n. Does it follow that {Xmn} is summable? ""HI

Problem 13. If F(m, n) = \{m + n — 2)(m + n — 1) + n^ then f is a one-to-one function on N x N into N. Hence the set of all pairs (m, n) is countable, and any countable union of countable sets is countable. Hint: If f(x) = | ( x —2)(x— 1), then F(m, n) = f(m + n) + n. Show that f(x + 1) - f(x) = x - 1, and conclude that iim + n = i + j + ljF (m, n) > F (/, / ) . Does F map N X N onto N? ""111

Problem 14. Let -< and 0 be two partial orderings, both of which make D a directed set. Suppose a < ^ implies a 0 ^ for all a, ^ e D. Let {Xa} be a net on D, and let lim Xa and limx^;

< 0 denote the limits with respect to the two orderings. Show that if lim Xa = I, then lim x« = £. ""HI

@ <

The next problem shows that the Riemann integral can be characterized as a limit of Riemann sums, where the partitions are not directed by refinement but by insisting that the length of the maximum subinterval tends to zero. We will use this result later to characterize the Riemann integrable functions as those which are continuous except on a set of measure zero.

Problem IS.liP = {XQ, . . . , X„} is a partition of [a, /?], define the norm of ? , denoted ||P||, by ||P|| = max(x/ — Xi-i). Let ^ 0 Q mean that || Q\\ < || P ||. Show that 0 makes the partitions P and the pairs (P, c) into directed sets. Let lim stand

. . IIP|l->o

for the limit with the direction 0 . Use Problem 14 to show that if lim P( /*, P,c) = L then f is Riemann integrable with

| |P||->0 ^' ' ' I

integral / . Conversely, lim R(f,P,c) = I implies lim R(f, P,c) = I.

P ||P||—>0

Prove this. Hint: Let PQ = {XQ, . . . , Xn} be a partition of [a, b] such that U( f, PQ) - L( /", PQ) < s, and so | R( /", PQ, CO)-I\ <8. Let Q = {yo, y i , . . . , y } be another partition with || Q\\ = 8 < min(x/ — x/_i). Let / be the set of all indices / for Q such that (yy_i, yj) is contained wholly within some (x/_i, Xi) of PQ. If Mj


and nij are the sups and infs of f for Q, then

^ ( M ; - m,) Ay,- < U{f, Po) - L(f, Po).

If K denotes the Q indices not in / (so all / such that (y/-i, y/) contains some x/), then

^(Mj — mj)Ayj < n(M - m)8 jeK

where M and m are bounds for f on [a,b]. Consequently, if

II Q\\ -8 < e/n{M - m), then

[/(/•, Q) - L(f, Q) < Uif, Po) - L(/*, Po) + s< Is,

andsolRC/*, Q , c ) - I | <l8. <

LEBESGUE MEASURE ON (0, 1)

Let f be the characteristic function of the rational numbers in (0,1); i.e., the function which is one on the rationals and zero elsewhere. Then for any partition P = {XQ, Xi,.,,, Xn} oi [0,1], nti = 0 and M/ = 1 for all /. Here, as usual, nti and Mi are the inf and sup of the function values on (x^-i, Xi). Since L(f, P) = 0 and (/(/", P) =: 1 for all partitions P, / is clearly not Riemann integrable.

Now recall the geometric interpretation of the integral as the area under the graph of f. If f is the characteristic function of the rationals in [0,1], then the region under the graph of f is a very simple one; it is just the "rectangle" Q x / where Q is the set of rationals in [0,1] and / = [0, 1]. The area of this rectangle obviously ought to be the length of Q times the length of / . Once we have a sensible definition for the length of Q we will have a reasonable value for the integral of f.

In this chapter we extend the idea of length from intervals to all subsets of R. This generalized length, called the (Lebesgue outer) measure of a set, will assign measure zero to Q (and all other countable sets) so that we will have no difficulty agreeing that ff = 0 when f is the characteristic function of the rationals. The difference between the definitions of the Riemann and Lebesgue integrals consists in just this fact: for the Lebesgue integral we allow partitions into sets more general than intervals^ and this requires that we can assign a length to these partitioning sets. With this one variation, the definition of the Lebesgue integral will be the same as the definition of the Riemann integral.

21


We now proceed with the definition of the measure function m. We restrict our attention initially to subsets of the open unit interval U = (0,1). This will ultimately give us also the measure of subsets of any interval (n, n + 1 ) , since we want measure to be invariant under translation. Countable sets will turn out to have measure zero, so we will finally define the measure of any set £ to be the sum of the measures of the sets En(n,n +l),n = 0 ,±1 ,±2 , . . . .

For any interval I we let 1{I) denote the length of I. An interval can be open, closed, or half-open, but not just a single point. Hence i(I) > 0 for every interval / . Roughly speaking, we define the measure of a set E to be the minimum of the sums of the lengths of families of intervals which cover £. To make this precise, we say a finite or countable family {Jy} of intervals is a covering of £ if £ c U/y. The family is an open covering if all the Ij are open intervals, and a closed covering if all the Ij are closed intervals. The total length of the family {Ij} is Y^l{Ij). Finally, we define the measure of £ to be the infimum of the total lengths of all coverings of £:

m(£) = i n f | ^ £ a , ) : £ c U J , | .

Let us check that it does not matter in this definition whether we use open intervals or closed intervals or a mixture. For each open covering {Ij} of £ there is a closed covering {Ij} of the same total length, so using closed coverings might conceivably give a smaller value for m(£). However, for each closed covering {Ik} of £ there is an open covering {Jk} whose total length is less than J2^(h) + ^- (Let Jk be an open interval containing 4 with ^Uk) < ^(h) + ^/2^.) Hence to each closed covering {//} of £ there are open coverings with total lengths arbitrarily close to the total length of {/y}, so the infima of total lengths of coverings are the same.

If £ is a compact set in (0,1)—i.e., a closed bounded set— then every open covering of £ has a finite subset which covers £; this is the Heine-Borel Theorem. It follows that if £ is compact, m(£) is the inf of the total lengths of finite open coverings.

3 LEBESGUE MEASURE O N (0, 1) 23

By the discussion above, we could also use finite closed coverings to find m(E) if E is compact. These remarks are perhaps unnecessarily complicated for defining measure on the line, since we could simply stick with coverings by open intervals. However, when we consider plane measure in Chapter 9 it will be convenient to know that the covering sets can be open or closed (rectangles), or anything in between.

The number m(E) is the Lebesgue outer measure of £ , although we will refer to m(E) simply as the measure of £ . Here are some immediate properties of m.

Proposition 1. (i) 0 < m(E) < 1 for all E C (0,1); (ii) m(E) < m(F) if E c F (m is monotone); (Hi) m(0) = 0; (iv) m ({x}) = 0 for all x.

Problem 1. Write out the very short proofs of parts (i), (ii)^ (iii), (iv) of Proposition 1. Note that subsets of sets of measure zero have measure zero. ""HI

Since m(E) is to be a generalization of length, we need to know that m(I) = i{I) for every interval I C (0,1). That is the content of the next proposition and problem.

Proposition 2. If J = [a, b] c (0, 1), or if J = (a, b) C (0, 1), thenm(J) =i(J).

Proof First assume / is a closed interval [a,b]. Clearly m(J) < I(/) since {/} is a one-interval covering of / of total length i(J). We show by induction that if / i , . . . , J„ is any finite covering of / by intervals, then €(/) < ELi^(^^)- ^ / ^ covered by one interval Ji, then clearly €(/) < €(/i). Suppose as our inductive hypothesis that whenever / is a closed interval covered by n or fewer open intervals I^,,,., I^ (m < n)^ i(J) < Y!k=i ^(h)- Let J = [a,b] C hU - 'U 4+1, and assume no n of these intervals cover / . If any 4 is disjoint from / we are done. Let us assume, to be definite, that /„+i = (c, d) with a < c < d < b. Let Ji = [a, c] and J2 = [d, b] be the two subintervals of / not covered by 7^+i. No interval Ik,k= 1 , . . . , w, can intersect both / i and / i , for such an interval would cover 4+i, and n of the Ik


would cover / . Therefore some of the intervals J i , . . . , I„ cover / i and the rest cover J2. By the inductive assumption applied separately to / i and J2 we see that

k=l

Since

we then have

This shows that m(J)=l(J) for every closed interval. The cases where 4+i = {c,d) with c<a or d>b are treated similarly. li J = (a,b)^ then we pick a closed interval [c,d]c(a, b) with d — o b — a — £. Hence, by monotonicity,

^(^, ) ^ ^[c, d] = d — c>b — a — 6,

We already know that m(I) < 1(1) for any / so m{a, b) = i(a,b). The remaining cases are left as an exercise, ill

Problem!. Show that m(0,1) = 1 and that m(^, b] = m[a, b) = b — a ior all half-open intervals in (0,1). ""HI

Proposition 3. m is countably subadditive; i.e., for any finite or countable family {EJ of subsets of(0,1),

m(\JEi) < ^ m ( £ , ) .

Proof. Let ^ > 0 and let {7 /} be a covering of Ej by open intervals so that

3 LEBESGUE MEASURE O N (0,1) 25

Then U£; C U/,y and hence

m(UEi) <J2^aj)

< ^ (m(E,) + ^ )

= Y^ m(Ei) + s. i

Since this holds for all £ > 0,

w(U£/) < ^ m ( £ , ) . ill

Problem 3. Show that countable sets have measure zero. ""HI

Problem 4. If Q is the set of rationals in (0,1) then we know from Problem 3 that m(Q) = 0, and hence for any 6: > 0 there are open intervals {7y} so that Q C UJy and J2i(Ij) < s. Show that if Q C /i U • • • U 4 with Ii,,.,, In open intervals in (0, 1), then 1(h) + h £(/«) > 1. Hence, although finite coverings suffice to approximate m(E) for compact sets £ , arbitrary sets require countable coverings. ""HI

Problems. Let £ = EiUEi with ^(£2) = 0. Show that m(£) = m(£i). Make precise and prove the assertion that if two sets differ by a set of measure zero, then the two sets have the same measure. ""HI

Problem 6. Let £1 C h and £2 C h where Ji and h are disjoint intervals in (0,1). Show that m(£i U £2) = m(Ei) + miEi). Generalize to a finite number of sets £ 1 , . . . , £«. ""HI

Problem 7. (i) Let £ C (0,1), and let £, = {x + r : x G £}. If Er C (0,1), then m(£) = m(£,).

(ii) Let r e (0,1). For x e (0,1), define

if x + r e(0,1)


We need not define x 0 r i f x + r = l , since we want x © r to lie always in (0, 1). Let £^ = {x © r : x G £}, so that Er is now the r-translate of E with the points that fall outside (0,1) put back at the left end of the interval. Show that m(E) = m{Er). ""HI

Note. We will show in the next chapter that the measure function m is countably additive on a usefully large family of sets (the measurable sets) but not on all sets. To construct a non-measurable set we will need the kind of translation invariance in part (ii) above.

MEASURABLE SETS: THE CARATHEODORY CHARACTERIZATION

The critical property of the measure function m is that it be additive. Ideally we should have an identity like

mi^U) =Y.^{E,) (1)

for all finite or countable disjoint families {EJ. Unfortunately, m is not countably additive over all sets, and we must sort out the so-called measurable sets on which (1) does hold.

If £ is any set in (0,1) and E' is its complement in (0,1), then a minimal requirement for additivity is certainly

m(£) + m(£0 = m(0,1) = 1. (2)

It turns out that this condition is sufficient to distinguish the sets £ on which m is countably additive, i.e., sets on which 1 holds. Hence we make the following definition:

A set £ C (0,1) is measurable if and only if

m(£) + m(£0 = l, (3)

where E' = (0,1) - £ . It is immediate from the definition that £ is measurable if and

only if £^ is measurable. Moreover, since m is subadditive, we automatically have

m(E)+m(E') > m(0,1) = 1.

Hence £ is measurable if and only if

m(E) + m(E') < 1.

27


The measurable sets include the intervals and are closed under countable unions and intersections. The measure m is additive on any finite or countable family of disjoint measurable sets. The verification of these facts is the program for this chapter.

A cautionary word about notation and nomenclature: most texts use m* for our function m and refer to it as Lebesgue outer measure. The unadorned letter tn is used by these authors for the restriction of m* to the measurable sets, and this restricted function is called Lebesgue measure. We v ill stick to m, defined on all subsets of (0,1), and call m(£) the measure of £ whether £ is measurable or not. In practice (i.e., after this chapter) we only consider m(£) for measurable sets since measurability is essential for m to have the critical property of additivity.

Proposition 1. (i) Ifm{E) = 0, E is measurable. (ii) Intervals are measurable.

Proof, (i) If m{E) = 0, then

m(£) + m(E') = m(E') < 1,

and this inequality is equivalent to measurability. (ii) Let J = (a,b) he a proper subinterval of (0, 1) and let

/ ' = J\ U Ji where / i and J2 are the two complementary intervals to / . (One of / i , / i will be empty if / = (0, b) or J = (a,l).) Since the measure of an interval is its length.

Therefore,

m(Ji) + m(J) + m(j2) = l.

nt(J') < m{]x) + m(/2) = 1 - m(/)

ni{]) + m{]')<l.

This argument works for closed and half-open intervals too. ill

We saw in Problem 6 of the preceding chapter that if ]\ and / i are disjoint intervals in (0, 1), then for any set £ ,

m(£ n (/i U /2)) = m(£ H / i ) + m{E n 72).

We now extend this to finite or countable families {//}.

MEASURABLE SETS 29

Proposition 2. If{Ji} is a finite or countable family of disjoint intervals in (0,1)^ then for any set E,

m(E n[jJ,) =Y.m(EnJi).

Proof If {/i, / i , . . . , Jn} is a finite family of disjoint intervals, and £ c / i U • • • U /„ , then m(E) = E m(E n //) by Problem 6, Chapter 3. Now let {Jk} be a countable disjoint family of intervals. Then using subadditivity in the first inequality below and monotonicity in the last we get

( CX) \ 0 0

k=l

= lim m(E n (/i U .. • U /„))

< m ( £ n U M .

Hence the first inequality is an equality and we are done, ill

The next result is a formalization of the statement that mea-surability is a local property.

Proposition 3. E is measurable if and only if for every interval J C (0,1),

m(E n / ) + m(£' n / ) = m(/ ) .

Proof Suppose, for example, that / = (a,b) with 0 < a < b<l. Let / i = (0, a), }2 = J = (a, b), / s = 0 , 1 ) . Since the two-element set {a, b] has measure zero, m{E — {a, b]) = m(E) and similarly for E\ so

m(£) = m{E n / i ) + m(E n J2) + m(E n J3)

m(E') = m(E' n / i ) + m(E' n J2) + m(E' n /s) .


Adding columnwise we get 3

'm(£n/,)+m(£^n/,) m(£) + m(£0 = Yl 3

i=i

Hence m(£) + m(E') = 1 if and only if

m ( £ n / , ) + m ( F n / , ) = m(/,)

for each /. If / is of the form (0, fc) or (a, 1), the same argument works by considering the single complementary interval, ill

Problem 1. Carry out the proof of Proposition 3 in the case } =(0,b). II

Our definition says £ is measurable if £ splits (0, 1) addi-tively. Proposition 3 shows that £ is measurable only if £ splits every subinterval of (0, 1) additively. We next show that £ is measurable if and only if £ splits every subset T additively. This result is due to Caratheodory and has become the modern definition of measurability in all general settings. That is, given any countably subadditive non-negative monotone function m defined on all subsets of any set, the function m will be countably additive when restricted to the sets £ which satisfy

m(E n T) + m(E' nT) = m(T)

for all subsets T.

Proposition 4. £ is measurable if and only if for every ''test set'' T c (0,1),

m(£ n T) + m(E' n T) = m(T).

Since m is subadditive, E is measurable if and only if for every set T

m(E n T) + m(E' n T) < m(T). (4)

Proof The condition (4) is clearly sufficient since we can take T = (0,1). To show that (4) holds for every measurable set £ ,

4 MEASURABLE SETS 31

let T be any set in (0,1), ^ > 0, and {Ij} a covering of T by open intervals such that

y^ w(/;) < m(T) + £.

Then

EnTc[j(En Ij), E'nTc [j(E' n J,).

If £ is measurable, then by Proposition 3

m(E n Ij) + m(E' n Ij) = m(Ij) (5)

for each / . Hence, using monotonicity, subadditivity, and (5),

m(E n T) + m(E' n T) < ^ m(£ n Ij) + ^ m(£' n Ij)

= ^ ^(i^/) < m(T) + £,

Since ^ is arbitrary we have the desired inequality (4). ill

Problem 2. If £1, £2 are measurable sets then m(£i — £2) = m(£i ) -m(£in£2)andm(£iU£2)=m(£i )+m(£2)-w(£in£2) . Do you need all the hypotheses? ""HI

We show next that m is countably additive on measurable sets, and that the measurable sets are closed under countable unions and intersections. The discerning reader will notice, with Caratheodory, that the next three propositions make no use of the fact that the sets are subsets of (0, 1), or of how the function m is defined. We use just these facts:

m(0) = 0

0 < m{E) < oc

m(£) > m(F) if £ D £

m ( | j £ / ) <^m(Ei)

and for measurable sets £ , and all T,

m(E n T) + m(E' n T) = m{T). (6)


Proposition 5. If{Ei} is a finite or countable family of disjoint measurable sets, then

Proof. Let £ i , . . . , £ „ be disjoint measurable sets. The set £i cuts the test set T = £i U • • • U £„ additively, so

m(£i) + m(£2 U • • • U £„) = m(£i U • • • U £„).

The measurable set £2 cuts £2 U • • • U £^ additively, so

m(£2) + m(£3 U • • • U £„) = ^(£2 U • • • U En),

and hence

m(£i) + m(£2) + m(£3 U • • • U £„) = m{Ei U • • • U £„).

In a finite number of steps we have

Now let {Ei} be a countable family of disjoint measurable sets. For each w,

and hence

( 00 \ 00

U£, >^m(£ , ) . i=i J i=i

The opposite inequality is automatic by subadditivity, so equality holds, ill

Problem 3. Show that if {£/} is a countable disjoint family of measurable sets and T is any set, then

m {Tn[JE,)=Y.m(TnE,). •'"Ill

Proposition 5 tells us what m([j Ei) is if the £/ are measurable (and pairwise disjoint), but does not say that U £/ is itself


El E2

Fig. 1

measurable. This we show next, proving first that £i U £2 is measurable.

Proposition 6. If £1 and £2 are measurable^ then E\ U £2 is measurable.

Proof. Let £1 and £2 be measurable sets and let T be any test set. Let T = Ti U 72 U T3 U 7] as indicated in Fig. 1.

What we must show is

m[(Ei U £2) n T] + m [(£1 U £2)' nT]= m(T);

or, in terms of Fig. 1,

m(Ti U T2 U T3) + m(T4) = w(T).

Cutting the test set Ti U 7^ with the measurable set £2 gives

m(Ti) + m(T2) = m(Ti U T2). (7)

Similarly, cutting T3UT4 with £2 gives

mm + m{%) = mm U T4). (8)

Cutting T with £1 gives

m(Ti U T2) + m(T3 U T4) = m(T). (9)


Combining (7), (8), (9) we can write

m(Ti) + m(T2) + miT^) + m{%) = m(T), (10)

Now cut Ti U Ti U 73 with £1 and then use (7):

w(Ti U T2) + m(T3) = m(Ti U T2 U T3)

m(Ti) + m(T2) + miT^) = m{Ti U T2 U T3). (11)

From (11) and (10) we have the desired equaUty

m(TiUT2UT3) + m(T4) = m(T). ill

Corollary. Finite unions and finite intersections of measurable sets are measurable. E\ — £2 is measurable if Ei, £2 are.

Proof. Notice that £ satisfies the characterizing equation

m(£ n T) + m(E' nT) = m(T)

for all T if and only if £^ does. Therefore

£1 n £2 = (£i u E'^y

is measurable whenever £1, £2 are. The inductive proof from two sets to a finite number is immediate. Since

£1 — £2 ^ £1 ^ £ 2 '

differences are measurable, ill

Proposition 7. If{Ei} is a countable family of measurable sets, then U£/ is measurable and (lEi is measurable. Open sets and closed sets are measurable.

Proof. We can assume the £/ are disjoint by replacing £2 by £2 - £1, £3 by £3 - (£1 U £2)5 etc. Let F„ = £1 U • • • U £„, so Fn is measurable and by Proposition 5

n

m(F„) = ^w(E,) .


For any test set T, by Problem 3,

m(T) = m(T n P„) + m(T n F^) n

= X^m(Tn£ / ) + m(TnP^) .

If £ = U Ei, then P„ C £ for all n so £^ D £^ m(T n £^) > m(T n E) and

mi (T) > ^ m(T n £/) + m(T n £'). / = i

This last inequality holds for all n^ so, again by Problem 3,

oo

ni{T) > J2 ^T n Ei) + m(T n £')

= m ( T n £ ) + m ( T n £ 0 ,

which shows that £ = U/^i £/ is measurable. The remainder is left as a problem, ill

A cr-algebra of subsets of any set X is a family of subsets which contains X and 0 and is closed under finite or countable unions, finite or countable intersections, and complementation. Since £i — £2 = £1 n£2, or-algebras are closed under differences.

Problem 4. Show that the measurable subsets of (0,1) form a (7-algebra, and that every open or closed set is measurable. ""HI

Problem 5. Let £ be a measurable subset of (0,1). Show that for each £ > 0 there is an open set U and a closed set F such that F c E cU and

m(E) — £ < m(F) < m(U) < m(E) + s.

Show that the existence of such F and JJ for every £ > 0 is also sufficient for £ to be measurable. ""HI

Now we extend the definition of Lebesgue measure to include arbitrary subsets of R. We use /x for the extended measure function, so /x is defined on all subsets of M. For a set £ C {n, n+1) define lui(E) = m(E — n), where E — n = {x — n:xe £}.


If £ C (0, 1), then of course /x(£) = m(E). For any set £ C M, define

oo

H(E)= Yl l^(En(n,n+l)). W = — 0 0

If £ contains integer points, that will not affect the value of /x(£) since we still want countable sets to have measure zero. Notice that /x(£) = oc is now a possibility.

We will say that £ is measurable if and only if £ Pi (w, w + 1) is measurable for each n^ i.e., if

/x(£ n(n,n+ 1)) + /x(F n(n,n + 1)) = 1

for each n. We could alternatively have used the Caratheodory criterion to define measurability, as the next proposition shows.

Proposition 8. A subset £ c M /s measurable if for every set T c R ,

/x(£ n T) + /x(F n T) = /x(T). (12)

Proof. By definition,

00

lx{T)= J2 M(Tn(«,n+l)) W = —OO

oo

/ i (Tn£)= Yl Ai(Tn£n(n,n+l)) « = —00

oo

fx(TnE')= Y t^(TnE'n(n,n + l)). n——OQ

If £ is measurable then

/x(Tn £ n (n, n+1)) + fi{Tn F n (w, w+1)) = /x(Tn (n, n+1) )

for each w, and addition gives (12). On the other hand, if (12) holds for all T, then we see that each £ Pi (n, n+1) is measurable by letting T = (w,w+l ) . ill

As we remarked earlier, the properties of m on the measurable subsets of (0, 1) which are proved in Propositions 5, 6, and 7


depend only on these facts:

m(0) = 0 (13)

0 < m ( £ ) < o c for all £ (14)

m(E) > m ( f ) i f £ D F (15)

m([jEi)<Y^m(Ei) forall{£,} (16)

£ is measurable if and only if

m(E n T) + m(E' nT) = m(T) for all T. (17)

Properties (13)-(16) obviously hold for /x and subsets of M, and Proposition 8 shows that (17) also characterizes measurable sets inR.

Therefore we have the properties of Propositions 5, 6, and 7 for /x and measurable subsets of R. Specifically,

/x f IJ £/j = ^ /i(£/) for disjoint families

£i U £2 is measurable if £1, £2 are

y £/ is measurable if all £/ are.

The measurable subsets of R are obviously closed also under complementation and so form a a-algebra, and /x is countably additive on this (T-algebra.

Problem 6. Show that /x is translation invariant; i.e., if E + x = {y + X : y e E}, then /x(£) = /x(£ + x) for all £ and x. ""HI

We finish by constructing a non-measurable subset £ of (0,1). To show that £ is not measurable we show that (0,1) is the union of a countable number of disjoint translates (modulo 1) of £ . If £ were measurable and m(E) = 0, then we could conclude that m(0,1) = ZZi ^ ( £ ) = 0- If ^ ( £ ) > 0, then we could conclude that m(0,1) = E ^ i ^ ( £ ) = 00.

The following "construction" of a non-measurable set £ depends heavily on the Axiom of Choice or some equivalent logical assumption, such as Zorn's Lemma. Although some mathematicians are not entirely comfortable with the Axiom of Choice, we will cheerfully accept it here as part of our common logic.


Pick any x e (0, 1), and then any y so y — x is not rational. Then pick z so z ~ x and z — y are not rational. Continue this process—uncountably many times—until there remains no number in (0, 1) which is not obtained by adding a rational to one of the already chosen numbers. The set E thus obtained is a maximal subset of (0,1) with the property that all differences X — y ioY x^ y e E are irrational. Hence for all ^ ^ £ , f = x + r for some x e E and some rational r. Let ri , ri? • • • be an enumeration of the rationals in [0,1) and let £„ consist of all numbers X + rn (modulo 1) for x e E. That is, if x € £, x + r„ € £„ if x + Tn < 1 andx + r„ —1 e En if x + r„ > 1. Since the sets £„ are essentially just translates of £ , all £„ have the same measure. (See Problem 7, Chapter 3.) The £„ are disjoint, for if x, y G £ and

x + Tn

or

X + rn — 1

y + rm

or

[y + ^m- 1

then X — y is rational, so x = y. Clearly (0, 1) = U £« since every t not in £ has the form x + Vn (modulo 1) for some x € £ , some rational r„. Thus (0,1) is a countable union of disjoint sets with the same measure. If £ is measurable then all En are measurable and m(0,1) = ^m(En)^ which is zero or infinity.

Problem 7. Use the Caratheodory characterization (6) to show that if for every s > 0 there are measurable sets A and B such that A c £ C B and /X(JB) - /x(A) < e, then £ is measurable, •"illl

Problem 8. (i) If {£/} is a sequence of measurable sets such that /x(£i) < oo and £i D £2 D £3 D • • •, then /x (fl £/) = lim/>6(£/). Hint: Let £ = R E,, so £1 - £ = (£1 - £2)U(£2 - £3)0 • • • and / x ( £ i - £ ) = E . ^ i / x ( £ . - £ m ) .

(ii) Show that /x(£i) < 00 (or /x(£„) < oc for some n) is a necessary assumption. ""HI

Problem 9. Let {£J be a sequence of measurable sets such that £1 C £2 C £3 C • • •. Show that /x (U £/) = lim A6(£/). '"Hll


Problem 10. For any two sets E and F^ define £ A JF by

EAF = (E- F)U(F - E).

E AF is called the symmetric difference of E and f. Agree to identify sets E and F ii E AF has measure zero. (Cf. Problem 5, Chapter 3.) Define a function J on pairs of subsets of (0,1) as follows:

d(E,F) = iJL(EAF).

Show that J is a metric on (equivalence classes of) measurable sets. Notice that the triangle inequality—the only non-obvious metric property—impUes that the relation E = f, defined by iJi(EAF) = 0, is an equivalence relation, thus providing the justification for identifying sets E and F ii E = F. Show that /x(£) = iJi(F) i( E = fjSO that /x does not object to the identification of equivalent sets. Show that /x is continuous with respect to the metric d; i.e., if d(En, E) —> 0, then fJi(En) —> l^iE). Is /x uniformly continuous.^ Is the restriction to subsets of (0, 1) necessary? ""HI

Problem 11. The operation A has some interesting properties which might appeal to those with an algebraic bent. For example, is A an associative operation.^ How does the operation A interact with n, U / ? Show that if intersection is interpreted as multiplication, and symmetric difference as addition, then the subsets of X (or the measurable subsets of X) form a commutative ring with identity. ""HI

Problem 12. The Cantor Set. Each number in [0, 1] can be written as a ternary series:

X = ai/3 + ^2/3^ + ^3/3^ H ,

where all ai are 0, 1, or 2. Some numbers have two such representations, e.g.,

^ = 2 / 3 + 0/9 + 0/27 + -..

= 1/3 + 2/9 + 2/27 + . . . .


Let Ui be open middle third of [0,1]; i.e., Ui = ( | , | ) . Let Ui be the two intervals which are the middle thirds of the two intervals in [0,1] - Ui; i.e., U2 = ( i , | ) U ( | , f). In general, let U„+i be the union of all open middle thirds of the closed intervals in [0,1] - ULi Ui. The Cantor set is [0,1] - U U„. Show:

(i) The Cantor set is a closed set of measure zero. (ii) The Cantor set consists of exactly those points in [0,1]

which can be written with a ternary expansion with all ai = 0 or 2. (For example, ^ f, | , | , | , | , ^ , ^ , ^ , ^ , . . .)• Equivalently, show that U Un consists of those points whose ternary expansion must have some ai = 1. Show that \ is in the Cantor set.

(iii) Show that all points of [0, 1] can be expressed as a binary expansion,

where each bi is 0 or 1. (iv) Since both the Cantor set and [0,1] can be put in a 1-1

correspondence with all sequences onto a two-element set, the Cantor set is an uncountable set, and thus is an example of an uncountable set of measure zero.

(v) Show that each point of the Cantor set is the limit of a sequence of distinct points of the Cantor set.

(vi) Show how to define a closed nowhere dense subset of [0, 1] with arbitrary measure between 0 and 1 by modifying the above procedure. For example, to get a set of measure 1/2 we remove open intervals with total length 1/2 as follows: Let t/i be the open interval of length 1/4 centered in [0,1]. Then [0,1] — U\ consists of two closed intervals whose lengths are less than 1/2. From these two closed intervals remove equal centered open intervals with lengths totaling | . Etc. '""H

Problem 13. Let £1 and £2 be disjoint measurable sets. Draw the appropriate figure similar to Fig. 1, showing £1, £2 and an arbitrary test set T. Label the subsets of T as follows: £1 n T = Ti, £2 n T = T2, (£1 U £2)' n T = T3. Write out the proof that m(Ti U 7 ) + m(T3) = m(T) which shows that £1 U £2 is measurable in this special case of Proposition 6. ""HI


Problem 14. Here is the historical definition of measurable set. An open subset of (0,1) is a countable union of disjoint open intervals. If U = [j{ai, bi)^ then define m{U) = Yîî —î)* If JF is a closed subset of (0,1), and U = (0,1) — P, then define m(F) = 1 —m(U). Define outer measure m* and inner measure m^ as follows:

m*(£) = inf{m(U) : £ C U, U open},

m^(E) = sup{m(P) : F C E, F closed}.

A set E is measurable if and only if m*(£) = m^(E). Show that m*(£) = 1 - fn'(E^) so m*(£) = m*(£) is the same as m%E) + nf(E) = 1. •"illl

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS

In this chapter we define the Lebesgue integral of a bounded function on a set of finite measure. The definition is very similar to the definition of the Riemann integral. We partition the finite measure set into a finite number of disjoint measurable sets. There is an upper sum and a lower sum for each such partition, and when the upper and lower sums come together the function is integrable. The only difference between the Riemann and Lebesgue integrals for bounded functions is that now we allow the domain to be a finite measure set rather than an interval, and the partitions consist of measurable sets rather than subintervals.

Let S be a finite measure set, by which we will always mean a measurable set of finite measure. A partition of S is a finite family {El,..., En} of disjoint non-empty measurable sets whose union is 5. If 5 is an interval and P is a partition in the earlier sense, P = {XQ, Xi , . . . , x„}, we will now understand that P denotes the partition of S = [xo, Xn] into the disjoint sets (xo, Xi), (xi, X2),... , (x^-i, x„), and the finite zero-measure set \ X o , X\, . . . , Xfif.

If P and Q are partitions of S, then Q is a refinement of P, denoted Q >P or P < Q^ provided each P € Q is a subset of some Ee P. If P = {£/} and Q is a refinement of P, we will write Q =={P//} to indicate that £/ = U/ Fij for each Ei e P. Notice that if P = {£/} is a partition of S, then fji{S) = E f^iEj).

If /* is a bounded function on a set S of finite measure, and P = {£/} is a partition of 5, we define upper and lower sums for

43


f and P exactly as in Chapter 1:

nii = 'mi{f(x) : x e £/}

Mi = sup {f(x) : X e £/}

n

u(f,p) = Y.^f^(E^y i=i

If S is an interval and P is a partition of S into intervals, then U{f, P) and L{f, P) have exactly the same meaning as in Chapter 1.

Proposition 1. If Sis a set of finite measure and m < f(x) < M for all X e S, and P, Qare partitions of S with Q>Py then

mniS) < L(f P) < L(f Q) < U(f Q) < U(f P) < M/x(5).

Proof Let P = {£,} and Q = {Fij} with U/ P// = £/ for each /. Let

nii = inf {/"(x) : x e £/}

Tftij = inf { /"(x) : X e Fij},

Then clearly mi < mij for all /, / , so

L(/-,P) = ^ m , / x ( £ , )

<Y^mij ^l{Fij)

= UfQ),

The proof that U( f Q) < U(f P) is similar, and the remaining inequalities are obvious. Hi

Problem 1. Show that every lower sum L(f P) is less than or equal to every upper sum U(f Q). ""HI

5 THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS 45

If 5 is a set of finite measure and f is bounded on 5, then f is (Lebesgue) integrable on S if and only if supLC/", P)==inf

p 2 U( f, Q). We write /^ /* for the common value if f is integrable.

We observe that f is integrable on S if and only if there is for every 6: > 0 a partition P such that U(f, P) — L(f, P) <e.

Proposition 2. If f is Riemann integrable on [a, b], then f is Lebesgue integrable on [a, b] and the integrals are the same.

Proof. If f is Riemann integrable then there is a partition P of [a, b] into intervals such that U(f P) - L(f P) < s^ so f is also Lebesgue integrable. The value of either integral lies between any lower sum and any upper sum, so the Riemann and Lebesgue integrals clearly coincide, ill

For the Riemann integral we partitioned the interval [a, b] into a finite number of subintervals (^, xi), (xi, Xi) , . . . , (Xn-i, b)^ and a zero-measure finite set { , xi, X2,. . . , x„_i,/?}. For the Lebesgue integral we partition an arbitrary measurable set S into a finite number of measurable subsets:

5 = £i U £2 U . . . U £„, £,- n £y = 0 if / ^ /.

Since /x is countably additive, we might ask whether we should not consider instead partitions of S into countably many disjoint sets:

5 = £1 U £2 U . •., EinEj = 0 a i ^ /.

The problem below asks you to show that countable partitions with their corresponding lower and upper sums would give an equivalent definition of integrable function. Since the sup of lower sums over all countable partitions is larger than the sup over finite partitions, and upper sums similarly could be smaller if countable partitions were allowed, it is a priori easier for f to be integrable if countably infinite partitions are allowed.

Problem 2. Show that the definition of integrability for a bounded function on a set of finite measure does not change if countable partitions are allowed. ""HI


The characteristic function of a set £, denoted XE? is the function which is one on the set and zero elsewhere. A simple function is a function cp defined on a set S of finite measure such that (p has a finite range {yi, yi. • • •. y«} and for some partition P = { £ i , . . . , £ , } o f 5 ,

n

Problem 3. If <p is the simple function above, then cp is inte-grable and Js(p = EU Ji f^(E,). II

The usual definition of "f is integrable over S" is

sup y (picp < f\ = i n f | y V :^> fj

where cp and if/ range over simple functions defined on S. This is just different terminology for expressing the same idea as our upper sum-lower sum definition.

If f is continuous on an interval [a,b]^ then f is Riemann integrable on [a,b]. The proof consists in showing that since f is uniformly continuous, each M/ — mi will be less than any given e > 0 provided P is any sufficiently fine partition of [a, b] into intervals. This implies

U( /*, P) - L( /•, P) = ^ ( M , - m,) Ax, < 8{b - a),

so f is Riemann integrable. A bounded function f will be Lebesgue integrable on a set 5 of finite measure, by the same argument, if there is a partition P = {£/} of S so that Mi —mi < s for each /. There will obviously be such a partition provided each set of the form {x e S : a < f(x) < ^ + 6:} is measurable; specifically, we can let

Ei = {xe S :m + i£ < f{x) <m+(i + 1)6:},/ = 0 , 1 , 2 , . . . ,

where m is a lower bound for f on S. These sets are disjoint, and a finite number of them will form a partition of S, since f is bounded, provided only that each of these sets is measurable. Accordingly, we agree that f is measurable on S provided

5 THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS 4 7

{x : a < f(x) < b) is measurable for all a^ b. Notice that if f is measurable on S, then S is necessarily a measurable set. The definition of measurable function applies to all functions f^ bounded or not, and all measurable sets 5, whether or not they have finite measure. In this chapter we are only concerned with the integrals of bounded functions on sets of finite measure, but in Chapter 7 we will consider unbounded measurable functions on sets with infinite measure.

Problem 4. If f is measurable on S and g = f except on a zero-measure subset of S, then g is measurable on S. ""HI

Problem 5. Every simple function is measurable. ""HI

Problem 6. If f is continuous on [a,b]^ then f is measurable on [a,b]. Hint: Show that {x e [a,b] : f(x) > a} is a closed set (and therefore measurable) for each a. Then use

{x e[a,b]:a < f(x) < y8} =

{x e [a, b] : fix) > a} - {x e [a, b] : f(x) > ^8}. ""HI

Problem 7. If f is measurable, then \f\ is measurable. ""HI

Proposition 3. If f is a bounded measurable function on a set S of finite measure, then f is Lebesgue integrable on S.

Proof Let - M < f(x) < M for all x e S. Let N be a large integer, and let

Ei = {x:~M+ (i - 1)/N < fix) <-M + i/N}

for / = 1, 2 , . . . , 2MN. Then P = {E/} is a partition of S and

U( f P) ~LifP)<Y: ^/x(£.) = /x(5)/N. I I

The converse of Proposition 3 is also true for bounded functions (Proposition 6 below), so that measurability is equivalent to integrability for bounded functions on sets of finite measure. The great virtue of this characterization is not the fact that more functions are Lebesgue integrable, but the fact that pointwise limits of measurable functions are measurable, as we show below. Hence ii f = lim f^ with each fn integrable, then


Jf makes sense provided only that f is bounded. It is this kind of result which makes it very much easier to deal with limits of integrals and integrals of limits in Lebesgue integration.

Problem 8. Show that there is a sequence {/"„} of Riemann integrable functions on [0,1], all with the same integral, such that fn(x) —> f{x) for all x e [0,1] and f is not Riemann integrable. '4

Now we proceed to show that every integrable function is measurable. We show that an integrable function is the point-wise limit of simple functions, which are necessarily measurable, and that every pointwise limit of measurable functions is measurable. We first introduce some useful alternative criteria for measurability.

Proposition 4. Each of the following conditions is necessary and sufficient for f to be measurable:

(i) {x: f(x) > a} is measurable for all a; (ii) {x: f(x) < a} is measurable for all a;

(Hi) {x: f(x) > a} is measurable for all a; (iv) {x: f(x) < a} is measurable for all a; (v) {x:a < f{x) < b) is measurable for all a, b.

Proof The sets in (i) and (ii) are complements, and similarly for the sets in (iii) and (iv). If f is measurable, then {x: f(x) > a} is the countable union of the measurable sets {x : a < f(x) < a + n}, n= 1,2, Conversely, if{x:a< f(x)} is measurable for all ^, then

{x : a < fix) < b} = {x : a < f(x)} - {x : b < f(x)}

is measurable for all ^, /?, and hence f is measurable. The other equivalencies are proved similarly, using the fact that measurable sets are closed under countable unions and intersections, and under complementation, ill

Problem 9. Complete the proof of Proposition 4. ""HI

In this chapter we consider the integral only for bounded functions. In later chapters, however, we will consider unbounded


functions, and indeed functions that take the values +oc or — oo, since these values can arise as limits of sequences of integrable functions. Accordingly, we agree that such an extended real-valued function f is measurable provided {x : a < f(x) < b] is measurable for all a^ b^ and the sets {x\f{x) = +00} and {x:f(x) = —00} are both measurable.

Proposition 5. If{ fn) is a sequence of measurable functions on a measurable set S, then sup fn^ inf fn, lim sup fn, and lim inf fn are measurable functions, / / l im fn{x) exists for all x, then the limit is a measurable function.

Proof. To show sup fn is measurable we verify condition (iii) of Proposition 4. Since

{x : sup fn(x) > a} = \J{x:fn(x) > a}, n

{x : sup fn{x) > a] is di countable union of measurable sets if each fn is measurable. Similarly,

{x : inf fn(x) < a] = |J{x: fn{x) < a], n

SO inf fn is measurable. If sup fn takes the value +oc, then

[x : sup fn(x) = +00} = P I | J { X : fn(x) > N}, N n

and this set is measurable. A similar equality holds for the set where inf fnipc) = —00.

Since

lim sup fn(x) = inf sup fk(x) ^ k>n

lim inf fn(x) = sup inf fk(x),

both lim sup fn and lim inf fn are measurable. If lim fn(x) exists for all X, then lim fn = lim sup fn = lim inf fn. •

We will use the phrase almost everywhere, abbreviated a.e., to mean "except on a set of measure zero." Hence " f = g a.e."


means that {x: f(x) ^ gix)} has measure zero, and " f >n a.e." means {x: f(x) < n] has measure zero.

Proposition 6. If f is a hounded function which is integrable on a set S of finite measure, then f is the ax. pointwise limit of simple functions, and hence f is measurable.

Proof. For each n we let Pn be a partition of S such that U( f, Pn) - L( /", Pn) < 1/n. We can assume that Pi>P2> P3 >- • • • by replacing each P„ by the common refinement of its predecessors. Let P„ = {Eni} and

mni = 'mi{f(x) : x e Eni}

Mm =sup{/"(x) : xe Eni).

Let (pn be the simple function which has the value m^ on E„/, and let V be M^ on Eni • Then \jfn — ^« is a simple function and

j {fn - (pn) = Yl^Mni " mni)fl(Eni)

= U(fPn)-L(fPn).

The functions cpn increase and converge a.e. to some measurable function g<f and the xj/n decrease a.e. to some measurable h>f. We will show that g=:h = f a.e., so f is measurable.

Suppose on the contrary that /z — g > 0 on a set of positive measure. Then (Problem 10) there is /? > 0 so that h — g > p on a set A of positive measure. For all n and all points of A,

fn-(Pn>h-g>p.

If Eni intersects A, then Mni — ni > P; these sets cover A and hence have aggregate measure at least /x(A). Thus for each w,

U(f Pn) - L(f Pn) = E ( ^ - - ^ - ) M ( £ - ) > P/ ( )-

This contradicts our assumption that U( f Pn) — L( f Pn) —> 0. ill

Problem 10. Show that if ^ is a measurable function and {x : k(x) > 0} has positive measure, then {x : k(x) > p} has positive measure for some p > 0. ""HI


Problem 11. If /" is a bounded function which is measurable on a set S of finite measure, and T is a measurable subset of 5, then f is integrable over T. ""HI

Problem 12. Show that if f is integrable over S, then Js f = lim LC/", P) = lim [/(/", P) where the partitions are ordered by refinement. ""HI

Problem 13. If f is bounded on [a, b] and continuous except at a finite number of points, then f is measurable and hence integrable. ""HI

PROPERTIES OF THE INTEGRAL

We will prove the linearity properties of the integral by showing that the integral is a limit of Riemann sums. First we need to know that kf and f -\- g are measurable if f and g are.

Proposition 1. If f and g are measurable on S, then kf is measurable for every constant k, and f + g is measurable.

Proof. It is clear that kf is measurable if f is, so we consider f + g. The inequality f(x) + g(x) > a is equivalent to f(x) > a — g(x)^ which holds if and only if there is a rational number r such that

f(x) > r and r > a — g(x).

Hence

{x : f(x) + g(x) > a} = [J{x : f(x) > r} n {x : g(x) > a -r}, r

where the union is over all rationals r. The right side is a countable union of measurable sets, ill

If f and g are bounded measurable functions on a set S of finite measure, then kf and f+g are integrable over S. Now we write / / as a limit of Riemann sums. If S is a set of finite measure and P = {£/} is a partition of 5, then a choice function for P is a finite sequence {c/} with q e Ei for each /. The Riemann sum for /", P, c is the usual sum

P(/-,P,c) = ^/'te)/x(£,).

53


The partitions P of S form a directed set under the partial ordering of refinement, and the pairs (P, c) SLYC ordered according to the ordering on the sets P; i.e., (P, c) > (P^ c^) means P > P ' (P is a refinement of P'). The Riemann sums R(f, P, c) are a net on the partially ordered pairs (P, c). In this context the limit condition for nets reads as follows: R(f, P, c) —> I, or lim R(f, P,c) = I^ii and only if for each ^ > 0 there is a partition Po such that \R(f, P, c) — I\ < s whenever P > PQ., and c is a choice for P. Although R(f, P,c)^ for fixed f^ is a function of the pair (P, c), the pairs are ordered only in terms of P, so we write lim R(f,P,c) instead of the correct but more cumbersome lim R(f, P, c).

Proposition 2. If f is a bounded function which is integrable on the finite measure set S, then R(f P, c) —> J^ f.

Proof. If P is any partition of S, and c is any choice function for P, then

UfP)<R(fP,c)<U{fP). (1)

If f is integrable then for any e > 0 there is a partition PQ SO that U(f, P) -L{f,P)<e for all ? ^ ?o- Hence

R(f.P.c)-lf < s

for all P > Po and all c; i.e., R{fP,c)-^ Js f. •

Notice that if {U{f P)}, {L(f P)}, {R(f P,c)} were nets on the same directed set, the consequence R(f P, c) —> Js f would follow immediately from the inequality (1) and the fact that lim L( f P) = lim U( f P) = Js f. As it is, the directed set for the Riemann sums is the much larger directed set consisting of all pairs (P, c) instead of just all partitions P.

The following proposition shows that the net of Riemann sums {R{f, PyC)} cannot distinguish between f and a function which equals f almost everywhere.

6 PROPERTIES OF THE INTEGRAL 55

Proposition 3. If f and g are arbitrary functions on a finite measure set S, and f = g a.e., then

limR(fP,c) = limR(g,P,c);

i.e., one limit exists if and only if the other does, and then the limits are equal. The functions f and g are not assumed to be bounded or measurable.

Proof. Let f = g except on A C S^ where /x(A) = 0. Assume lim R(f P, c) = L. Let ^ > 0 and choose a partition PQ so that \R(f, P,c) - L\ < s ii P > PQ. Let Pi be the refinement of PQ obtained by replacing each set £/ of PQ by the two sets Ei n A and £/ n A. Many of these sets may be empty, but that doesn't matter. Let P >- Pi. If P = {P/}, then each P/ is a subset of some Ej n A, so P/ C A and fji(Fi) = 0, or P/ is a subset of some Ej n A , so f = g on P/. Therefore, if P : Pi : PQ,

P(/-,P,c) = ^/-(c,)/x(P,)

Since \R( f P, c) -L\ <6ifP >Po, it follows that |P(g, P, c)-L\ < 6 ii P > Pi; i.e., if lim R(f P,c) exists, then so does limP(g, P, c) and the limits are equal. The situation is symmetric in f and g, so we are done, ill

Notice from Proposition 3 that Riemann sums can converge for an unbounded function, which is unlike the situation for the Riemann integral. For example, let g be a bounded measurable function on 5, so R(g, P, c) —> /^g. Let f = g except on some countable set {x„}, and let f(Xn) = n^ so f is unbounded, but f = g a.e. By Proposition 3, lim R(f P,c) = lim P(g, P,c) = J^g. The next two problems point out that the only way Riemann sums can converge is for the function to be essentially equal to a bounded integrable function.

Problem 1. If g is a bounded function on a set S of finite measure, and R(g, P, c) —> / , then g is integrable (hence measurable) and J^g = I. Hint: Cf. Proposition 3, Chapter 2. ""HI


Problem 2. If /" is any function on a set S of finite measure, and R{f, P, c) —> J, then there is a bounded function g such that f = g a.e. and R(g, P, c) —> I. Hence R( /", P, c) —^ J if and only if /" = g a.e. for some bounded integrable function g. ""HI

The following proposition is now a simple consequence of the fact that the integral is a bona fide limit.

Proposition 4. If f and g are bounded measurable functions on a set S of finite measure, and k is a constant, then

(i) Iskf = k!sf;

(ii) Is(f + g) = Isf + Isg;

m \ki\<k\f\-Proof, (i)

f kf = lim R{kf, P, c) Js P

= lim^k f(Ci)ix(Ei)

= l i m ^ ^ f(Ci)fi(Ei)

= limkR(f,P,c)

= klim Rif,P,c)

111 = ^ / /

Problem 3. (i) Verify that the net {R( f + g, P, c)} is the sum of the nets {R{f P,c)} and {R(g, P,c)} and use this to prove part (ii) of Proposition 4.

(ii) Verify that R(\ f\, P, c) > \R{f P, c)\ for all (P, c), and use this to prove part (iii) of Proposition 4. Why does the net {R(|/"I, P,c)} converge? '"HII

Proposition 5. If f is a bounded measurable function on a finite measure set S, and T is a measurable subset of S, then f is integrable over T, and

/ r ^ = / / ^ '

6 PROPERTIES OF THE INTEGRAL ^7

Proof. The function fxT is clearly bounded and measurable on 5, so fxT is integrable over S. Let Q be all partitions Q of 5 such that every £ G Q is either a subset of T or disjoint from T. Every partition P of 5 can be refined to get such a partition Q e Q, so all integrals over S can be expressed as limits of sums R( /*, Q, c) as Q ranges over Q. Hence if Q € Q and Q = {Ei n T, £/ n T}, then

| ^ A X T = U m R ( / x T , Q , c )

= l i m ^ / ' ( c , ) / x ( £ , n T ) . ^ i

= / / • < '

The last equality holds because every Riemann sum for fj f is one of the sums in (2). ill

Corollary. If f is a bounded measurable function on AVJB, where A and B are disjoint measurable sets, then

JAUB J A JB

Proof f=f.xA+f'XB. ill

Problem 4. If f is bounded and measurable on the finite measure set £5 and £ = U/^i £/? where the £/ are disjoint and measurable, then j ^ f = TZi k f' '"""

Most of the functions one wants to integrate are continuous— perhaps even analytic. For such functions there is no difference between the Riemann and Lebesgue integrals over a bounded closed interval. The Lebesgue integral, however, is much more accommodating in the matter of limit theorems. For the Riemann integral one must generally know that fn —> f uniformly to conclude that jfn —> Jf. For the Lebesgue integral, pointwise convergence is enough provided the functions stay uniformly bounded. The reason for this is that pointwise convergence is nearly uniform on finite measure sets. We make this idea precise in the next two propositions.


Proposition 6. If {fn} is a sequence of measurable functions on a finite measure set S, and fn —> f pointwise on S, then given 8 > 0 and 8 > 0, there is a measurable set E of measure less than 8 and a number N so that \fk(x) — f{x)\ < s for all k> N and all x e S ~ E.

Proof Let

Fn = {x e S : \ fk(x) - f(x)\ > £ for some k > n}.

The sets P„ are measurable, and decreasing, and f]Fn = 0 because fn{x) —> f{x) for all x. Since /x(Fi) < oc, lim/x(f J = 0. Let /x(F]v) < 8. For x e S - FN? I fki^) - f{x)\ < ^ foi" all k>N. Illllll

Problem 5. If fn —> f a.e. the same result holds. ""HI

The next proposition gives the form in which it is easiest to remember and apply the above result.

Proposition 7. (Egoroffs Theorem) If {fn} is a sequence of measurable functions on a finite measure set Sy and fn —> f pointwise on S, then for every 8 > 0 there is a measurable set E c S of measure less than 8 so that fn —> f uniformly on S — E.

Proof For each n we find a set E„ of measure less than 8/2^ and a number N„ so that \fk{x) — f{x)\ < ^ior k> Nn and x e S — En. Let £ = U £«5 so that /i(£) < 8.lix e S — E then given £ > 0 there is N (any N„ with ^ < 6:) so that | fk(x) — f(x)\ < £ iik>N. Illllll

For uniformly bounded sequences, the limit of the integrals is the integral of the limit. This follows directly from Egoroffs Theorem, as we show next.

Proposition 8. (Bounded Convergence Theorem) If {fn] is a sequence of measurable functions on a finite measure set S, and the functions fn are uniformly bounded on S, and fn(x) —> f{x) pointwise on S, then

\imlj„ = l^\imf„ = lj.

6 PROPERTIES OF THE INTEGRAL 59

Proof. Let s > 0. Let £ be a measurable set of measure less than s such that fn —> f uniformly off E.

Let \fn(x)\ < M for all n and all x e S. Then

\l(fn-n\<l\f^-f\

<[ \f„-f\ + 2Ms. JS-E

Since /„ —> f uniformly on S—£, there is Nso that \fn— f\ < £ on S — E a n > N. Hence if « > N,

/ . (fn- f) s

< siJi(S - E) + IMs

= s[fi(S-E)+2M] <s[fji(S) + 2M],

Since 6 is arbitrary and /x(5) < oc, the result follows, ill

The hypotheses of the Bounded Convergence Theorem require that fn —> f pointwise with all fn remaining in some fixed finite area. Specifically, we require that all fn lie in the rectangle S x [—M,M]^ where fi(S) < oc. If the functions are allowed to wander outside a fixed finite area the result can fail, as the following problem shows.

Problem 6. Let S = (0, 1). Give an example of bounded measurable functions fn on S so that J^ fn = 1 for all n and fn(x) —> 0 for all xeS. <

Problem 7. (i) If f is a measurable non-negative function on [0,1] and If = 0, then f = 0 a.e.

(ii) If f and g are bounded measurable functions on a set S of finite measure, and f < g a.e., then Js f < Js g- """I

Problem S. If 0 < fn < h < gn < M on [a, b] for all w, where {/„}, {gn} are respectively increasing and decreasing sequences of measurable functions with lim Jfn = lim /g„, then h is measurable and Jh = lim Jfn. ""HI


Problem 9. Show that almost everywhere convergence is sufficient in the Bounded Convergence Theorem. ""HI

Problem 10. Let fn(x) = nx/(l + n^x^) for 0 < x < 1.

(i) Show that fn{x) —> 0 for all x G [0,1], but the convergence is not uniform.

(ii) Does the Bounded Convergence Theorem apply? '"Hl

Problem 11. If f and g are measurable functions, then fg is measurable. Hint: fg= \[(f -\- g)^ — if — g ) ]? so it suffices to show that h is measurable if h is measurable. ""HI

Problem 12. li f is bounded on [a,b]^ then f is Riemann inte-grable on [a, b] if and only if f is continuous almost everywhere, i.e., the set where f is discontinuous has measure zero. Verify the following arguments. Let {?„} be a sequence of partitions of [a, b] such that ||P„|| < ^ and Pi > Pi > P3 ^ - •-. Let Ini be the /th subinterval of P„. Let M„/, mni be the sup and inf of f on Int. Let gn be the step function which is mni on /„/, and let hn be Mni on Ini. Then gn and hn are measurable, and gn < f < hn for all n. The sequence {g„} increases to a measurable function g < f^ and {hn} decreases to a measurable function h > f. Show that if f is continuous at ^ , then g(xo) = h(xo). Show conversely that if/z(:> )—g(;C()) > 6:,thenM„/—m„/ > 6: whenever XQ e Ini J and consequently there are points Un,Vn G Ini such that f(Un) - f(Vn) > s* Since ||P„|| —> 0, Un —> xo and Vn —> XQ, and f is discontinuous at XQ.

Now we know that f is continuous at XQ if and only if g(xo) = h(xo). Use the fact that J gn = L( /", P„), Jhn = U( /", P„) to show that / is Riemann integrable if and only if the Lebesgue integral I(h- g) = O.BY Problem 6 we know J{h- g) = 0 ii and only if h = g a.e., so f is Riemann integrable if and only if f is continuous a.e. ""HI

Problem 13. Let C be the Cantor set (Problem 11, Chapter 4) and let D be a Cantor-like set of positive measure (i.e., a nowhere dense closed subset of [0, 1] of positive measure). Are the characteristic functions xc and XD Riemann integrable? ""HI

THE INTEGRAL OF UNBOUNDED FUNCTIONS

In this chapter we extend the definition of /^ f to cases where f is unbounded or S has infinite measure. These are situations where the Riemann integral would be called improper. For the Lebesgue integral the extensions to unbounded functions and infinite measure sets is more natural, and we will not need to stigmatize that situation with the "improper" terminology. We will now also consider extended real-valued functions, which may occasionally take the values +00 or — oc. Such functions f will arise naturally as limits of sequences, and indeed as limits of sequences {fn] such that jfn converges to //". For integrable functions the sets where f = ±00 will of course have measure zero.

The general definition of /^ f will still coincide with the geometric idea of the net area under the graph, i.e., the area above the X-axis minus the area below the x-axis. Both these areas will be required to be finite, in contradistinction to the improper Riemann integral. For example, the function which is (—IT^ on the interval (n,n + 1) is improperly Riemann integrable over [1, cx)) since

. 1 1 1 - 1 + 2 - 3 + 4 - -

converges, and consequently

lim / /

61


converges. This function is not Lebesgue integrable since the positive area is

1 1 1 X + 7 + 7 + ---, 2 4 6

which is infinite. A Lebesgue integrable function is always absolutely integrable in the sense that if f is measurable, then f is integrable if and only \i\f\ is integrable.

We use the same upper sum-lower sum approach to define /5 f for f unbounded or S of infinite measure, only now we allow countable partitions. We saw earlier (Problem 2, Chapter 5) that countable partitions make no changes for bounded functions and sets of finite measure, so none of our earlier results change. As before, integrability will be equivalent to measurabil-ity, provided both the positive and negative areas are finite. The integral is a limit of Riemann sums as before, and the linearity properties follow speedily from that fact.

Upper and lower sums are defined as usual, but these sums are now generally infinite series, and we require that all such series converge absolutely. It would not do if a lower sum changed its value just because the partition sets were listed in a different order. For the partition P = {£/} we again let m/ and Mi be the inf and sup of f on £/, and now define

Mi = sup {| f(x)\ : X e Ei),

If I /*! has a finite upper sum,

^ M , / x ( £ , ) < o o , (1)

then we again let

L( f,P) = J2 ^^^^(E^), U( f.P) = Y. M.M(£ )- (2)

Since

-Mi < nii < Mi < Mi,

the series in (2) will converge absolutely if | /"I has the finite upper sum (1). If (1) holds for some partition P of S, then we say f

7 THE INTEGRAL OF UNBOUNDED FUNCTIONS 63

is admissible over 5, and P is an admissible partition for 5. The existence of an admissible partition implies that the graphs of f and I f\ are contained in a countable union of rectangles Ei X [—Mi, Mi] of finite total area. We can write the sums Y^miiJi(Ei) and J] MiiJi(Ei) whether or not f is admissible, but we write L(f, P), U(f, P) only if P is admissible for f.

Problem 1. If P is an admissible partition for f over 5, and Q> P^ then Qis admissible and

L(f, P) < L(f, Q) < U(f, Q) < U(f, P).

All lower sums are less than or equal to all upper sums. ""HI

A function f is integrable over the measurable set S provided supL(f, P ) = 'miU(f, P). The sup and inf are over all admissible partitions of S. An integrable function is necessarily admissible, and has finite integral.

Since functions are no longer required to be bounded, some nii or Mi may be infinite. If £/ is a set in an admissible partition with Mi = 00, then necessarily /x(£/) = 0 so MiiJi{Ei) = 0. As usual, we define

0 • 00 = lim 0 ' n = 0. n—>OQ

Problem 2. Show that fix) = ;^ is admissible on [0, 1]. Hint: Let P = {Ei\ with Ei = ( i , ^pr) for / = 1, 2, 3 , . . . , and £o = {0, 1, | , J , . . . } . U( f, P) is a geometric series. ""Ill

Proposition 1 and Problem 3 below show that integrability and measurability are again equivalent, given that the function is admissible.

Proposition 1. An admissible measurable function on S is integrable over S.

Proof Let P = {Ei] be an admissible partition of S so that J2Mi[i(Ei) <oo. We can assume that all zero-measure sets of P have been lumped together as EQ, SO M/ = oc only if i = 0. There is N so that E^^jv+i M / M ( £ / ) < 5 ^nd hence E £ N + I ( ^ / - mi) X tJi(Ei) < 28. The set T = ^ 1 U^- • U E^has finite measure and f is bounded on T by max {Mi, M2 , . . . , MM}.


Hence (Prop. 3, Ch. 5) /* is integrable over T and there is a partition ?T of T with U(f, PT) - L{f, PT)<S. The sets of Pj together with EQ, £ N + I , £N+2. • • • form a partition Q of 5 with U{f.Q)-Uf^Q)<?>s. ill

Problem 3. If /"is integrable, then /" is measurable. Hint: See the proof of Proposition 6, Chapter 5. ""HI

Now we show that the general integral is a limit of Riemann sums, and prove the linearity properties.

A Riemann sum for f defined on the measurable set S is a sum Y.f{Ci)iJi{Ei)^ where P = {£J is a partition of S, and Q e Ei for each /. We will use the notation R(f, P, c) for such a sum only if P is an admissible partition of S. Riemann sums are now generally infinite series, and the admissibility condition ensures that the series converge absolutely.

The pairs (P, c) are directed as usual, with (P, c) >- (Q, cO if P > Q^ and with this agreement {R(f, P, c)} is a net on the pairs (P, c) with P admissible. Recall that lim R(f, P,c) = I means that given s > 0 there is PQ SO that \R(f, P,c)-I\ < s ii P > PQ.

Proposition 2. f is integrable over S with J^ f = I if and only ifR(fP,c)-^L

Proof. Assume f is integrable, so given s > 0 there is a partition Po with t7( /", Po) - L( f Po) < e. If P ^ Po, then for any choice c for P,

L(f Po) < Uf P) < R(f P, c) < U(f P) < U(f Po). (3)

From (3) it follows that \R{f P, c) - J^ f\ < e ii P > Po, so R(fP,c)^Jsf.

Now assume that {R(f P,c)} converges, with R(f P, c) —> I. The admissibility of f is implicit in the notation R(f P,c). Let £ > 0 and let \R(f P,c) - I\ < s if (P, c) > (Po, co). In particular, since (Po, ^) >- (Po, ^o) for any choice c, any two Riemann sums for Po are within Is of each other. If Po = {£/}, we find for each / with />6(£/) :^ 0 a c/ € £, so that

f{c^) > M, - s/2'fi(E,)

f{c,)iJi{E,)>Miix{E,)-8/2\


For this pair (PQ.C)^

U(f,Po)<R(f,Po,c) + 8.

One can similarly choose c- for PQ SO

L(f,Po)>R(f,Po,c')-8,

Now we have

U(f, Po) - L(/*, Po) < PC/*, Po, c) + ^ - (PC/*, Po, c')-s)

<R(f,Po,c)-R(f,Po,c') + 2e

< 46,

Hence f is integrable, and from the first part of the proof we know the integral is the limit of the Riemann sums, ill

Notice that nothing in the argument above precludes the possibility that some Mi or m/ or f(Ci) might be infinite, provided the corresponding Ei has measure zero.

Proposition 3. If f and g are integrable over S, then kf and f + g are integrable over S, and

Proof. The partitions which are admissible for f may not be admissible for g^ so we need to restrict our attention to those partitions which are admissible for both f and g. If Po is such a partition (Problem 4) then all the nets {R( f P, c)}, {P(g, P, c)}, {R(kf P, c)}, {R(f + g, P, c)} are defined for P >- Po, and

R{fP^c)—> j j

R(g^P^c)-^ l^g.

It follows from general net theorems that

kR{fP^c)^kjj


and

R(f, P, c) + R{g, P,c)-^ j j + l^g.

Since

R{kf,P,c) = kR(f,P,c)

R(f + g, P, c) = Rif, P, c) + Rig, P, c),

we have immediately that

J kf = Urn R{kf, P, c)

= limkR(f,P,c)

The convergence of {R(f,P,c)} implies the convergence of {kR(f, P, c)}, and hence of {R{kf, P, c)}, which says that kf is integrable.

Similarly,

/ . ^(f + g)^limR(f + g,P,c)

= lim[R(f,P,c) + R(g,P,c)]

= \im R(f, P, c) + limR(g, P, c)

Lf^L The limit of a sum is still the sum of the limits, ill

Problem 4. Show that if f and g are admissible over S, then so is f + g. ""Ill

Problem 5. If f and g are integrable over S, and f < g on S^ then Js f < Is g- Be explicit about what facts you are using. ""HI

Problem 6. Define f'^(x)= rmx{f(x),0}, f~{x)= mm{f(x), 0}, so /*+(x) > 0, f-(x) < 0, and f(x) = /*+(x) + f-(x). Show that f is integrable if and only if /"+ and f~ are integrable, and in this case If = !f^ + Jf-. <


Proposition 4. If f is integrable over S and T is a measurable subset of S, then f is integrable over T, fxr is integrable over S,andJjf = JsfxT'

Proof See the proof of Proposition 5, Chapter 6. ill

If f is integrable over T and T C 5, we will now feel free to consider f as defined on all of 5, with /" = 0 off T, and know thatfsf = JTf.

Problem 7. If f is integrable over A and integrable over B, where A and B are disjoint, then f is integrable over AU JB and

JAUB J A JB II

The usual approach to defining / f for unbounded functions or sets of infinite measure is to consider first f > 0 and define / /" to be the sup of integrals fj g where g is a bounded function under f and T has finite measure. Simultaneous approximation by areas inside and outside is not required. Conceivably this could yield more integrable functions, but the next problem shows it does not.

Problem 8. Show that if / is a positive measurable function on 5, and f has no finite upper sums, then f has infinite lower sums, and there are bounded integrable functions 0<g<f with g = 0 off a set of finite measure and / g arbitrarily large. Hint: Note that we necessarily consider inadmissible partitions here. Let ? = {£,} where Ei = {x : 1' < f(x) < 2^+^} for / = 0, ± l , d = 2 , . . . . Show that E | / | < N ^ / M ( £ / ) ^ | E|/|<NMA^(£i)-You had better also consider what happens if iJi(Ei) = oc, since that isn't allowed in partitions. ""HI

Proposition 5. If f is integrable over S, then there is a finite measure subset T of S such that f is bounded on T and

Is-T \f\<^'

Proof. Let P = {Ei} be an admissible partition of 5, so E Mifji(Ei) < 00. Then E . ^ N + I Miiui(Ei) < s for some N, and f is bounded on T = £i U • • • U Ejv (by max {Mi , . . . , Mjy}), and Is-T\f\<^' •


The approximation result of Proposition 5 allows us to prove the general Lebesgue convergence theorem below as an easy consequence of the bounded convergence theorem (Proposition 8 of Chapter 6).

Proposition 6. (The Lebesgue Dominated Convergence Theorem) If {fn) is a sequence of measurable functions and fn —> f a,e, on a set S and there is an integrable function g on S such that \fn\ < g for all n, then f is integrable and J^ fn —> Is f-

Proof. Notice that all fn are integrable since \fn\ < g and g is integrable. Since fn —> f^ f is measurable, and f is integrable since \f\ < g. Let 6: > 0 and let T be a finite measure subset of 5 such that g is bounded on T and JS-TS < ^- Clearly {/*„} is uniformly bounded (by a bound for g) on T, so Jj fn —> Jj f by the bounded convergence theorem. Since \fn\ < g and \f\ < g,

JS-T I \JS-T < 8.

Therefore, if we choose N so that \ Jj fn- h f\ < ^ forn > N, then for n > N^

f fn- f f \ < \ f fn- f f\ + \f fn- f f JS Js I \JT JT I \JS-T JS-T

< 36, ill

The dominated convergence theorem says in essence that pointwise convergence of {fn} implies convergence of the integrals, provided only that the graphs of the fn all lie in some fixed finite area. If the fn are positive and their graphs are not all in some fixed finite area, then the integrals ffn may get too big, but they are always at least as big as Jf in the limit. That is what Fatou's lemma says.

Proposition 7. (Fatou's Lemma) If {fn} is a sequence of non-negative integrable functions on S, and fn —> fax., and f is integrable over S, then

lim inf / ' Jf-


If f is not integrable (i.e., not admissible)y then

lim inf j fn = lim fn = oo.

Proof. Let /z be a bounded measurable function, with 0<h< f and h = Ooiia finite measure set T C 5, and fjh > Js f — - Let g^ = f^Ah^ so the gn are uniformly bounded on T, and gn —> h on T. By the bounded convergence theorem,

Since s is arbitrary, lim inf/^ ^ > /^ f. The second part of the proof is the following problem, ill

Problem 9. Let {fn} be a sequence of non-negative measurable functions such that fn —> f a.e. Assume that /", which is measurable, is not admissible. Show that lim J f^ = oo. Hint: See Problem 8. •""Ill

Problem 10. State and prove a Fatou-type theorem for negative fn. <

Here is a slight generalization of the Dominated Convergence Theorem.

Problem 11. Let {/^}, {gn} be sequences of non-negative measurable functions such that fn < gn for all n^ fn —> f a.e., and gn —> g a.e. If g is integrable, and Jgn —> Jg^ then Ifn —> If' Hint: Use Fatou's lemma. ""HI

Problem 12. (The Monotone Convergence Theorem) If {fn} isasequenceof non-negative measurable functions, with fn<f for all n, and fn —> f a.e., then / fn —> J f H f is integrable. Note: The usual situation is the case where {fn} increases to f; hence the name. ""HI

We have seen that some functions which are not Lebesgue integrable are nevertheless improperly Riemann integrable because of a cancellation of positive and negative areas. If a non-negative function is Riemann integrable, however, then it certainly is Lebesgue integrable. Improper Riemann integrals


are of two basic types: either f is unbounded at one end of a finite interval, or f is bounded on an infinite interval. We will consider the second case next and leave the first case as a problem.

Proposition 8. If the improper Riemann integral f^f converges, and f >Q on{Q, oo)^ then f is Lebesgue integrable on [0, oo)^ and the integrals are the same.

Proof. We are assuming that f is Riemann integrable, hence bounded, on each interval [0, w], and that lim JQ f converges. It follows that f is integrable, in both senses, on each interval [n, n + 1], and

N POO rl\

/ / •= lim / f Jo N-^ooJo

N-1 /.«+!

= lim V / f

oo .n+\

n=0

CX)

(4)

where the last integrals are in the Lebesgue sense. For each of the Lebesgue integrals /[„,„+i] f there is a partition P„ = {£„/} of [n, n+1] such that U(Pn, f) — L(Pn, f) < 6:/2". For convenience assume that the singletons {n} and {w + 1} are elements of each P„. Let Q be the partition of [0, oo) consisting of all the sets in all the Pn. We have

/ f-8/r<L{fPn)<u{fPn)< I f + s/r.

Now add and get

CX) ^ CX) «

E / ^^f-^<Uf,Q)<U{f,Q)<Y.l ^,J + ^-


Using (4) we conclude

/ f-s< f< / f + s, Jo J[0,oo) Jo

and the integrals are the same, ill

Problem 13. Let f be non-negative and Riemann integrable on [6:, 1] for each £>0 and assume lim f^f=L Show that f is

£ ^0+

Lebesgue integrable over [0,1] with integral / . ""HI

Problem 14. Let fn(x) = n^^^x/(l + n^x^) for x e [0,1]. (i) Show that fn{x) —> 0 for all x G [0, 1].

(ii) Show that {fn] is not uniformly bounded. (iii) Show that g{x) = 1/v^ is Lebesgue integrable on [0,1],

and fn<g for all n. (iv) Conclude that J fn—^ 0. II

Problem IS. (i) Let {£„} be a sequence of measurable sets in [0,1] with/x(£„) < ^ ,andle tg^ = XE,. Show that g^ —> Oa.e. Hint: Let FM = U / ^ N ^ « ^^^ ^ = ONPN- Show />6(P) = 0 and gn —> 0 off P. (ii) Show that assuming /x(£„) —> 0 is not sufficient to conclude that gn —> 0 a.e. Hint: Let £i = [0, | ] , £2 = [i, 1], £3 = [0, i ] , £4 - [|, i ] , £5 - il | ] , £6 = [f, 1], £7 = [ 0 , i ] , . . . . -iill

Problem 16. Ii f is integrable on 5, then for every s > 0 there is 5 > 0 so that /E I / j < ^ whenever E C S and />6(£) < 3. Hint: Assume without loss that f > 0 and suppose the statement false. Then there is £ > 0 and a sequence {En} of measurable subsets of S with lui(En) < 1/2" and /^^ / > ^. Use Problem 15 and the Lebesgue Dominated Convergence Theorem with gn = XEn ' f to reach a contradiction. ""HI

Problem 17. (Differentiation under the Integral Sign) Let f(x, t) be defined on [a,b]^ {c, d), with to G {c, d). Assume that for each fixed t e {c, d)^ f(x, t) is a measurable function of x on [ , /?], and for each fixed x e [a,b]^ f(x, t) is a differentiable function of t on (c, J) ; i.e., ft{x, t) exists for (x, 0 ^ [ . ^] x (c, J) . Assume that f{x, to) is integrable on [a, b]^ and that there


is an integrable function g on[a, b] such that | ftix, t)\ < g(x) for all (x, t) e[a,b] X (c, d). Show that:

(i) ft(x, t) is a measurable function of x for each ^ e (c, J) , and hence f^(x, 0 is integrable on [i , &].

(ii) For each t e (c, d)^ /"(•, 0 is integrable over [a,b]. (iii) If F(t)= J^f{x,t)dfi(x) for te{cj), then f^(^o) =

Xf /^(^' to)dfJi(x) for each ^ G (c, J) . Hints: (i) Let tn —> t € (c, J) , so for x e [a,b]

—— > ftix, t). tfl t

(ii) For each t e (c, d) there is s G (c, d) such that f{x, t) = f(x,to) + ft(x,s)(t - to), and hence \fixj)\ < \f(x,to)\ + \ft(x,s)\\t-to\.

(iii) Let tn —> ^ , so

The difference quotients are dominated by g(x) and converge pointwise to ft(x, to). "4

8 DIFFERENTIATION AND INTEGRATION

In this chapter we make some connections between differentiation and integration. In particular, we see how the Fundamental Theorem of Calculus fares with respect to Lebesgue integration. The results of this chapter are necessarily germane only to real-valued functions defined on the real line. This is in contrast to our earlier work, where the results are stated for functions and sets on the line, but the techniques and proofs have quite general application. The theorems of this chapter are standard useful results, but the proofs tend to be difficult and of limited applicability. The reader may be well advised to study the statements of the theorems and problems, and leave the proofs for a long rainy day.

The two forms of the Fundamental Theorem are

dx rf=f(x) (1)

Ja

t f = fib) - fia). (2) Ja

In calculus one shows that (1) holds for a Riemann integrable function f which is continuous at x. This is also true for a Lebesgue integrable function, with essentially the same proof (Problem 1). It is considerably more difficult (Proposition 5) to show that if f is bounded and measurable, and so Lebesgue integrable, then (1) holds a.e.

Problem 1. Show that if f is Lebesgue integrable on an interval containing a and Xo, and f is continuous at XQ, and a < XQ:,

73


and

Ja

then F'ixo) = f(xo). <\

We will now show that (2) does not hold in general by constructing a continuous increasing function fon [0,1] with /"(O) = 0, f(l) = 1, and fix) = 0 a.e. Thus

Jo r = o^ fiD- f(0) = i.

Define f on [0,1] as follows: let f = j on [|, | ] ; then let f = ^ on [|, | ] and /" = | on [|, | ] . In the next step, take the middle third out of each of the four remaining intervals, and define f to be | , | , | , and | on these middle thirds. The picture thus far is Fig. 1.

Continuing in this way we get f defined on a union of disjoint closed intervals with total length

1 2 4

3 + 5 + 2 7 + - = l-

-2. Z _§. 3 9 9

Fig. 1

8 DIFFERENTIATION AND INTEGRATION 75

Clearly f'{x) = 0 on the union of the interiors of these intervals, so f\x) = 0 a.e. The inductive scheme outlined above defines f on a dense subset of [0,1]. To show that this definition can be extended to the rest of [0, 1] with the result being continuous, it suffices (Problem 2) to show that f is uniformly continuous on the union of the closed intervals. Notice that

1 1 if \xi - X2\ < - , then | f(xi) - f(x2)\ < -

1 1 if \xi - xil < —, then | f(xi) - f(x2)\ < - ,

and so on, so f is uniformly continuous on its domain. The function f is known as the Lebesgue singular function.

Problem 2. Let f be defined and uniformly continuous on a dense subset £ of [0,1]. Specifically, given s > 0 there is S > 0 so that if xi, X2 e E and \xi — X2\ < 5, then | f(xx) — f(x2)\ < s. Let X G [0,1] and let [Xn) be a sequence in £ such that Xn -^ x. {x may or may not be in £.) Show that {f{Xy } converges, and the limit is independent of the sequence {Xn\ converging to x. If X G £, show that lim /^(x) = f{x). If x ^ £, define f{x) = lim ^(x) , where Xn ^ E and x„ -^ x. Show that f is continuous on [0,1]. II

A Vitali covering of a set £ is a family V of proper intervals (open or closed or half-open, but not points) such that every point of £ lies in intervals of V of arbitrarily small length. That is, given x e E and t] > 0 there is / G V such that x e I and 1(1) < r]. The proof that an increasing function has a derivative a.e. depends on the following proposition.

Proposition 1. (VitaWs Theorem) If iJi{E) < oo and V is a Vitali covering of E, then for each 8>0 there is a finite disjoint set {/i,...,/^} of intervals of V such that IJL(E—(II U • • • U7]v)) < £y and hence iJi{E n U Jy) > /x(£) - s.

Proof. Let JJ be an open set containing £ with /x(U) < oc. Discard from V all intervals not contained in U, and all intervals which do not intersect £ . What remains is of course still a Vitali covering of £ . We can assume without loss that


all intervals of V are closed, because closing them would affect neither the hypothesis nor the conclusion. Suppose we let 7i be any interval from V of maximum length, and h any interval of maximum length which does not intersect Ii. Of course there may not be such maximum length intervals, but for the moment we assume there are to see how the argument goes. Later we will correct the argument by replacing these by intervals which are nearly of maximal length. Define inductively a disjoint sequence 7i, / 2 , . . . of intervals of V such that each J„+i is of largest possible length and disjoint from /i , . . . , / „ . If at any stage E is covered by /i U • • • U J„ we are done. Otherwise, since /i U • • • U I„ is closed, each x e £ — (/i U • • • U /„) is in some intervals of V which are so small they miss /i U • • • U I„. Thus we have a sequence {/„} of disjoint intervals, all contained in U, with Y^i(In) < oo and i(In) -^ 0. Let £ > 0 and pick N so large that E N + I ^(^n) < ^. Consider any point x of £ — (/i U • • • UI^) and any interval J of V which contains x and is disjoint from /i U • • • U /]v. The interval I must intersect some I^ with m > N^ for otherwise since I^ is of maximal length among unchosen intervals, i(Im) ^ ^(I) for all m, and this is impossible since l(Im) -^ 0. Let /^ be the first interval that / intersects, so m > N. Since £(J^) > £(J), x lies in the interval / ^ which has the same center as I^ but is three times as long (Fig. 2).

Hence every point x of £ - (Ji U • • • U EM) is in Um=N+i /^^ which has measure less than 3^.

*Jm

Fig. 2

'm


Now instead of assuming that each 4+i has maximum length, let Ln = sup{£(/) : / G y and / n (/i U • • • U 7 ) = 0}. Then pick 4+1 so that l(In+i) > jLn. The arguments above all proceed as before except we must make Jm five times as long as I^, so that £ — (Ji U • • • U Ijq) has measure less than Ss. ill

Problem 3. Let q be any function and assume that

{x : liminf ^(0 < lim sup q(t)}

has positive (outer) measure. Show there are rationals r and s such that

{x : liminf^(^) < r < s < limsupq(t)} t-^x-\- t^x+

has positive (outer measure). Hint: The measurability of the sets is not relevant here. ""HI

Proposition 2. If f is increasing on [a, b] then f\x) exists a.e.

Proof. The four Dini derivates of / at x are

n + / / ^ r f(x + h)- fix) D^ f {x) = lim sup

p ^ - . , , y f(X + h)- fix) U iix) = lim sup

' h-^0- h

The derivative f\x) exists if and only if the four derivates are equal and finite at x. The proof consists in showing that the set where any two derivates differ has measure zero. We illustrate by showing that the set {x : D+ fix) > D+ fix)} has measure zero. If this set has positive measure, then there are (Problem 3) rational numbers r and s such that

A={x: D^fix) <r <s < D""fix)}


has positive measure. Assume /x(A) = /? > 0, and let U be an open superset of A so that /x(U) < p + 6. For each x e A there are, by definition of D+ f(x), arbitrarily small numbers h such that [x, X + h] c U and

f(x + h)- fix) < rh.

These intervals [x, x + h] form a Vitali covering of A, so there is a finite disjoint collection [xi,Xi +hi]^i = 1, 2 , . . . , N, with

fji(An[j[xi,Xi +hi]) > p- 6.

Since U[^M / + / i] C U,

I J2hi </x(U) < p + £.

Let B = A n U/li(^o / + ^z)? so that /x(B) > p - s and each point of i5 is the left endpoint of an interval [y, y+k]^ contained in some (x/, Xi +hi)^ such that

f(y + k)- f(y)>sk.

This last inequality follows from the fact that D^ f(x)>s for each X e B. Invoke Vitali's Theorem again to get a finite disjoint family [yy, yj + kj]^ / = 1 , . . . , M, such that

JBH \J[yi,yi + ki]\ > fi(B)-s> p-ls,

M

It follows that J2 kj > p — Is. Now we have 7=1

N N

J2(f(x,+h,)- f(x,))<r^h,<r(p + 8)

and

M M

E(/"(y/ + ^;) - /*(y/)) > 5 E ; > s(/7 - 2s).


Since each interval [yy, yj+kj] is contained in some and / is increasing,

M N

X:(/'(y; + ;) - f(yi)) < Y.^f{x, + hi) - fix,)), i=l i=l

Thus, for every s > 0

s{p-l£) < r(p + s),

so

sp < rp

s <r,

and we have a contradiction, ill

Proposition 3. If f is increasing on [a, b], then f is measurable and j ^ f < f(b)- f(a).

Proof Let

fn(x) = n fi^+D-M where we interpret f(x+j^) = f(b)iix+^ > ^. Then /*„ is measurable for each n^ and fn(x) -> f\x) a.e., so f is measurable. Each fn is non-negative, so Fatou's Lemma gives

rb rb

/ f'= lim/; Ja J a

rb

< lim inf / fn Ja

= limmi {nf"f-nf"f^

= f(b) — lim sup n f Ja

< fib) - f{a).


The last inequality follows from the fact that f is increasing, so for all w,

/*^+^ 1 n f> nf(a)- = f(a),

Ja n lllllll

Proposition 4. If f is integrable on [a, b] and J^ f = 0 for all X e[a, b], then f = 0 a.e.

Proof Suppose J^ f = 0 for all x, so jff = 0 for all c, J G {a,b). Suppose, to be specific, that f is positive on a set E of positive measure. Then there is a closed subset F c E so that /x(f) > 0 and Jp f > 0. Let U be the open set (a, b) — P , and write U in terms of its components: U = [jiai, bj). Since UUF = (a, b),

Therefore

/ f = - f f<0. Ju JF

E / f<o,

and so X?' f ^Oior some interval (ai, bj)^ which contradicts the assumption, lllllll

The following result is true for functions which are integrable on [a, b]^ and so not necessarily bounded. The proof of the more general theorem requires additional machinery, and so we stick with the version below, which contains the essential ideas.

Proposition 5. If f is bounded and measurable on [a, b] and

'(x)= r f Ja

then F is continuous and F(a) = 0 and F\x) = f{x) a.e.

Proof. Let f-^ = fvO and /*" = (-/") v 0 , so /*+ and f~ are non-negative functions and f = f^ — f~. Hence

F{x) = r f^- r f- = F,(x) - f2(x), Ja Ja


where Fi, Pi ^^e increasing functions. It follows from Proposition 2 that F\x) exists for almost all x. Moreover, if M is a bound for f^

\F(x2)^F(x^)\ =

so F is continuous. Let

f„(x) = n P \x + n,

< M|:x'2-:x;i|,

F(x)

= n /, Jx

SO that I fn{x)\ < M for all x, and fn{x) -^ F\x) a.e. By the Bounded Convergence Theorem and the continuity of P,

rx nx

/ F = lim/ U J a « J a

lim

= lim

rnF {t+-\dt- rnF{t)dt Ja \ nj Ja

/ nF{t)dt- / nF{t)dt Jx Ja

= F(x)-F(a)

= F(x)= r f. Ja

Now we have

f\F' - /•) = 0 Ja

for all x, and consequently F' = f a.e. by Proposition 4. Illllll

For convenience we will call the points x where F\x) = f(x) the L-points of f. The L-points play critical roles in many theorems on integral representations. We give one example from potential theory. The problem is the so-called Dirichlet problem— namely, to find a harmonic function u{r,6) on the open unit disc in the plane with a specified boundary value f{(p) on the


unit circle. Here we will let f he a bounded measurable function on [0, ITT]^ which we regard as the unit circle. The Poisson kernel is the function

1 +r^ -2r cos(6 - (p)'

defined for (r, 6) in the open unit disc (polar coordinates), and (p on the unit circle. For each fixed cpo, P(r,6; (po) is a harmonic function on the open disc ( 0 < r < l , 0 < ^ < 2 : / r ) . For each fixed (r, 0), P(r,0; cp) is a continuous function of ^ e [0, ITT]. Hence f{(p)P(r,6', cp) is an integrable function of (p for each (r, 6) in the disc. We let

u{r,e) = [ " f((p)P(r,e; cp)--dfi((p), Jo LTt

The ^ is so the total measure of the circle is 1. We can regard the integral as a limit of Riemann sums, so

u{r, 6) = l i m j ] P(r, 6; cp,) f{ip,)—ii{E,). ^^^ In

The sums have the form

N Y.^^Mr,e) (3) i=l

where p/(r, 6) = P(r, 0; cpi) and Ui = ^ f{(pi)iJi{Ei). The sums (3) are linear combinations of harmonic functions, and hence are harmonic. The limit of these sums, u{r, 0), is also harmonic, although not obviously so. The function u(r,6) solves the Dirichlet problem in the following sense: if cpo is an L-point for f (and that includes all points where f is continuous) then

lim u(r,(po) = ficpo)- (4)

In particular, (4) holds for almost every cpo.

Problem 4. Let F(x) = JQ f. Show that F'(0) can exist even though f is not continuous at 0. Hint: Let f be defined on [0, 1]


as follows: f = 1 on [i , | ] and f =-1 on (f, 1], so 0 < /f /* < I for i < X < 1. Divide [\,\) up into four equal intervals, and let f be alternately +1 and - 1 on these, so 0 < /f /" < ^ for I < X < I , etc. Show that f is integrable—even properly Riemann integrable since the set of discontinuities has measure zero—and that (F(x) - F(0))/x -^ 0 as x -> 0+. """11

PLANE MEASURE

In this chapter we develop Lebesgue measure for sets in the plane, R^. Our purpose is threefold. First, by developing another specific example of a measure we show how the same techniques used on the line can be used to define countably additive measures in quite general situations. Our second purpose is to have two-dimensional measure—area—defined so we can show that jfdiJi really is the area under the graph of f. Thirdly, the development of plane measure provides a template and example for defining general product measures.

We take the rectangle as our basic plane figure, with the rectangle playing the role played by the interval on the line. Here we will use "rectangle" to mean rectangle with sides parallel to the axes and having positive length and width. Thus a rectangle is a set / X / with / and / intervals, which may be open or closed or half-open. Single points and line segments are not rectangles. If R = J X / , then the area of R is a{R) = l{I)t{]).

For any set £ c M ,

X{E) = mi{Y,ot{R,) : E C[}R^}

where [Ri] is a finite or countable family of rectangles. A(£) is the Lebesgue outer measure of £ which we will call simply the measure of £ .

The same arguments used when we defined m{E) on the line will show that it is immaterial whether we use coverings of £ by open rectangles or closed rectangles or partially closed rectangles (Problem 2). We can also invoke the plane version of

85


the Heine-Borel Theorem and notice that if £ is a compact set, then A.(£) is the inf of sums Ylot(Ri) for finite coverings of E by rectangles of any sort.

Problem 1. (i) Show that hne segments have plane measure zero; for example, the segment from (0, 0) to (1, 2) has measure zero.

(ii) Shov^ that lines have plane measure zero. ""HI

It v ill occasionally be convenient to use coverings by squares of the form d x e, v^here d = [^, ^ ] , e = [j^, ^ ] . We will call an interval of the form [^, ^ 1 , with or without endpoints, a dyadic interval of length ^ . A square of the form d x e will be called a dyadic square of side ~.

Any rectangle R can obviously be covered by a finite number of dyadic squares. Si, S2 , . . . , S„, with

Any countable covering {R„} of any set can be covered by count-ably many dyadic squares {S„y}, where S„i, Sni,... cover R^, and

Y,ct(Sni)<a{Rn)+s/2\ i

so

^a (S„y) < Y,a{Rn) + s,

Thus we could replace coverings of E by rectangles in our definition of k with countable coverings of £ by dyadic squares. Notice that a finite covering of £ by non-overlapping dyadic squares can be replaced by a finite covering of non-overlapping dyadic squares all of the same size, with the same total area. (Rectangles are called non-overlapping provided their interiors are disjoint.)

Problem 2. Show that the infimum of sums Y^a{Ri) for families {Rj} of closed rectangles covering £ is the same as the

9 PIANE MEASURE 87

infimum of sums J2a(Qi) for coverings {Q/} of E by open rectangles, and the same as the infimum of sums Yloi(Si) for coverings {Si} of E by dyadic squares. ""HI

Problem 3. Xis translation invariant. ""HI

The following elementary properties are immediate from the definition.

Proposition 1. (i) A(0) = 0; (ii) A(C) = 0 for every countable set C; (m)X(E) >OforallE; (iv) ifE C F, then X{E) < k(F); (v) k is countably subadditive; i.e., if {Ei) is any countable or

finite family, then >^([j Ej) <^X(Ei).

Problem 4. Prove parts (ii) and (v) of Proposition 1. ""HI

Since X(E) is to represent the area of E we need to know that X gives the right answer for rectangles.

Proposition 2. If Q is a rectangle, X(Q) = af(Q).

Proof First assume that Q is a closed rectangle. Clearly HQ) < oi(Q) since {Q} is a 1-rectangle covering of itself. Fix 6 > 0 and let {Sk} be a finite covering of Q by dyadic squares of the same size so that I ] a(5^) < X(Q) + s. We may assume all Sk intersect Q. Since Qis a rectangle, say Q= I ^ J, the squares Sk will consist of all squares di x ej(i = 1,,,, ,n; j = 1,,.. ,m) where the di form a non-overlapping covering of I and the Cj form a non-overlapping covering of / . Hence

[JSk = \Jdi xcj k i,j

= (Jl U • • • U (4) X ( 1 U • • • U Cm),

and

i j


Since this holds for all £ > 0, and we already have A(Q) < /x(J)/x(/)5 we conclude that X(Q) = /x(/)/x(/). The remaining cases, where Q is not necessarily closed, are left to the reader as an exercise, ill

Problem 5. Complete the proof of Proposition 2 by showing that A(Q) = Qf(Q) for an open rectangle Q, and consequently, by monotonicity, for rectangles containing some but not all of their boundary points. ""HI

We saw in Chapter 4 that the Caratheodory criterion for mea-surability, while not in itself very intuitive, does get us speedily to the additivity properties that outer measure has when restricted to the measurable sets. We therefore adopt this condition forthwith as our definition of measurability.

A set £ C M is measurable if and only if

k{E n T) + X(E' n T) = X{T) (1)

for every set T. Here F = R^ - E. Since k is subadditive, we always have

X(E nT)+ X{E' n T) > A(T),

so E is measurable if and only if, for all T,

X{E n T) + X{E' n T) < A(T). (2)

Clearly we need only consider sets T of finite measure.

Problem 6. (i) Sets of measure zero are measurable. (ii) If £2 = El U £0 with £1 measurable and X(£o) = 0, then

£2 is measurable. ""HI

Problem 7. Translates of measurable sets are measurable. ""HI

Proposition 3. E is measurable if and only if

X(£nR) + A(FnR) = A(R)

for every rectangle R.

Proof. (Cf. the proof of Proposition 3, Chapter 4.) The condition is obviously necessary, so assume that E splits all rectangles

PIANE MEASURE 89

additively, and let T be any set of finite measure. Let {Rj} be a covering of T by rectangles with

^a(Rj)<HT)+s.

Then £ n T c U(£ n R,) and E'nT c [J(E' n Rj). Hence, by monotonicity and subadditivity,

HE n T) + X(E' nT)<Y,x{En Rj) + Y.x(E' n R,)

= Y,[X(E n R;) + k(E' n R;)]

= ^ M i ^ , ) < M T ) + £.

Therefore X(Er]T)+ k(E' n T) < A(T) and E is measurable, ill

Proposition 4. Rectangles are measurable.

Proof. Let Q be the rectangle that we want to show is measurable. We show that Q splits any rectangle R additively. R is the union of Q n R and at most eight other non-overlapping rectangles S-[,..., Sg whose areas total a(R). (See Fig. 1.)

Since Q ' fi R c Si U • • • U Sg,

X(Q'nR)<a(Si) + ---+a(Ss).

Therefore,

XiQ nR) + X(Q'n R) < a{Q n R) + a(Si) + •••+ a^Ss) = a(R) = X(R).

S3

S2

S4 [ S E

, j

^ .

Ss S7

Fig. 1


Subadditivity gives the other inequaUty, so Q spUts any R addi-tively, and hence Q is measurable, ill

Proposition 5. If{Ei} is a finite or countable family of disjoint measurable sets, then X(U Ei) = YJHEJ).

Proof The proof of Proposition 5, Chapter 4, appUes verbatim to this case, ill

Problem 8. If T is any set and {£J is a finite or countable family of disjoint measurable sets, then

x{Tn[jE,)=J2UTnE,). II

Proposition 6.IfE\,.,.,En are measurable, then £iU- • • U £„ is measurable, £i Pi • • • n £„ /s measurable, and E\ — £2 is measurable.

Problem 9. Prove Proposition 6. Hint: Cf. Proposition 6, Chapter 4. •""Ill

Proposition 7. If{Ei} is a countable family of measurable sets, then U Ei is measurable.

Proof. We can assume the sets are disjoint because the differences £2 - £1, £3 - (£1 U £2), etc., are measurable by the preceding proposition. If £„ = £1 U • • • U £„ and £ = U/^i ^M then for any T,

A(T) = M T n f „ ) + M T n £ ^ )

= f:MTn£,) + MTnf^)

n

>^A(Tn£,) + x(Tn£0. i=\

Since this holds for all w,

00

>A(Tn£) + x(Tn£'). ill

9 PLANE MEASURE 91

Problem 10. If A and B are measurable, AcB^andX(B) < oc, then B - A is measurable and HB - A) = X(B) - k(A). <\

Proposition 8. Every open set in R^ is a countable union of open squares, so every open set and every closed set is measurable.

Proof. Every point (x, y) of an open set U lies in arbitrarily small dyadic squares. Since some open disc around {x, y) is contained in [/, a sufficiently small dyadic square containing {x, y) will lie in U. The union of all these squares is obviously U, and there are only a countable number of dyadic squares altogether. Squares are measurable, so open sets are measurable. If F is closed, then F^ = U is open, so measurable, and hence F = U^ is measurable, ill

Problem 11. For any measurable set £ ,

X(E) = mi{X(U) :EcU, Uopen}

= sup{A(F) : F c E, F compact}.

Hint: Show this first for a set £ C [0, 1] x [0, 1] and then write £ as a countable union of sets £ n 5^ where the Sn are non-overlapping unit squares. '•"HI

Any measure v on a topological space X is called regular provided that all open and closed sets are measurable, and for every measurable set £ ,

v(£) = inf{v(U) : £ C U, Uopen}

v(£) = sup{y(£) : F c E, F compact}.

Problem 11 is the statement that k is regular on R x R. We have already seen that /x is regular on R.

Problem 12. If for each s > 0 there are measurable sets A and B such that A c £ C B and X(B) < X(A) + s, then £ is measurable. Show that the condition k{B) < k(A) + s^ which allows A(A) = k(B) = oo, does not suffice. ""HI

10 THE RELATIONSHIP BETWEEN fji AND A

Since R^ == E x E there is naturally a relationship between plane measure and linear measure. In this chapter we will investigate the connection, and thereby preview the development of general product measures.

Proposition 1. If A and B are subsets of E of finite outer measure (but not necessarily measurable sets), then A( A x B) < /x(A)/x(B).

Proof. Let s > Q and let {Ij}., {]k] be coverings of A and B by intervals, with

5]€( / , ) <^Ji{A) + 8

^l(Jk)<tJL(B)+8,

Since {/y x Jk} is a covering of A x B by rectangles, we have

X(AxB)<Y,a(IfxJk)

<(/x(A) + ^)(/x(B)+^).

This holds for all 6: > 0, and /xCA), /x(B) are finite, so X{A xB) < /x(A)/x(B). ill

Problem 1. Let A, B be subsets of E with /x(A) = 0. Show that k(A X B) = 0. Hint: If /x(B) < oc this is clear. Otherwise, write B = [jBn where B„ = B n [w, n + 1), w = 0, ± 1 , . . . . ""H

93


Proposition 2. If A and B are compact subsets of M, then X{A X B) = ^l{A) ^l{B).

Proof. Since A and B are closed and bounded sets they are automatically measurable with respect to /i, and have finite measure. By Proposition 1 we need only show that A(A x B) > ^Ji{A)^l{B).

Since A x B is compact we can approximate A( A x B) by the total areas of finite coverings by dyadic squares all of the same size. Let {Sij} = {4 x ej} be such a covering, where all ei and dj are dyadic intervals of length 1/2^, and

Y,a(Sij) <X(A X B) + 6,

We may assume that {Sij} is exactly the set of all dyadic squares of side 1/2^ which intersect A x B, and hence that

y S/y = Ji X ( 1 U • • • U ^^) U • • • U (i„ X ( 1 U • • • U em),

where {di,,,.,dn} is a non-overlapping covering of A and {^1,..., em} is a non-overlapping covering of B. Thus

X(A X B) + 6>^a(Sii)

>/x(A)M(B).

The use of dyadic squares of the same size is necessary to ensure that we have a finite covering of A x B of the form {Ij x /y : / = ! , . . . , « ; / = ! , . . . , m}. ill

Proposition 3. If A and B are measurable subsets ofR, then

A(A X B) = /x(A)/x(B).

Proof First assume that A and B have finite measure, so there are compact sets F and G, with F C A, and G C B, and

/ x ( A ) - ^ < / x ( P )

A6(B)-^</x(G).

10 THE RELATIONSHIP BETWEEN /X AND A 95

Hence

A6(A)/x(B) >X(A X B)

> X(F X G)

= /x(f)/x(G)

> ( /x (A) -^ ) ( /x (B) -8 ) .

Therefore the desired equaUty holds if A and B have finite measure.

If /x( A) = 00 or /x(B) = oo, we consider two cases: (i) /x( A) = 00 and 0 < /x(B) < oo, with />6(A)/x(B) = oo, and (ii) IJL(A) = oo and /x(B) = 0, with /x(A)/i(B) = 0. In case (i), for each n there is a compact set f C A with /X(JF) > n. Therefore, for every n^

k(A xB)> k(F xB) = /x(P)/x(B) > ^/x(B),

so k(A X B) = oo = /x(A)/x(B). If /x(A) = /x(B) = oo then clearly X(A x B) = oo.

In case (ii), we let A^ = An[n,n+ 1) for —oo < w < oo, so liJi(An) < oo and

UA xB)< ^ A ( A , X B) = 5^/x(A,)/x(B) = 0,

so k(AxB) = 0 = /x(A)/x(B). ill

Proposition 4. If A and B are measurable subsets of R, then A X B is a measurable subset o/* E x R.

Proof. We check that

X{{A X B) n R) + A((A X B) n R) < A(R)

for every rectangle R = I x J which is sufficient by Proposition 3, Chapter 9. Since A and B are measurable,

/ x ( A n / ) + / x ( A ' n / ) = fJi(I)

n(Bnj) + fji(B^nj) = tz(j).


Hence, since X(R) = /x(J)/x(/)

x(R) = [^l(AnI) + M(AnI)][^l(BnJ) + ^l(B^nJ)]

= /x(A n l)fi(B n / ) + fjL(A n /)/x(B' n / )

+ fJi{A n I)^l(B n / ) + /x(A' n J)/x(B^ n / ) . (1)

Let

Ri = (AnI) X ( J3n / ) - (A X B)nR

R2 = (Anl )x(B^n/ )

R3 = ( A n 7 ) x ( B n / )

R4 = (A'n J) X (B'n/) .

R15 R25 35 R4 are disjoint, their union is R, and R2 U R3 U R4 = (A X B)' n R. By Proposition 1,

/x(An J)/x(Bn/) > A[(An/) x (Bn/)] = x[(A x B)nR],

with similar inequaUties for R2, R3, R4 in place of Ri. Hence from (1),

X(R) > X[(A X B) n R] + k(R2) + HR3) + HR4)

>A[(A X B ) n R ] + A[(A X B / n R ] . ill

In the general study of measure theory one starts with a a-algebra of subsets of some set X, and a countably additive measure function v on these specified "measurable sets." To define the product measure v x v on X x X one starts with the basic sets of the form A x B^ where A and B are measurable subsets of X. The product measure of such sets is of course v x v ( A x B ) = v(A)v(B).

Our plane measure X is of course the same as the product measure /JLX /JL^ although we did not define it that way. There is a fundamental intuitive idea of area, based on the area of a genuine rectangle, which is more or less independent of linear measure. We therefore used a(I x / ) = i(I)l(J) as our basic notion in defining X on plane sets. This procedure was also designed to provide a second example of how an outer measure

10 THE RELATIONSHIP BETWEEN jJi AND X 97

is defined in terms of some more basic notion. In the following proposition we show that X could equally well have been defined starting with A( A x J5) = /x( A) /x(B) for measurable sets A and B rather than intervals.

Proposition 5. For any set E c M ,

k(E) = inf {E A(A X B/) : £ c U A x B,}

= inf {E/x(A)/x(B,) : £ C U A X B-}

where {Ai}and {Bi}are countable families of measurable subsets ofR.

Proof. Let X^(E) denote the right side above; i.e., A^(£) would be the outer measure of £ starting with measurable sets A X B rather than rectangles I x / . Clearly A^(£) < A,(£) since the inf is over a larger collection of coverings. On the other hand, £ C U A x Bi and subadditivity and Proposition 3 imply that

A(£) < 5]A(A X B,) = ^ / i (A) /x(B, ) .

Hence A(£) = A^(£). ill

Now we have the following: (i) If A and B are measurable subsets of R, then A x B is a

measurable subset of R^ and k(A x B) = /x(A) /x(B). (ii) If £ c R^ then

M£) = inf{^/x(A)/x(B,) : £ C U ( A x B,)}

where the {A} and {BJ are countable families of measurable subsets of R.

We finish this chapter by showing how plane measure is related to linear measure through integration. In calculus one defines the area of the plane region

S = {(x, y) : a < X < b, 0 < y < f(x)}

to be the integral /^ fd/uL. We now have a second definition of this area as k(S). It is necessary, and not hard, to show that these definitions agree.


Problem 2. If f is measurable and non-negative on [a,b]^ and 5 is the region under the graph of f as given above, then S is measurable and X(S) = Xf fdfi. <

We v^ant to establish the connection between area and integration for more general plane regions, and thus pave the way for multiple integrals and the Fubini Theorem.

Let £ be a measurable subset of the plane, and to start we will assume that £ c [0,1] x [0, 1]. Let Ex be the cross section of £ over the point x e [0, 1]; i.e.,

Ex = {y: (x, y) e £}.

Let f(x) = iJi(Ex)^ so f(x) is the "length" of the x cross section of £ . We proceed to show that f is measurable and, as one would expect.

X(E) = / ' fdfi. Jo

Problem 3 is an essential lemma for Proposition 6.

Problem 3. Let £ be a compact subset of [0,1] x [0,1], and let f; be the x cross section of £ for each x e [0,1]. For x e [0, 1], let L„(x) be the total length Y. ^(^i) of all dyadic intervals of length 2" ^ which intersect F^. Let a„(£) be the total area Y.a(di X ej) of all dyadic squares of side 2"" which intersect £ .

(i) £ is A-measurable, and each F^ is /x-measurable. (ii) Qf„(£) decreases to X(£) as n —> oo. (iii) Ln(x) decreases to /x(£x) as n —> oo. (iv) What goes wrong if £ is not compact? ""HI

Proposition 6. IfF is a compact subset of [0,1] x [0,1] and F^ is the X cross section, and f(x) = fiiFx), then f is measurable and

A(£) = / ' f(x)dfM(x). Jo

Proof. Let {4 x ej} be all the dyadic squares of side 2~^ which intersect £ . For x in the interior of 4 , let (Pn{x) be the total length Y.l{ej) of all ej such that di x ej intersects £. That is, cpni^) is

10 THE RELATIONSHIP BETWEEN /X AND X 99

the total length of the column of dyadic squares in the covering which lie over 4 . If x is an endpoint of 4 there may be two columns over x, so define (Pn(x) only for interior points of the 4 , and let (pn(x) = 0 elsewhere. Thus cpn is a simple function and

Moreover, (Pn(x) —> iJi(Fx) for all x which are not dyadic points; thus (pn(x) —> IJi(Fx) a.e. It follows that f(x) = iJi(Fx) is measurable, and

lim / (pndii= / iJi(Fx)diJi(x) = / fd/n. « Jo Jo Jo

If n is so large that

^a(di X ej) < X(F) + s

for the dyadic covering sets 4 x Cj of side 2~~^^ then

X(F)<J2o^(diXei)

Jo

<X(F) + £.

Hence

X(F) = lim / (pndfji = f fi(Fx)dfM(x), ill ^ Jo Jo

Recall that A is a regular measure, so for any measurable subset £ of finite measure there is an open set U and a compact set F so that F c E cU and

A(£) -£ < X(F) < X(E) < X(U) < X(E) + e.

If £ C (0,1) X (0,1) then of course U and F can also be taken as subsets of (0,1) x (0,1).

Proposition 7. If E is a measurable set, then iJi{Ex) is measurable and J^ lii(Ex) dfiix) = k(E).


Proof. First assume that £ C (0,1) x (0,1). Let P„ C £ C U , with each £„ compact and each U„ open, and X(Un — Fn) < ^• Assume that £i C £2 C • • • and Ui D U2 3 • • •. Let iC„ = [0,1] X [0,1] - Un, so Kn is compact, and iJi(Fnx), l^i^nx) are measurable functions, with

X{Fn) = I ii{Fnx)dii{x) —> A(£) Jo

X{Un) = 1 - HKn)

= 1 - / fl{Knx)dfl(x) Jo

= 1 - [\l-tz{Unx))dfl(x)

Jo

= I iJiiUnx) dfi(x) —> A(£). JO If fn(x) = iJi(Fnx), gnix) = fJi(Unx), then { Q and {g„} are mono

tone sequences of measurable functions with

fnix) < /X(£;c) < gn(x)

for all x, and

lim / frid/ji = lim / gndfi = k(E). ^ Jo " J o

It follows that

lim fn(x) = limg„(x) = fi(E^) a.e., n n

and so />6(£jc) is a measurable function and

/ iui(E^)dfi = lim / fnd/ii Jo ^ Jo

= X(E), ill

Problem 4. Extend the above proof of Proposition 7 so that it applies to general measurable sets. ""HI

Notice that the preceding results are proved first for subsets of the unit square. The final result is obtained by adding up

10 THE RELATIONSHIP BETWEEN / i AND A 101

what happens on a finite or countable number of squares. This is possible because both the line and the plane can be written as countable unions of finite measure sets. A general measure v with this property is called a-finite. Thus v is a a-finite measure on X if X = U Ei with v(£/) < oo for each /. The hypothesis of or-finiteness is essential in the general discussion of product measures.

Integrals with respect to X are defined just as they are for /x, and the same integration theorems hold. Integration for general measures will be treated in Chapter 12, but for now the reader should take on faith that A-integrals have the standard properties.

Proposition 7 provides the basis for the proof of the Fubini Theorem, which states that A-integrals ("double integrals") can be evaluated as iterated /x-integrals. In practice, there is no other way to evaluate most double integrals. One of the most useful aspects of this connection between A-integrals and /x-integrals is the fact that generally speaking the order of integration in iterated integrals can be reversed. The "Fubini Theorem" is the name commonly associated with the results of the next proposition, but the reader is cautioned that the name Tonelli is also given to this form.

Proposition 8. (FubinVs Theorem) Let f(x, y) be a non-negative measurable function, and define fxby fx(y) = f(x, y). Then f^ is a measurable function on R for almost all x, and the function F (x) defined by

F(x)= I fx{y)dfi(y) = I fix, y)dfi(y) JR JR

is measurable and non-negative, and

( f(^^ y)df^(y)) dfi(x) = fix, y)dXix, y). (1) JR \JR J JR^

If f is integrable, fx is integrable for almost all x and F is in-tegrable; conversely, if the iterated integral is finite, then f is X-integrable. Of course the same statements apply to the iteration in the other order, so the order of integration for iterated integrals (of a positive function) is immaterial.


Proof. First we give an outline of the proof. For characteristic functions the resuh follows directly from

Proposition 7. By linearity of the integrals—all three of them— the result then holds for simple functions. A non-negative measurable function f is the limit of an increasing sequence cpn of simple functions. This fact depends on the a-finiteness of R^. (If R^ were not a-finite we could not guarantee the existence of such functions cpn which are zero off a set of finite measure; cf. Problem 8 below.) The theorem extends from simple functions to f by the Monotone Convergence Theorem.

Now we fill in the details. Let f{x, y) = axni^^ y) where E is a measurable set and X{E) < oc. Then

fx(y) = ^X£.(y).

jfAy)diJi{y) = aiJi{E,). (2)

By Proposition 7,

j aii{Ex)diJi{x) = aX{E). (3)

Integrating both sides of (2), and using (3),

i.e., for f(x, y) = axE{x, 3/), we have the result:

J U f^^^ y)dfji(y)] dfi(x) = aX(E) = J f(x, y)dX(x, y).

If (p(x, y) is a simple function, then massive applications of linearity give the result. We illustrate the argument for the sum of two very simple functions. Let Ei and E2 be disjoint measurable sets of finite measure, and let f = fi+fi where /i = ^IXEI?

10 THE REIATIONSHIP BETWEEN / i AND A. 103

fi = aiXEi' Then

jf{x, y)dk(x, y) = J(fi+ f2)dX

= JfidX + JfzdX

= J U fi(x, y)dfi(y)] dii{x)

+ J Uf2(^^ y)^M(y)j dfi(x)

= J U fi(x, y)dix(y)

+ J fiix, y)dfM(y)] dix{x)

= J{J(M^^ y) + /2( ' y))dfj^(y)) dii{x).

Now let f be measurable and non-negative on E^. There is an increasing sequence {(p„(x, y)} of simple functions so that (pn —> f a.e. By the Monotone Convergence Theorem

(p„dX —> fdX.

That is,

J(^J<p„(x, y)d/x(y)^ dfx(x) -^ Jfdk. (4)

The right side of (4) could be +00. For each fixed x, (p„(x, y) is an increasing sequence of simple

functions, and

(<Pn)x(y) = (pn(x, y) —> fxiy) = fix, y).

Therefore,

^n{x) = / (pnix, y)dfi(y) —> I fix, y)d/xiy) = Fix).


For each w,

c&„(x) = j (pn(x, y)dfi{y)

N

N

— / ^^nit^y^niJx'

i=l

By Proposition 7, it follows that each 0„ is a measurable function of x, since iJi{Ex) is measurable if E is. Since <E>„(x) increases to P(x), F is measurable. (F might take the value +00.) Hence the integral

j f(x,y)dfi(y)

is a measurable function of x. Finally, by the Monotone Convergence Theorem,

/o„(x)(i/x(x) = U (Pn{x, y)dfi(y)] dfiix)

iF(x)diji(x)

X, y)^M(y)) J/x(x).

/ '

= / ( / '

By (4) we have the desired result for non-negative measurable functions f. The A-integral is finite if and only if the iterated integral is finite.

If /" is a not necessarily non-negative function, but is inte-grable, then the result holds for f^ and f" and hence for f. ill

The following problem illustrates the way the Fubini Theorem is usually used to justify changing the order of integration in an iterated integral.

Problem 5. Let f(x, y) be measurable. If

jj I fix, y)|J/x(x)J/x(y) < oo

10 THE RELATIONSHIP BETWEEN /X AND X 105

then f{x, y) is A-integrable and the order of integration can be reversed. ""HI

Problem 6. Let Xbe an uncountable set, and S the a-algebra of all subsets of X. Let /x be counting measure on X, so /x(£) is the number of elements in £ if £ is finite, and /x(£) = oc otherwise. Show that the function f which is identically one on X is measurable, but there is no sequence {cpn} of simple functions which increases to f a.e. ""HI

11 GENERAL MEASURES

The basic additivity property of Lebesgue linear or planar measure holds when the outer measure is restricted to the cr-algebra of measurable sets. We axiomatize the general theory of measure by starting with a or-algebra S of subsets of some set X, and a countably additive non-negative extended real valued function v defined on the sets in S. This triple (X, S, v) is called a measure space. In practice, one says "v is a measure on X." The a-algebra S generally disappears from the discussion, and we write "E is measurable" to mean v(E) makes sense—i.e., that E e S.

Measures are automatically monotone in the sense that if A and B are measurable and Ac B^ then v(A)<v(B). This is clear because B — Ais measurable and

v(B) = v(A) + v(B -A)> v(A).

Countable subadditivity is also an automatic property of measures, as we show next.

Proposition 1. If {Ei} is a finite or countable family of measurable sets, then

Proof Let P„ = £„ — (£i U • • • U E^-i), so the F^ art measurable and disjoint, and {jF^ = \J En. Clearly v(Fn) < v(En) for all n, so

V (U E„) = V (U f«) = E ^(Pn) < E ^(E„). ill

107


Problem 1. Let X be any set and for £ C X, let v (£) be the number of elements in £ if £ is a finite set, and v(£) = oc otherwise. Show that V is a measure on the a-algebra of all subsets of X. This measure is called counting measure. ""HI

Problem 2. Let Xbe an uncountable set. Let S be all subsets of X which are either countable or have a countable complement. Show that iS is a a-algebra. Let v(£) = 0 if £ is countable and v(£) = 1 if £ is uncountable. Show that v is a measure on X. Hint: "Countable" means "countably infinite or finite." ""HI

Problem 3. Let / be a non-negative function on a set X. Let ^(£) = YlxeE f(^)'> where the sum is the unordered sum in the sense of Problem 9, Chapter 2. Show that v is a measure on all subsets of X. If Xis uncountable and f is strictly positive, show that y(X) = oo. [If f is the characteristic function of a single point X, then v is called point-mass at x, and f gdv = g(x) for allg.] ""Ill

Proposition 2. Let {£„} be a sequence of measurable sets.

(i) If El ::> El D " ', and v(£i) < oc, then

v(p|£„) =limv(£„).

(ii) IfEiCEiC", then

v ( | j £ „ ) =l imv(£„) .

Proof (i) Let £i D £2 D • • • and let £ = fl En. Then

£1 = £ U (£1 - £2) U (£2 - £3) U . . . ,

and the summands are disjoint. Hence

y(£i) = (v(£i) - v(£2)) + (y(£2) - v(£3)) + • • •

+ (y(£„_i)-y(£„)) + . . . + v(£)

= v ( £ i ) - l i m v ( £ „ ) + v(£).

Since v(£i) < oc, subtraction is legitimate and gives the result. (ii) Let £1 C £2 C • • • and let £ = U £«• Then

£ = £1 U (£2 - £1) U (£3 - £2) U . . . ,

11 GENERAL MEASURES 109

and hence

v(£) = v(£i) + (v(£2) - v(£i)) + (y(£3) - v(£2)) + • • • = limv(£„). ill

The two measures /x on E and A on M x R already studied have the property that every set of (outer) measure zero is a measurable set, and consequently every subset of a set of measure zero is measurable. Measures with this property are called complete. In a general measure space (X, <S, v) it is possible to have subsets A and B of Xwith AcBeS, and v(B) = 0, but A ^ <S, so v(A) makes no sense. However, for any measure space (X, S, v) we can extend 5 to a larger or-algebra iSo by throwing in all subsets of zero-measure sets. The measure v can then be extended to a complete measure VQ on SQ by defining vo(A) = v(B) ii B e S and A differs from B by a subset of a zero measure set; i.e., if A A B c Cwithv(C) = 0.

Problem 4. Define a non-complete, non-trivial (i.e., not identically zero) measure on a a-algebra of subsets of the three-element set X= {a,b,c}. ""HI

Proposition 3. Any measure can be extended to a complete measure.

Proof. Let (X, «S, v) be a measure space. For brevity let us say that a subset A of X is a null set provided A is a subset of some B e S with measure zero. Let So consist of all sets of the form (£ U A) — J3 where E e S and A and B are null sets. Clearly S C So since 0 is a null set. We can assume that the sets in So have the form (£ U A) — B where A and B are null sets such that £ D A = 0 and B c £. In this case,

( £U A ) - B = ( £ - B ) U A

[ ( £ U A ) - B y = ( F u B ) - A ,

so So is closed under complementation. Let F^ = (£wU A„) — B^E 5o, with En e <S, and A„, B„ null sets. Then

[JFn = \J[(En U An) " B,] = U(£, U A,) - C,


where C c U S^, and C is consequently a null set. Therefore

and U A„, C are null sets. Hence So is a a-algebra. We define

v o [ ( £ u A ) - B ] = v(£).

To show that VQ is well defined, let

(£i U Ai) - Bi = (£2 U A2) - B2

with £1, £2 G 5 and Ai, A25 Bi, B2 null sets. Let Ai be a zero measure set in S which contains Ai. Then

£ i U Ai C £2U A2UB1,

£1 UATi C £2 U A2 U J5i UATi.

Since A2 U Bi UATi is a null set, there is C e S with v(C) = 0 and C D A2 U Bi U A"i. Hence £1 U Xi C £2 U C, and both of these sets are measurable. Hence

y(£i) = v(£i U ATi) < v(£2 U C) = v(£2).

The argument is symmetric, so y(£i) = v(£2) and VQ is unambiguously defined. It is left as an exercise to show that VQ is countably additive on SQ. •

Problem S. (i) Show that every set in SQ can be written £ U C with F e S and C a null set.

(ii) Show that VQ is complete and countably additive on So. ""Ill

Problem 6. (i) What are So and VQ for the completion of your example in Problem 4?

(ii) Are the measures of Problems 1 and 2 complete? ""HI

Since any measure can be extended to a complete measure, v is henceforth assumed to be a complete measure. We will say that V is finite if v(X) < oc, and v is or-finite if X is a countable union of finite-measure sets.

If vi and V2 are two measures on the same a-algebra of subsets of X, then it is clear that vi + vi is also a measure on X.


More generally, any finite or countable sum JâjVi with all Ui > 0 is again a measure on X if the V/ are measures on the same or-algebra. If we consider a difference vi — V2 of two measures we again get a countably additive set function provided both measures are finite. However, if there is a set E such that vi(£) = V2(£) = oo, then (vi —V2)(£) makes no sense. More complicated atrocities can happen. Since vi — Vi could take both positive and negative values, it would be possible for J2î(î — ^2)(£/) to converge conditionally for some disjoint sequence {£J. Conditional convergence means that the sum depends on the order of the summands. Since U £/ obviously does not depend on the arrangement of the £/, the countable addi-tivity condition

would be an impossibility in this case.

Problem 7. Show that if vi and vi are finite measures on the same or-algebra of subsets of X, then vi — V2 is countably additive on this a-algebra. ""HI

Now let us consider a countably additive set function v, defined on some (j-algebra, with v taking both positive and negative values. We assume that v(0) = 0, and allow the possibility that some sets have measure +oc, but allow no sets to have measure —oo. The other choice of allowing — oc but not +00 would work equally well. We call such a function v a signed measure.

The countable additivity assumption requires that if {£/} is a disjoint family, then vQJEi) = Z;v(£/) and the sum of the negative terms v(£/) is finite. Otherwise the union of these £/ would be a set of measure — oc. Thus for any disjoint family {£/} the sum Y^v(Ei) converges absolutely, or diverges to +00 as an unordered sum.

From Problem 7 we know that some signed measures arise as the difference of two positive measures, and we show next that in fact every signed measure is of this form.

A measurable set A is called a positive set provided v(£) > 0 for every measurable subset £ of A, including A itself.


A measurable set B is called a negative set provided v(£) < 0 for all measurable £ c B. If A is both a positive set and a negative set, so that v(£) = 0 for all £ C A, then A is called a null set. Notice that "null set" in this context is different from "null set" in the earlier completion argument.

Problem 8. If v is a signed measure on X and A is a positive set for V, and vi(£) = v(£ n A), then vi is a positive measure on X. If B is a negative set for v and V2(£) = - v ( £ Pi B), then V2 is a positive measure on X. ""HI

We proceed to show that every signed measure v can be written as the difference of positive measures as indicated in the preceding problem, where J5 is a "largest" negative set for v and A = X- B. The problem is to show that there is a "largest" negative set B, so that X — B is necessarily a positive set. We find B first rather than A because of the possible complications stemming from the fact that v(A) might be +oc.

Proposition 4. Let v be a signed measure on X. Ifv(E) < 0, then E contains a non-null negative set.

Proof. If £ contains no subsets of positive measure, then £ is a non-null negative set, and we are done. Otherwise let

pi = sup{v(£) : f C £},

so 0 < /?! < 00. Let ^ be the largest of the numbers 1, i , | , • • •

such that ;^ < pi5 and pick £i C £ with

1 — < v(£i) < pi. ni

If ni > 1, then 1/wi < pi < l/{ni — 1). If £ — f i has no subsets of positive measure, then £ — £i is a negative set, and

v(£ - Fi) = v(£) - y(£i) < 0,

so £ — £i is non-null. If £ — £i is not a negative set, we let

p2 = sup{v(£) : £ c £ - £i}


SO 0 < p2 ^ oc. Again let — be the largest of the numbers 1, i ,

| , • • • such that ^ < p2? and pick Fi C E — Fi^ with

1 — < v(F2) < p2. Hi

Continuing this way we either find a non-null negative set of the form £ — (Pi U • • • U F^)^ or we find a disjoint sequence {Fk} such that for all k

1 — < v{Fk) < ph fik

and iirik > ly then

1 — < v(Fk) <pk< 1

Hk-l

The Fk are disjoint, so

v{E-lJFk) + J2v(Fk) = v(E),

Since v(E — \J Fk)y^-'Oo and v(E) < 0, the positive series Y^ v (Fk) must converge to a finite number. Therefore Yl ^ converges, so rik —> oc, and hence pk —> 0. We claim that E— \JFk is a negative set. Otherwise, there is G C E— \JFk with y(G) > 0. For some N, p]v < v(G), and this contradicts the definition of pM- The relation

v(E-[jFk) + ^v(Fk) = v(E)<0

shows that v(E — [jFk)< 0, and hence E — [jFkis non-null, ill

Problem 9. Does the proof of Proposition 4 with the appropriate changes work to show that if v(£) > 0, then E contains a non-null positive set. '"Nil

Proposition S. If v is a signed measure on X, then there is a positive set A and a negative set B so that A and B are disjoint and AUB = X.

Proof. If there are no measurable sets of negative measure, then we take B = 0, A = X, and we are done. Otherwise, there


are sets of negative measure, and hence negative sets of negative measure. Let

q = inf{v(£) : £ a negative set}.

Let {B„} be a sequence of negative sets such that v(B„) —> q <0. The union of negative sets is a negative set, so we can assume that Bi C Bi C B3 C • • •. Let B = U B«5 so that B is a negative set and

v(B) = hmv(B„) = q.

By our assumption that v does not take the value —oo, we have v(B) > —cx), so there is no set of negative measure disjoint from B, and X —B is a positive set. ill

A decomposition of X into disjoint sets A and B, with A a positive set and B a negative set, is called a Hahn decomposition of X. Such a decomposition is not unique since any null set can be thrown into either A or B. However, apart from null sets, "the" Hahn decomposition is unique.

Problem 10. Let v be a signed measure on X and let Ai and Ai be positive sets for X and Bi, B2 negative sets for X, with X = Ai U Bi = A2 U B2 and Ai n Bi = A2 n B2 = 0. Show that Ai A A2 is a null set, which proves that the Hahn decomposition is unique except for null sets. ""HI

If A and B form a Hahn decomposition of X for the signed measure v, then we know from Problem 8 that v can be written as the difference of two measures, v = v^ — v~, where

v+(£) = v ( £ n A), v-(£) = v ( £ n B ) .

The two measures v" and v~ are supported on the disjoint sets A and B in the sense that v+(B) = 0 and v"(A) = 0. We will say that two (positive) measures vi and vi on X are singular (or mutually singular, or singular with respect to each other), provided there are disjoint sets A and B with A U B = X and vi(B) = V2(A) = 0. The canonical representation V = v+ —v~ of a signed measure v as the difference of two singular measures is


called the Jordan decomposition of v. The Hahn decomposition of X into a positive set A and a negative set B is only unique up to null sets. However, the null sets do not affect the Jordan decomposition v = y+ — y~ofv into singular measures, because null sets do not affect the definition of v" and v". Therefore the Jordan decomposition v = v" — v~ is unique.

If y is a signed measure, then we define the absolute value of V or total variation of v by

|v|(£) = v+(£) + v-(£) .

Observe that |v| is again a measure on X, and |v(£)| < |v|(£) for all £ .

12 INTEGRATION FOR GENERAL MEASURES

In this chapter we consider a general measure space (X, S, v), and define the integral of an extended real-valued function with respect to v. Measures are non-negative and all measures are complete. Measures are at least cr-finite, which means that X can be written as the countable union of finite-measure sets, but this does not rule out the possibility that v(X) < oo. Earlier we defined the integral in two steps—first bounded functions on finite measure sets, then general functions and sets. Here we do it in one swift swoop—the functions are not necessarily bounded and the sets may have infinite measure.

A function f is measurable provided {x : a < f{x) < b} is measurable (i.e., a set in <S, and therefore a set in which v is defined) for all a and b. Since 5 is a cr-algebra, all the usual equivalent conditions for measurability still hold. (See Proposition 4, Chapter 5.) For functions whose graphs lie in a finite area, U £/ x [ - M/, M/] with Yl MiV(Ei) < oc, we again show that measurability is necessary and sufficient for integrability.

The reader is encouraged to notice that the development here is basically the same as for integrals on the line (Chapter 7), and to consider specifically what each statement means for integrals with respect to plane measure X. Most of our earlier theorems, and their proofs, will carry over verbatim to the present setting.

Let 5 be a measurable set. A partition P = {£/} of 5 is a finite or countable collection of disjoint measurable sets, all of finite measure, whose union is S. Measurable sets, including X, have partitions because Xis or-finite. Another partition Q= {Fj}

117


is a refinement of P, denoted P < Qor Q> P, provided each set Fj of Q is a subset of some set Ei in P. The partitions of S form a directed set under the partial ordering <.

Problem 1. Show that all measurable sets have partitions. ""HI

For the partition P = {£/} of S, v e again let m/ and M/ be the inf and sup of the function values on E/, and let

Mi = sup{| f(x)\ : X e Ei).

If J2 Miv(Ei) < cx), then P is an admissible partition for f on S, and f is an admissible function on S.

If P is an admissible partition, and only then, we write

L{f,P) = Em,v(E,)

U{f,P) = EMME,),

For admissible partitions, the lower and upper sums (1) are either finite sums or absolutely convergent series. Since f can be unbounded, and even occasionally take the values dzoc, some mi and Mi may now be infinite. If nti or Mi is infinite, then necessarily v(Ei) = 0, so MiV(Ei) = 0. When convenient we can lump all zero measure sets in a given partition together in a single set £o, and effectively ignore it.

The lower sums L(f, P) and the upper sums U(f, P) are nets on the admissible partitions P. Lower sums increase and upper sums decrease when a partition is refined, and all lower sums are less than or equal to all upper sums. The lower sums {L( /*, P)} form an increasing net which is bounded above by any upper sum, and the upper sums {U(f, P)} form a decreasing net which is bounded below. Both nets {L{f, P)} and {U(f, P)} converge, and

l imL(/, P) = sup Lif, P) < inf U(f, P) = limU(f, P). 1 p F 1

If the limits are equal, f is integrable over S, and we write /^ f for the common limit.

Problem 2. If f and g are integrable over S and f < g a.e. on S,thcnJsf<Jsg. I

12 INTEGRATION FOR GENERAL MEASURES 119

Problem 3. If g is constant, kj on the finite-measure set T, and g = 0 on S — T^ then Jsg = kv(T). Hint: The sets T and S -T may not form a partition of S since sets of a partition have finite measure. ""HI

Proposition 1. If f is admissible and measurable on the measurable set S, then f is integrable over S.

Proof. Let P = { £ J be an admissible partition of 5, so J2MiV(Ei) < oo.Lets > 0 and choose Nso that E. '^N+I î^(Ei) < 6. The set T = £/ U • • • U E N has finite measure, and \f\ is bounded on T by X = max {Mi , . . . , Mjv}. Let {J i , . . . , J^} be a covering of the interval [—K.Kjhy disjoint intervals each of length less than 8/v(T). Let Fj = {x e T : f(x) e I/}, so that Q = {Py} is a partition of T, and

U(f, Q) - L(f, Q) <J2^^v(Fi) = s.

The sets of Q, together with EN+I? EN+2, * • • form a partition Po of S and

00

U(fPo)-L(fPo)<U(fQ)-L(fQ)+ Yl (M,-m,)v(E,) i=N-\-l

< 8 +26 = 36. ill

Next we show that measurability is also necessary for integra-bility. It is convenient to modify the definition of simple function to include now all functions which can be written cp = YjîXEi'> where {EJ is a finite or countable family of disjoint finite measure sets, and J2Wi\^(Ei) < oo. li P = {E/} is an admissible partition for f on S and m/, Mj are as usual, and

(pp(x) = EîXE,(x)

i^p(x) = EMiXE,(x),

then (PP and Vp are simple functions with cpp < f < ^p-

Problem 4. Show that if (pp and x/fp are as in (2), then (pp and i/rp are integrable and their respective integrals are L(f P) and U(fP). <


Problem 4 shows that approximating by upper and lower sums is equivalent to approximating by the integrals of upper and lower simple functions. Thus a function is integrable over S if and only if for each e > 0 there are simple functions cp and ifr on S with (p < f < ilr and J x// — J (p < s.

Problem 5. Show that a simple function is measurable. '""H

Problem 6. Show that the a.e. pointwise limit of a sequence of measurable functions is measurable. Hint: The proof hasn't changed—see if you remember it. Start with sup ^ . ""HI

Problem 7. Let {g„} be a decreasing sequence of measurable functions which are non-negative and integrable on S. If Is Sn —> 0, then g„ —> 0 a.e. Show that "a.e." is necessary. ""HI

Proposition 2. If f is integrable over S, then f is measurable.

Proof. Let Fi ^ P2 -< P3 ^ • • • be a sequence of admissible partitions of S, with U(/", P„) - L(/*, Pn) < \> Let cpn^ i/n be simple functions such that J (pn = L( f P„) and J '{J/^ = U( f Pn). Then [cpn] is an increasing sequence of measurable functions since Pi -< P2 ^ • • •, and (pn S f for all n. Similarly, {yj/n} is a decreasing sequence with yj/n > f. The functions {V„ — (pn) are non-negative and decreasing, and measurable by Problem 5, and

jifn - H>n) = (/(/•, Pn) " L(/*, P J - ^ 0.

ByProblem7, V „—<p„—> Oa.e.5so<^„—> / a.e. (and V^„—> f a.e.), and f is measurable. Hi

The integral is of course a linear functional, and we want to prove that next. Linearity is not particularly obvious from our definition, so we again show that the integral is the limit of Riemann sums. Linearity is then obvious since R{f + g, P, c) = R(f P,c) + R(g, P, c), and the limit of a sum is the sum of the limits.

If f is an admissible function on S, and P = {£/} is an admissible partition for S, we define R(f P,c) = Yl f(Ci)v(Ei)^ where Q e £/ for each /. We will write R{f P, c) only for admissible partitions, so the notation implies the admissibility of P.


The pairs (P, c) form a directed set as usual with (P,c) < (P\c^) meaning the same as P -< P^ For all P and c

L(f,P)<R(f,P,c)<U(f,P),

It follows almost immediately that the net {P( f, P, c)} converges to the integral if f is integrable.

Problem 8. Show that if f is integrable over S then {R( /", P, c)} converges to /^ f. Hint: Take note of the fact that {jR( /*, P, c)} is a net on a different directed set than the nets {L(/*, P)} and {U(f,P)}. I

Problem 9. If f and g are integrable over 5, then f + g and kf are integrable over 5, and

Hint: An admissible partition for f may not be an admissible partition for g. What is the domain of the net {R(f + g,P,c)}} •"III

Proposition 3. If f is defined on S, then lim R(f P,c) = I

if and only if f is integrable (hence measurable) over S and

Isf=I-Proof The "if" part is Problem 8 above, so we assume that

R(f P, c) —> I. The notation R{f P, c) presumes that there are admissible partitions for f. Let ^ > 0 and let PQ = {£/} be a partition such that \R(fP,c) — I\<6 whenever P > PQ, so in particular any two Riemann sums R( f PQ, C) and R{ f, PQ, C') for Po are within Is of each other. We can choose Ci G £/ with fid) > Mi - 8/Tv{Ei), so R{f, P, c) is within s of U(/*, PQ). Similarly, if f{c[) is within e/Tv{Ei) of m/, then R{f, PQ, d) is within 8 of L(/*, PQ). It follows that U{f, PQ) - Uf PQ) < 4^, and f is integrable. By Problem 8, lim R(f P,c) = Jf. The details are the same as in Proposition 2 of Chapter 7. ill

The limit theorems for the Lebesgue integral depend basically on these facts:


I. If fn —> f uniformly on a finite-measure set T, then /T fn —> IT f- This is the convergence theorem for Riemann integrals, and of course it still holds here.

11. If fn —> f pointwise on a finite measure set 5, then fn —> f uniformly off sets of arbitrarily small measure.

III. If g > 0 is integrable over S^ then there are finite-measure subsets T of S such that g is bounded on T and Jjg is arbitrarily close to J^g. It follows that ii \fn\ < g and fn —> f, then

and all these T integrals are close to integrals over S.

Proposition 4. (Egoroffs Theorem) If {fn} is a sequence of measurable functions and fn —> f a.e. on a finite-measure set S, then given 5 > 0 there is an exceptional set E C S with v(£) < 8 such that fn —> f uniformly off E.

Proof We can assume (Problem 10) that fn(x) —> f(x) for all X G S. For a given s let

A]s[ = {x e S : \ fn{x) — f{x)\ > 6 for some n> N}.

Clearly Ai D Ai D A3 D • • •. Since fn(x) —> f(x) for all x, for each x there is some N with x ^ AM; i.e., fl Ajv = 0. Since v(Ai) < oo, and the A^ are nested, lim V(AN) = 0, and for any given 8 there is Nwith V(AN) < 8. Thus we have for each pair £ > 0, 5 > 0, an integer N and a set A^ of measure less than 8 such that I fn(x) - f(x)\ < s for all x ^ A^ and n > N.

Now let {sk} be a sequence of ^'s with Sk —> 0. Let 8 > 0. For each pair Sk-, 8/2^ we find as above an integer Mk and a set Tk of measure less than 8/2^ such that | fn(x) — f(x)\ < Sk if n > Mk and x ^ Tk. We can assume the Mk increase. For any given £ > 0 there is A^ (corresponding to some Sk < s) such that I fn{x) - f{x)\ < £k < £iin> Mk and x ^ Tk. Let T =\JTk^ so v(T) < 8 and if x ^ T, then x ^ Tk. Hence ii x ^ T and k > Mkj I fn{x) — f{x)\ < £. That is, fn —> f uniformly off T and v{T) <8. ill


Problem 10. Show that Egoroff's Theorem holds if fn —> f a.e. on S (instead of for all x G S as in the proof). ""HI

Problem 11. Prove the Bounded Convergence Theorem. If {^} is a uniformly bounded sequence of measurable functions on the finite measure set 5, and fn —> f a.e. on 5, then

The next proposition takes care of the cases where the function is unbounded or the set has infinite measure.

Proposition 5. If g is integrable on S and s > 0, then there is a finite measure set E such that j ^ _ ^ \g\ < ^ ^^d g is bounded on £.

Proof. Let P = {£/} be an admissible partition of S for g, so EM,v(£. ) < oo. \iT.T=N+i^i<^t) < ^. ^ n ^ £ = U ^ i £ / , then /5_£ \g\ < £ and \g\ < max {Mi , . . . , M^} on £ , and v(£) < 00. ill

Problem 12. Prove the Lebesgue Dominated Convergence Theorem: If {f } is a sequence of measurable functions on S, and \fn\ < g on S, and g is integrable on S, and fn —> f a.e. on 5, then /^ fn —> Js f* Hint: Do not forget to show that f is integrable. ""HI

The Dominated Convergence Theorem says that integrals converge nicely provided all the functions stay in a fixed finite area (between —g and g). If they don't—i.e., if the fn are allowed to wander out into pastures of infinite area—then Fatou's Lemma says what can go wrong. If 0 < fn —> f-, then the integrals Jfn can of course approach Jf^ but if they don't it's because they have included too much area outside f.

Proposition 6. (Fatou's Lemma) If{ fn} is a sequence of non-negative integrable functions, and fn —> f a.e. on S, then lim'mf Jfn > Jf if f is integrable, and Xixnjfn = oo if f is not integrable.

Proof. We suppose f is integrable over S with /^ /" > 0, and leave the other case as a problem. Let P = {£/} be a partition


such that L(f, P) = Y.mv(Ei) is within s of Jf. Pick Nlarge enough so E,^i miV(Ei) is within 2s of Jf. Let h = Y^fLi ^IXE, be the corresponding simple function. The function h is bounded and non-zero on the finite measure set T = U/li £M and h < f. Since /*„ —> f^fn^h —> h^ and {/"„ A /i} is a uniformly bounded sequence on a finite-measure set. Therefore

and lim i n f / / " „ > / / " . ill

The "monotone" convergence theorem is immediate from Fatou's Lemma: if 0 < /"„ < f and fn —> f, then jfn —> jf.

Problem 13. Give two examples where f is integrable and liminf/^ f^ > J^ f: one example where S has finite measure so the fn can't be uniformly bounded, and one example where the fn are uniformly bounded, so v(S) must be infinite. ""HI

Problem 14. {Second Part of Fatou's Lemma) Let {fn} be a sequence of non-negative integrable functions on S such that fn —> f on S but f is not integrable. Show that lim /^ fn = oc. Hint: Use the partition £„ = (x : 2 ^ < f(x) < 2^^+^ for n = 0, ± 1 , ± 2 , . . . to find a simple function hjq < f such that h^ is bounded and non-zero on a finite measure set and fhN>N. II

Problem 15. Define f^{x) = max{/'(x),0} and f~(x) = min{ fix), 0} so f^(x) > 0 and f-(x) < 0.

(i) Show that f~^ and f~ are measurable if and only if f is measurable.

(ii) If f is measurable, then f is integrable if and only if /"+ and f~ are integrable, and then Jf = Jf~^ + Jf~. ""HI

It is conventional to write J^ f = oo ii f is non-negative and measurable on S, but not integrable (i.e., not admissible). With this convention Fatou's Lemma can be stated thus: // {fn} is a


sequence of non-negative measurable functions and fn —> f a.e., then

lim inf jf,>jf. This is now correct even if f is not integrable.

Problem 16. Show that if 0 < fn —> f and the fn are measurable, then

limini f fn> flim'mifn, <

13 MORE INTEGRATION:

THE RADON-NIKODYM THEOREM

We continue to consider a complete, non-negative measure v on a (T-algebra S of subsets of X. X will always be at least a-finite, and sometimes finite. Since X is or-finite it can be written as a countable disjoint union of finite-measure sets, so Xand its measurable subsets have partitions.

We have seen that if f is integrable over Xand T C X, then f is integrable over T and Jj f = J^ f . XT- Hence if f is integrable over T we can extend f to X, with f = 0 off T, and write

iTf-Ixf-We will show next that if f is integrable over X, and ^(E) =

/E fdv^ then ^ is a new (signed) measure on X.Ii f > 0 then ^ is a positive a-finite measure on X. We will then characterize those measures ^ which can be represented as integrals.

If A and B are disjoint measurable sets and f is integrable over AUB, then XAUB = XA + XB and hence

/ f= ff(XA + XB)= [f+ If' JAUB J J A JB

Thus if ^ (£) = /^ fdv^ ^ is finitely additive:

for disjoint sets Ei,,.., En.

Problem 1. (i) If f is non-negative and integrable over X, and )S(£) = /^ fdv for every measurable set £ , then )S(£) = 0 whenever v(£) = 0.

127


(ii) Is P necessarily complete? Hint: Subsets of v-measure zero sets are measurable, but are subsets of j6-measure-zero sets measurable? ""HI

Proposition 1. If f is a non-negative integrable function on X, and ^(E) = J^ f for each measurable subset E of X, then ^ is a finite measure on X.

Proof From Problem 1 we know that ^(0) = 0. The monotone property of ^ is clear since f is non-negative. Since ^ is defined on the same sets as v by definition, ^ is defined on a a-algebra. We need only show that ^ is countably additive.

Let {Ei} be a countable disjoint family of measurable sets, w i t h £ = U£/.

By finite additivity we have

= lf(XEr+"' + XE^)dv. (1)

Welet/Z]v = f'{XEi-\ ^XEN) SO {/IN} is an increasing sequence of non-negative measurable functions, and {h^} converges to /"XE, v^here £ = U £/. From (1) we have

N

Y,^{E,)= / hndv^

and therefore by the Monotone Convergence Theorem

= lim / hhidv N Jx

= ^([JE,).

Problem 2. Suppose f is non-negative and measurable on X, but not integrable. If ^(E) = J^ fdv for measurable sets

13 MORE INTEGRATION 129

E on which f is integrable, and ^(E) = oo otherwise, is ^ a measure? ""HI

For any two finite measures ^ and v on X we will say that ^ is absolutely continuous with respect to v, written ^ <^ v, provided ^(£) = 0 whenever v(£) = 0. Problem 1 shows that if )6 is an integral with respect to v, then ^ is absolutely continuous with respect to v. We now show that absolute continuity characterizes measures ^ which are v-integrals. That is, if ^ attaches measure only to sets £ where v(£) is positive, then there is a weighting function f for v which gives ^.

Problem 3. Show that if y6 and v are finite measures on X, then ^ <^v if and only if given s > 0 there is (5 > 0 so that v(£) < 8 implies y6(£) < 8. Hint: The "if" part is just logic, since if v(£) = 0, then v(£) is less than any S. To show "only if," assume the contrary condition that yS <$C v and there is a sequence {En} with v(En) < XIT" and ^(£„) > £. If £ = n^(£„ U £„+i U • • •) then v(£) < 1/2 - for all ^. If £^ = £^ U En^\ U • • •, then £i D £2 D • • • and n £. = £, so ^(£„) —^ iS(£). •""I

Proposition 2. If ^ and v are finite measures with ^ <^v and ^ ^ 0, then there is a set A with v(A) > 0 and a positive number 8 such that 8v(E) < ^ (£) for every measurable £ C A

Proof. For every n let An^ Bn be a Hahn decomposition of X for the signed measure B — -v. Thus B > ^v on all subsets of An and ^ < ^v on subsets of B„. The An increase and the Bn decrease; let B = fl Bn and A=[j An. Then for all n,

^(B) < ^(Bn) < -v(Bn) < - v ( X ) . n n

Since v(X) < 00, (B) = 0. Since 6 # 0, ^(A) > 0, and hence v(An) > 0 for some n. Thus we have a set An of positive v-measure and a positive number 8(= ^) such that ^v(£) < )S(£) for all £ c An. ill

Corollary. I/" and v are finite measures with ^ <^ v, there are non-negative functions g such that for all E

L gdv < P(E). (2) E


Proof. The function g = \xAn with An as in Proposition 2 is such a function, since for any set £ ,

/ -XAA^ - / -XA„^v < ^ ( £ n A,) < ^ (£ ) . II VE W JE^An n

Proposition 3. (The Radon-Nikodym Theorem) If ^ and v are finite measures on X and ^ <^ v, then there is a non-negative v-integrable function f on Xsuch that

^(E) = 1^ fdv

for every measurable set E.

Proof Assume yS and v are defined on the same a-algebra in X, with ^ <^ v. We know there are non-negative functions f >0 such that for all £, /^ fdv < ^{E). If /i and fi ^re two such functions, then /i V fi is again such a function (Problem 4). Let T be the set of all such functions, and let

p=:sup< I fdv : f e J='\.

Clearly p > 0, and since ^(X) < oo, /? < CXD. Let fn^^ with jfndv —> p. Let gn= /i V 2 V • • • V ^ , so the g„ increase and Jxgndv —> p. Let f = sup fn = limg„, so

/ fdv = p, (3)

and for any set £ ,

/ fdv = lim / gjv < ^ (£ ) . (4) JE JE

Now we have to show that equality holds in (4). Let a be the measure defined by

a (£) = yS(£) - / fdv.


Clearly ot is finite, and positive, and a <^ v. Either a = 0, which is what we want, or by Proposition 2 there is a set A and 6: > 0 such that A is a positive set for a — 6:v. If a ^ 0, and E C A,

OL{E) > £v(E)

yS(E) ~ [ fdv> sv(E) JE

P(E)> Uf + e)dv, JE

Hence for any set E,

)6(E)> [(f + sxA)dv. JE

Thus f + SXA ^ ^, and

J (f + ^XA)dv > J fdv = p,

which is a contradiction, and hence a = 0. ill

Problem 4. Show that fiV fieJ^ii fu (2^ T. ""HI

Problem 5. Let ^ be Lebesgue measure on [0,1], so 6 is defined on the (7-algebra of Lebesgue measurable subsets of [0,1]. Let V be counting measure on the same or-algebra; i.e., if E is measurable, v(E) is the number of elements in £ , so v(E) is usually +00. Obviously ^ <^v. Show there is no measurable f on [0,1] such that ^(E) = /g fdv. Where does the proof of the Radon-Nikodym Theorem fail? ""HI

When there is more than one measure in sight, the concepts of "zero measure" and "almost everywhere" become ambiguous. We therefore write a.e.[v] for "almost everywhere with respect to y." So " f = g a.e.[v]" means that f = g except on sets of v-measure zero.

Problem 6. Show that the f of Proposition 3 is unique; i.e., if f and g are two such functions, then f = g a.e.[v]. ""HI


The weighting function f for v which gives ^ is called the Radon-Nikodym derivative of ^ with respect to v, and we write dfi=fdv,o^%=f.

Problem 7. Extend the Radon-Nikodym Theorem to the case where ^ and v are a-finite, and jS <§C v. Hint: Recall the convention that we write /g fdv = oc if /" is non-negative and measurable but not integrable over £. ""HI

The Radon-Nikodym theorem characterizes measures ^ which have positive mass only where v has positive mass. The antipodal relationship would be a measure ^ all of whose mass is carried on some set of y-measure zero. For example, Lebesgue measure /x on [0,1] is a uniform mass distribution where mass is the same as length. A point mass at :xb G (0, 1) would be the measure ^ on measurable sets such that )6(£) = 1 if XQ G £ and ^(£) = 0 otherwise. This ^ is clearly a finite measure, and all of ^'s mass is carried on the set {XQ] which has /x-measure zero.

Recall that two measures ^ and v are singular with respect to each other if there exist disjoint measurable sets A and B such that AUB = Xand v(A) = ^(B) =^ 0. We say that ^ is singular with respect to v or v is singular with respect to ^ depending on the emphasis. The notation is 6 ± v.

Proposition 4. If ^ and v are two a-finite measures on X (i.e., on the same a-algebra of subsets of X), then ^ can be written as a sum ^ = ô + )8i with ô <^v and î i . v.

Proof. Let A = 6 + v, so that X is a-finite and both ^ and v are absolutely continuous with respect to X. Let / be a non-negative measurable function on X such that

y(£) = / fdX

for all measurable sets £ . Let

A = {x : f{x) > 0}, B = {x : f{x) = 0},

so A and B are disjoint measurable sets and X = AUB. We define


Clearly ^ = Po + ^^.To see that î ±v observe that

v(B) = f fdX= f OdX = 0.

JB JB

To see that fio <^ v, let v(£) = 0. Then f = 0 a.e. [X] on £; i.e.,

X{xeE: fix) ^ 0} = 0,

X{xeE: fix) > 0} = 0,

A(£ n A) = 0. Since

ô(£) = îE n A) < A(£ n A) = 0

)6o(£) = O i f v ( £ ) 3 3 0,

a n d ^ o « v . ill

Problem 8. If j6 and v are finite measures on X, and ^ <^ v and ^ ± V, then )S(£) = 0 for all £ . ""Ill

Problem 9. Suppose ^ and v are finite measures on X, and )S <?C V. Let A = )6 + V, so )S <$: A, and hence )6(£) = /^ gÂ for some g > 0 and all measurable sets £. Show that 0 < g < 1 almost everywhere with respect to v; i.e., gix) < 1 except possibly on a set £ of v-measure zero. ""HI

Problem 10. Let S and T be or-algebras on the same set X, with T c S. Let V be a finite measure on S and let / be a non-negative v-integrable function. Let VQ be the restriction of v to T. The function f is necessarily <S-measurable but not necessarily T-measurable, and therefore not necessarily vo-integrable. Show that there is a T-measurable, vo-integrable function g on Xsuch that f^gdvo = Jj^ fdv (or ^\l A eT. ""Ill

14 PRODUCT MEASURES

Let (X, S, jji) and (Y, J, v) be two measure spaces. We want to define the product measure A on Z = X x Y so that sets A x B have measure X{A xB) = /x(A)v(B). The process is roughly parallel to that used to develop plane measure A in Chapters 9 and 10. Perhaps the most useful result of this chapter is the theorem which allows one to change the order of integration in an iterated integral:

Ix (/Y ^ ' > ) (y)) ^^^ ) = ly ( / , /*( ' y)dt^(^)) My)-

The most effective path to this result involves showing that both iterated integrals equal the "double" integral with respect to the product measure A:

i f(x, y)dk(x, y).

In many cases the product integral above is little more than a curiosity except for its relationship with the iterated integrals.

Plane measure A on R^ provides our model in a limited sense. Recall that the outer measure of a set £ C M is the inf of all sums Y.oi(Ri)^ where {R/} is a sequence of rectangles which cover £ , and a(Ri) is the area. One of the first problems with plane measure was to show that the outer measure of a rectangle is its area; i.e., that outer measure really extends the basic idea of area for rectangles. The same problem had to be faced for linear measure: namely, to show that the outer measure of an interval is

135


its length. For both hnear and plane measure we could use compactness to reduce some countable coverings to finite coverings, and thereby simplify the arguments. In the general setting of this chapter we have no such simplification, which accounts for the distressing amount of set theory which is necessary.

Our plan in outline is as follows. We will call a set A xB, with A a measurable subset of X and B a measurable subset of Y, a rectangle. We will define the "area" of a rectangle by

a{A X B) =/x(A)v(B),

and the area of a finite disjoint union of rectangles as the sum of their areas; thus, if £ = Ai x Bi U • • • U A„ x B„ and the A/ x Bi are disjoint, then we define

a{E) = X^a(A, X B,) = X^/x(A,)v(B,). (1)

The problem immediately arises to show that (1) gives a consistent definition, since the representation of £ as a finite disjoint union is not unique; indeed, a rectangle can be written as a finite disjoint union of rectangles, and in many different ways.

Once we have shown that a is consistently defined by (1) on the set R of all finite disjoint unions of rectangles, we show that JR is an algebra of sets; i.e., a family of sets which is closed under finite unions and finite intersections, and differences and complementation. Since a is consistently defined on R, a is finitely additive on R. R is not necessarily a or-algebra, so countable unions of sets in R need not belong to R. However, a countable disjoint union of sets of R might be in R, and if so, say E = \JEi e R^ we need to know that a is countably additive for this family:

a{\jE^)=E^(E,). (2)

Equation (2) is essential if \J Ei e R because our final measure A is to be an extension of a, so of course (2) must hold for a if it is to hold for X.

Once we have a defined on the algebra R, with a countably additive on R in the above sense, we define A. on all subsets of

14 PRODUCT MEASURES 137

Z in terms of countable coverings by sets in R:

HE) = mi{j^a(E,) : E c[j £/, £. G R} ,

Measurability with respect to A is defined by the Caratheodory criterion, and as before the measurable sets form a or-algebra. The sets in R must be shown to be measurable, and then we know that the a-algebra of measurable sets contains the or-algebra generated by the rectangles Ax B.

Now the details. Throughout this chapter (X, <S, /x) and (Y, J, v) will be measure spaces, and Z = X xY. All subsets AxB oi Z with A and B measurable subsets of X and Y respectively are called rectangles, or sometimes measurable rectangles. We let R denote all sets in Z which are finite unions of rectangles. We define a on rectangles by a(A x B) = /x(A)v(JB).

Proposition 1. Every set in R can he written as a finite disjoint union of rectangles, R is an algebra; i.e., R is closed under finite unions and finite intersections, and differences and complementation.

Proof. The two-dimensional picture (of genuine rectangles) shows that the union, intersection, or difference of any two rectangles can be written as a finite disjoint union of rectangles. The complement of a rectangle is a finite disjoint union of rectangles, since

(A X By = (A X J30 U (A X B) U (A X B'),

To show by induction that any finite union of rectangles is a finite disjoint union, suppose this is true for any n rectangles. Let jRi,. . . , Rn-\-i hen+ 1 rectangles. Let

Ri U • •. U R„ = Si U •.. U S^

where the Sj are disjoint rectangles. Then each set Si — JR +i is a finite disjoint union of rectangles, so

Ri U . . . U R^+i = Rn+i U (Si - R^+i) U'^'UiSm- R^+i)

and the right side is a finite disjoint union of rectangles. Of course any finite intersection of rectangles is again a rectangle.


I f £ i , £ 2 e R.

£ i = Ri U . . . U R„

£2 = Si U . . . U S;,,

with the Ri and the Sj disjoint rectangles, then the Ri n Sy are disjoint rectangles and

ErnE2 = [JR,nSi, hi

so R is closed under finite intersections, ill

Problem 1. Show that the intersection of any family of algebras is again an algebra. Hence there is a smallest algebra of sets containing the rectangles; this is called the algebra generated by the rectangles. ""HI

Now we want to show that a is countably additive on R. The critical step is the next proposition.

Proposition 2. If Ax B is a finite or countable union of disjoint rectangles, A x B = U A/ x Bi, then

a ( A x B) = 5 ]a (A , X Bi),

Proof. The characteristic function oi E = [j Ai x Bi = Ax B is

XE(X, y) = XA(x)xB{y) = J2xA,{x)xB,(y)' (3)

Integrate both sides of (3) with respect to v, using the Monotone Convergence Theorem in the case of a countable sum:

XA(XMB) = '£XA,(XMB,). (4)

Now integrate both sides of (4) with respect to /x, again using the Monotone Convergence Theorem for a countable sum:


This is the same as

a(AxB) = J2^(A^Bi), ill

Proposition 3. If E e R and E is written as a finite or countable disjoint union of rectangles in two different ways,

thenj:aiR,) = Ec((Si).

Proof. Each Ri and Sj can be written as a disjoint union of rectangles as follows:

Ri = [JRir) Sf i

Sj = \jR,nSj.

By Proposition 2,

aiRi) = J2c^(R,nSi) i

a(Sj) = Y,<^(R,nSj).

Therefore,

Yâ(R,) = ^ a ( 5 ; ) = Jâ(R, n 5,). ill

Now we can define a unambiguously on the algebra R by

a(E) = Yâ(R,)

for any finite disjoint family {R/} of rectangles whose union is £ . Moreover, using Proposition 3, it is easy to verify that a is countably additive on R.

Problem 2. Show that a is countably additive on R; i.e., if £• = [j. R-y G R, where {R/y} is a finite collection of disjoint rectangles for each fixed /, and E = \JEi e R^ where {£/} is a countable disjoint family, then a(E) = â(Ei). ""'HI


The outer measure k is defined on all subsets of Z by

X{E) = inf { ^ a ( £ , ) : £ C U £ - ^^ ^ ^ } ^

where the inf is over all countable coverings {£/}. This differs formally from the definition of plane measure in Chapter 9 in that here we use coverings by sets which are finite unions of rectangles (i.e., sets in R) rather than coverings by rectangles. The difference is not material because a is additive on the sets in R. The formal change to Rrather than sticking to rectangles AxB is desirable because R is an algebra of sets, and we will use this fact.

Problem 3. Verify that A is an outer measure; i.e., that X(0) = 0, and A is a monotone, non-negative, and countably subadditive function defined on all subsets of Z. ""HI

As before, a set £ C Z is measurable provided

A ( £ n T ) + A ( F n T ) =A(T)

for every set T. The proofs of Chapter 4 apply without change to show that the A-measurable sets form a a-algebra, and that k is countably additive on this a-algebra.

Proposition 4. The sets in R are measurable; in particular, rectangles are measurable.

Proof, (cf. Proposition 4, Chapter 9) Let E e R. We must show that

X{EnT) + X{E' r\T) <X(T)

for every set T. Assume that A.(T) < oc and let {£J be a countable covering of T by sets from R, with

Y,a(E,)<X(T)+s.

Then

£ n T c [J{E, n £)

F n T c U ( £ / n £ 0 ,

14 PRODUa MEASURES 141

and since R is an algebra, the sets E n E/, E' Pi Ej are all in R. Therefore, since a is additive on R,

A(E nT) + X(E' nT)<^a(En Ei) + ^ a ( E ' n Ei)

= 5 ^ [ a ( E n E , ) + a ( F n E , ) ]

Since ^ is arbitrary, E is measurable by the Caratheodory criterion, ill

We will let Ra denote all sets which can be written as countable unions of sets in R, and R^d will denote all countable intersections of sets in R^. We will find for every measurable set E a set A in R^ which includes E and has approximately the same measure. We then find a set B in R^ which includes E and has the same measure as E. We can show that sets in R^, and then those in R^ , have the basic property relating A to /x and v:

HA) = Jv{A,)df^(x)

Here A is of course a subset of X x Y and for each x e X^ Ax is the x-section of A:

Ax = {yeY: (x, y) e A},

We then get the result for arbitrary measurable sets E by limit theorems on the integrals, and from this we can show the iterated integrals are equal. First we need the following lemma.

Proposition 5. Every set AeR^ can he written as a union of increasing sets in R, and every set B e R^s can be written as the intersection of a decreasing sequence of sets in R^.

Proof Let AeR^, with A = [jEi and each Ei e R. If Fn = El U •.. U E„, then E„ € R and Fi C f 2 C • • •, and A = U Fn.

Now let Ai and A2 be sets of R^, with

AI = U E I , - , A2 = U ^ 2 / ,


and all the sets E^^ £2/ ^ R- Then

Ai n A2 = [jiE^, n £2;), ^/

and all the sets En D Eij are in R, so Ai n A2 G R^. Hence R^ is closed under finite intersections. If B G Ras-, with B = fi A„ and each A„ G R^? then let

A* = Ai n . . . n A„,

so each A* G R^, A* D A* D • • •, and B - fl A*, ill

Proposition 6. / / £ /s any measurable set with k(E) < 00, and 8 > 0, there is a set A e R^ with E c A and X{A) < k(E) + s. There is also a set B e R^s tvith E c B and X(B) = X(E).

Proof. Let {£/} be a countable covering of £ by sets in R with

J2X(Ed<HE) + e,

If A = U £0 then A G R^, £ C A, and

HA) <^X(E,) <X(E) + 6.

Now let An G Ra with £ C A„ and A(A„) < X(E) + ^. We may assume that Ai D A2 D • • •. Let B = fl A„? so B G R ^ and £ C B. Since A(Ai) < 00 and the A„ are nested, k{B) = limX(A„) = k(E). ill

Proposition 7. All sections of sets in R^ are measurable; i.e., if B G Rcrs then B^ = {y : (x, y) G B} is a measurable subset of Y for each x and B^ = {x : {x, y) ^ B] is a measurable subset of X for each y. Since R C R^ C R^s, the sections of R and R^ sets are measurable.

Proof. If £ is a rectangle, or a finite union of rectangles (i.e., a set in R) then sections are obviously measurable. If A G R^, with A = U £/5 Ei G R, then for each x

A,= (]jE,)^ = [j(E,),,


SO Ax is measurable. If B G Ras with B = f]An and each An G jR r, then each (A ); is measurable so

B,= (f]An)^ = f](An),

is measurable, ill

Proposition 8. If B e R^s and k(B) < oo, and

f(x) = v(B,), g{y) = f^(By),

then f and g are measurable functions, and

jf{x)dii{x) = jg{y)dv{y) = HB),

Proof If JB is a rectangle, the result is clear. If B e i^^, then B is a countable disjoint union of rectangles, B = [jRn. Hence

f(x) = v{Ex) = Y,v(Rn)x,

so f is the sup of a sequence of measurable functions. Moreover, by the Monotone Convergence Theorem,

j f(x)dfi(x) = ^ jv{Rn)xdii{x)

= YimJ2KR,) n ^—^

\i=i

Now let B G RfxB, with A,(B) < oo. Then, as in the proof of Proposition 6, we can write B = [^A„ with A„ e Ra, X{Ai) < oo, and Ai D Ai D • • •. We know

JviAU^^ix) = MA,) < 00,


SO v(Ai)x is measurable and finite-valued for almost all x. For these X, v(An)x is a decreasing sequence of integrable functions and, since v(Ai)x is integrable,

I v(An)xdiJ.(x) —> v(Bx)dix(x).

But

and

so

/

To show that

v(An)xdti(x) = X(An),

X(An) -^ X(B),

X(B) = Jv(Bx)dfi(x). I

X(E) = jv{E,)dix{x)

holds for all measurable sets £ of finite measure, we surround £ with a RCJS set B with the same measure. The result follows once we show that v{Bx) = v(Ex) a.e. [/x].

Proposition 9. IfE is measurable and X{E) < oc, then E^ and E^ are measurable for almost all x and almost all y, and

fix) = v(Ex), g(y) = fi(Ey)

are measurable functions defined ax., and

X(E) = jf{x)d^l{x) = jg{y)dv{y).

Proof. Let X(£) < oo and use Proposition 6 to select B e R„s with E c B and A.(B) = X(E). Let C = B—£, so C is measurable and A(C) = 0. For all x, Q = B^ — Ex. Let H e R^s where H D C and X(H) = X(Q = 0. Then by Proposition 8

Jv(H,)dl^{x) = k(H) = 0.


It follows that v(Hx) = 0 a.e., and since Hx D Q , v(Q) = 0 a.e. Therefore Ex = B^ a.e., so v(Ex) is measurable and

A.(£) = jv{Ex)diJi{x).

The companion result for y-sections is proved similarly, ill

Proposition 10. (Fubini) Let (X, S, JJL) and (Y, J, v) be two measure spaces, and let X be the product measure on XxY. For any integrable function f{x, y) on XxY let

fxiy) = fix, y), fy(x) = fix, y).

Then fx is an integrable function on Y for almost all x and f^ is an integrable function on X for almost all y, and

Fix)= / fix, y)dviy), Giy) = / fix, y)diiix)

are integrable functions on X and Y, respectively, and

/ Fix)dixix) = / Giy)dviy) = / fix, y)dXix, y); JX JY JXxY

i.e.,

Jx UY * ' ^^^ * v '^'^^^ ^ JY UX * ' y ^ * ^ ) ^^^y^ = / fix, y)dkix, y),

J XxY

Proof. By symmetry we need only show that fx is integrable for almost all x^ and JF(X) = Jfxiy)dviy) is integrable, and

/ Fix)diiix) = / fix, y)dXix, y). Jx JXxY

It is sufficient (Problem 4 below) to consider only non-negative functions fix, y).

Since fix, y) is integrable and non-negative, there is an increasing sequence {cpn} of simple functions on X x Y so that (pnix, y) —> fix, y) a.e. [k]. It follows that

i(Pn)xiy) —> fxiy) a.e.[A].


If (p is simple, ^ is a finite or countable sum of functions V (x, y) of the form

. f{x,y) = axE{x,y)

where E is measurable and X(E) < oc. Notice that

i^x(y) = axESy)'

Since Ex is a measurable subset of Y for almost all x, XE^ is a measurable function of y for almost all x, and each summand V x(y) of (Px(y) is measurable. Therefore (px is measurable for almost all x if ^ is a simple function. If (pn(x, y) increases to f(x, y), then ((Pn)x increases to fx^ so fx^ is measurable for almost all X. By the Monotone Convergence Theorem,

j((Pn)x{y)dv{y) -^ Jfx(y)dv(y) = F(x)

for almost all x. Since each cpn is simple, each integral on the left is a measurable function of x by Proposition 9, so f is a measurable function. By the Monotone Convergence Theorem (twice), and Proposition 9 again,

/ ( / f(^^ y)dv{y) j (i/x(x) = lim / ( (pndv j dfx

= lim / (pn(x, y)dX(x, y)

= I f(x, y)dX(x, y), ill JXxY

Problem 4. Show that it suffices to prove the Fubini Theorem for non-negative functions. ""HI

Problem 5. Let X and Y be uncountable sets, for instance X = Y = [0,1]. Let fi{x} = 1 for all x G X and v{y} = 1 for all y G Y, so both /x and v are counting measure. If A. = /x x v, then H(x^ y)} = 1 for all (x, y). Show that no sequence of simple functions increases to the function f which is identically one on X X Y, so the proof of Fubini's Theorem fails. (Of course this f is not integrable.) ""HI


Proposition 11. (TonelWs Theorem) If(X, S, JJL) and (Y, J, v) are a-finite measure spaces and f is a non-negative measurable function on XxY, then the necessary measurability conditions hold as in the Fubini Theorem, and

J (j f(x^ y)dv(y)] dfiix) = J (J A^ ' y)dlii(x)] dv{y)

f(x, y)dX(x, y). = / JXxY

Proof. If X = [jXn.Y = UYn with each X^ and Yn of finite measure and Xi c X2 C • • •, Yi C Yi C • • •, then Xx Y = U X„ X Y„ and X„ x Y„ has finite A-measure. There is a sequence {(pnj} of simple functions supported on X^ x Y„ which increases to / on X„ X Y„:

hm (pni(x, y) = f{x, y) on X„ x Y .

Therefore there is a sequence i/ „(x, y) of simple functions which increases to f{x, y) on X x Y. The point is that the fn are zero off finite-measure sets, which was the critical point where Proposition 9 is used in the proof of Fubini's Theorem. The rest of the proof of Proposition 10 proceeds as before, ill

To use Tonelli's Theorem to change the order of integration in an iterated integral

/ / JXJY

f(x, y)dix(x)dv(y), XJY

you first check that /x and v are a-finite measures. If you are an analyst, X and Y are probably both the real line or the plane, so this is no problem. (If you are a probabiUst, then you need more help than you can get here; I suggest prayer.) Given that X and Y (or /x and v) are or-finite, you check that f is measurable. Generally this is because f is continuous. If all you have is facts such as /"(x, y) is continuous in y for each x and measurable in X, then you are in the realm of Polish topology, which is beyond the scope of a primer. If you know that f is measurable, and either iterated integral of | f{x, y)\ is finite, then you can change the order of integration.


Problem 6. If f{x) is integrable on X and g(y) is integrable on Y and h(x, y) = f{x)g(y)^ then h is integrable on X x Y and

/ h(x, y)dk(x, y)= f{x)dii{x) / g{y)dv{y).

Do you need to assume that h is non-negative or that X and Y are a-finite? ""HI

15 THE SPACE L^

The study of Fourier series in the early 1800s gave rise to many fundamental advances in analysis. The question was this—what functions can be represented by Fourier series? It soon became clear that this question could not be answered without a better understanding of the basic ideas of analysis, including what is meant by "function" and "represent." After the Lebesgue integral was introduced in 1904, the space L^ of square-integrable functions and the space i^ of square-summable sequences emerged as heroes.

We will start with £^, which we introduce as the infinite dimensional analogue of Euclidean space.

We are familiar from elementary physics and calculus with the treatment of R^ and R^ as vector spaces. li x = (xi, X2) and y = (yi^ yi) are elements of R^, and a is 2i real number, then we define vector sum and scalar multiplication as follows:

x + y = (xi + yi, X2 + yi) (1)

ax = {ax\, ax2). (2)

The sum, +, of (1) makes R^ an abelian group, and the scalar multiplication (2) then makes R^ a vector space, or linear space. Similarly, R^ is a linear space under coordinate-wise addition and scalar multiplication. \ix= (xi, Xi.Xy) and ^=(^1,^2.^3)? then

x+y={xi + yi, X2 + yi, X3 + 3/3)

ax = {ax\,ax2,axy).

149


The absolute value of a vector (point) of R or R^ or R^ is its distance from the origin. For x G R, |x| is the distance from the origin, and |x — yl is the distance between x and y. For X = (xi, X2) in R^, \x\ = yx^ + xf is the distance from x to the origin, and

\x-y\ = ^J{xl - yiY + (X2 - yi)^

is the distance from x to y. Each R" becomes a linear space v^ith coordinate-wise addition and scalar multiplication. For X = ( x i , . . . , x„) G R" with n>3 the distance from the origin is called the norm instead of the absolute value, and the notation acquires two additional vertical bars:

ll ll = V i + * • • + ^l

11^- y\\ = \l{x\ - yi)^ + '" + {xn- ynY'

For any R" the norm satisfies the following relation with scalar multiplication:

IMx|| = |^|||x||.

The dot product or scalar product of vectors in R^ or R^ is familiar, and we make the analogous definition for vectors in R"" : if X = (x i , . . . , x„) and y = (y i , . . . , y^), then

(x, y) = xiyi + --- + x„)/„.

Problem 1. Verify that the following identities hold in R"".

(i) a(x+ y) =ax + ay; (ii) 11 x11 = 11 llxll;

(iii) (x,y) = (y, x); (iv) (x + y,z) = (x, z) + (y, z)\ (v) (ax,y) =a(x,y).

(Parts (iii), (iv), (v) say that the inner product is a linear function of each variable.) ""Ill

15 THE SPACE£2 151

In M, M , and R^ the angle 0 between two vectors x and y is determined by the formula

(x,y) = | |x|| | |y||cose. (3)

We can define the angle 6 between vectors in any E" by (3) provided |(x, y)\ < \\x\\ \\y\\ for all x and y. The definition of 6 is not important, but the inequality is.

Problem 2. (Cauchy's Inequality.) If x = (x i , . . . , x„) and y = (yi , . . . ,y«) , then

\(x,y)\ < \\x\\

i.e.,

Yl^iyi i=l

n \ 1/2 / „ X 1/2 ,2 \ / V ^ ..2 \ ns^j [^/-

Hint: First show that it is sufficient to prove this when \\x\\ = \\y\\ = 1. Notice that \ab\ < (a^ + b^)/l for all a, b, since {a — b)^ > 0, and therefore |x/y/| < (xf + yf)/2 for each /. Summation, using ||x|| = \\y\\ = 1, gives the result. ""HI

The other elementary but important inequality for norms in M!^ is Minkowski's inequality, which gives us the triangle inequality for the metric ||x — yll.

Proposition 1. If x = (xi,,.,, Xn) and y = (y i , . . . , y^), then

l|x + y | | < | | x | | + 113/11;

i.e..

Proof. E(xi + jif = T.xf ^Tyf + 2E^,y,- By Cauchy's inequality,


SO

Efe + y.)' < E ^ + 2 (E ^)'" (E 3^)"'+E yf

= ((E=^)"^(E)f)"T-Taking the square root of both sides gives us the inequahty we wanted to prove, ill

Problem 3. Let d{x, y) = \\x — y\\ he the distance between x and y in R". Verify the three conditions which characterize a metric:

(i) d(x, y) >0 and d(x, y) = 0 if and only ii x = y; (ii) d(x, y) = d(y, x) for all x, y;

(iii) d(x, y) + d(y, z) > d{x, z) for all x, y^ z. ""HI

The family of all sequences x = (xi, Xi , . . . , x„,. . .) again forms a linear space under coordinate-wise addition and scalar multiplication. To get a norm analogous to that for R' we restrict our attention to sequences x such that Y^xf < oo. This space is called £^, and in i^ we define the norm

11=11 = (E^)"'

and the inner product

(x^y) = '^^iyi' (4)

Problem 4. (i) The sum in (4) converges absolutely and | (x, y) \ < \\x\\ \\y\\ for all X, y G £^

(ii) Given that x, y e £^, show that x + y e l^ and that l|x-fy||<||x|| + ||y||. I

Now fix a measure space (X, S, v), and let L^ be the space of all square-integrable functions on X; i.e., f e L^ ii and only if Jf^dv < 00. We will primarily be interested in just two measure spaces: (i) X = [—7t,7t]^ which is the natural home for the trigonometric functions cos kx^ sin kx; and (ii) X = N, with

15 THE SPACED 2 I53

V equal to counting measure. Notice that L^[N] is exactly the space (^ of square-summable sequences.

For our first few results we let (X, S, v) be any measure space and define the norm and inner product in L^ as follows:

(f,g) = lfgdv.

Proposition 2. If f,g e V-, then (i) f+ge V-, and \\f+g\\ < II f\\ + \\g\\;

(ii) fg is integrable, so (f, g) makes sense, and ! \fg\ < II f\\ \\g\\. Hence \{f,g)\ < jj / | | ||g|| for all f, g.

Proof. The function x^ is convex, which means that

{Xx^ + (1 - X)x2f < Xx] + (1 - X)o^ (5)

for all numbers x\ and X2 all X € (0,1). Assume fgGL^ and let fo = f/\\ flgo = g/\\g\\, so II foil = llgoll = 1. Estimate as follows:

\f(x) + g(x)\^<(\f(x)\ + \g(x)\f

2 = (II f\\ fo(x) + WgWgoix))

The second parenthesis in the last line above is a convex combination of fo(x) and go(x), of the form (5), so

\f(x)+g(x)\^

= (II f\\ + \\g\\)(\\ f\\ foixf + ||g||go(^)').

The right side above is integrable, so f + g € L^. Integrating both sides above and using || /o||^ = Hgoll = 1» we get

ll/' + gll '<(ll/ ' l l + llgll)'.


Part (ii) is proved just as before. Using

\f{x)g{x)\<{f{xf + g{xf)/l.

we see that fg is integrable, and

/iAgi<^(ii/' i i ' + ii^ii')-

Replace f by f/\\ f\\ and g by g/||g|| in (6) and get

(6)

/• | | | |gl | / '^^'-2y 1 ^ 1 il f\\ 1

|2 1

+ g 1 Inf i l l

= 1. ill

The nice connection between the functions in L^[—7t,n] and the Fourier series with coefficients in (} = L^[N] depends on the fact that every L^ space is complete in its norm. The following apparent digression is necessary for the completeness proof. In any normed space the convergence of infinite series is defined just as for series of reals. If {x„} is a sequence of vectors (points in a normed space) then X] x„ = x means as N —> oo. The series J^ Xn is absolutely convergent provided the real series J2 W^nW converges.

Proposition 3. A normed space is complete if and only if every absolutely convergent series converges to a point of the space.

Proof First assume that {x„} is a sequence in a complete normed space and Yl W^nW converges. If 5„ = Xi H h x^, then

l|5„ - SmW = WXm^i + Xm+2 + '" + X„||

< ||x^+ilH---- + ||x„||.

Since Y. II n II converges, the right side above is less than any given 6 for all sufficiently large m and n. Thus {5„} is a Cauchy sequence in a complete space, so s„ converges; that is, Y ^n converges.

Now assume that every absolutely convergent series converges. Let {Xn} be a Cauchy sequence; we must show that {x„} converges. For each k pick nk so that \\Xi — Xj \\ < 2~^ if /, / > nk. We can assume that the nk are strictly increasing. Consider the

15 THE SPACE£2 155

series

whose ^th partial sum is x„ . This series is absolutely summable since \\Xn,^^-^ — ^ J l < 1/2^. Therefore the series (7) converges; i.e., there is some x such that Xn^ —> x as ^ —> oc. Since {Xn} is Cauchy and has a subsequence which converges to x^ the sequence {x„} itself converges to x, and the space is complete, ill

Proposition 4. (The Riesz-Fischer Theorem) The space L^ is complete. Since (} is L^ for X = N and v equal to counting measure, l^ is complete.

Proof. We show that every absolutely convergent series in V-is convergent. Let {fn] be a sequence in L^ such that Y. II 11 = M < oc. Let gn{x) = YJi^i I fi{x)\', so gn is the sum of a finite number of L^ functions, and hence gn e L^. Moreover,

iig.ii<Eii/^ii<M-

For each fixed x, {gn(x)] is an increasing sequence of real numbers, so {gn} converges pointwise to an extended real-valued function g, which is measurable if it is finite almost everywhere. For all w,

\\gnf = lgl<M'

SO / g^ < M^ and g e L^. Since g^ is integrable, g is finite almost everywhere. For those x for which g(x) = J2\f(x)\ is finite, the series ^ fi(x) converges; let

n

Sn(x) = Y^ fix) i=l

00

i=l

The function s is measurable. Moreover, for all w, n

\s„(x)\ < ^\fi(x)\ =g„ix),


SO \s(x)\ < g(x). Therefore s e L^, and the absolutely convergent series Yl fi converges pointwise a.e. to s G L^. We must show that the series converges to s in the L^ norm; i.e., \\Sn - s\\ —> 0. Notice that

\Sn{x) - 5(x)|^ < {lg{x)f = 4g(x)\

The functions (s„(x)—s(x))^ converge pointvîse a.e. to zero, and they are dominated by the integrable function 4g(x)^. Therefore

/ (Sn - sY -^ 0;

i.e., \\Sn — s\\^ —> 0 and the absolutely summable series Ylfi converges to s in the L^ norm, ill

Now we return to Fourier series and give one pleasant answer to the question of what functions can be represented by Fourier series. A trigonometric series is a series of the form

i OO

~ao + âk cos kx + bk sin kx, (8) ^ k=i

If Sn(x) is the nth partial sum of (8), then of course Sn G L^ [—7t,Tr]^ and we will need the formula for the L^ norm calculated in the following problem.

Problem 5. If s„(x) = |ô + E L i ^kôskx + bkînkx^ then ||s„||^ = Tc[\al + Y!k=:\ ^k + ^ 1 ^ i- -5 ^^^ ^^ norm of s„ is essentially the same as the l^ norm of the sequence of coefficients: jUo, ^1, ^1, ^2, ^2i •. • Hint: The requisite orthogonality relations are:

r 1 . T . X (7t a k = i / cos kx cos ixdLI (X) = < , ,

/ smkxsm£xdfji(x) = < ^ . , [y) J-7T 10 if k^ i

f sin kx cos i xdiJL(x) = 0. ""U

15 THE SPACE [2 157

If the trigonometric series (8) converges pointwise to an inte-grable function /", and if the partial sums are dominated by an integrable function so that the Lebesgue convergence theorem appHes, then the coefficients {an}^ [bn] are determined by f as indicated in the next problem.

Problem 6. Let Sn{x) be the ^th partial sum of the trigonometric series (8). If there is an integrable function g on [—it.n] such that \Sn{x)\ < g(x) for all x and n^ and if {s„} converges pointwise a.e. on [—7t, 7t] to the integrable function f^ then

1 r^ ak = — f(x)coskxdfx(x)

71 J-7t

bk = - r f{x)smkxdix{x). II (10) 7t J-71

Unfortunately, the kind of bounded or dominated convergence required in Problem 6 is not easy to come by, so pointwise convergence of Fourier series is less than ideal. However, the formulas (10) make sense for any integrable function f on [—n, 7t]^ and hence for any L^ function. If f is an integrable function which happens to have a trigonometric series that converges to it nicely, then we know what the series must be. Accordingly, the numbers {uk]^ {bk} are called the Fourier coefficients of / . The trigonometric series with these coefficients is the Fourier series for / , and no assumption is made about the convergence of the series. The mapping from integrable functions (or L^ functions) to sequences of Fourier coefficients is linear; this is a simple consequence of the fact that the integral formulas (9) are linear functions of f. The mapping from integrable functions, and therefore from L^-functions, to sequences of Fourier coefficients is also one-to-one; i.e., if /* 7 g, then f and g have different sequences of Fourier coefficients. The proof of this last fact is unfortunately out of our path, so this will have to be an article of faith. Given this fact, we will show that the mapping from L^-functions to their sequences of Fourier coefficients is a one-to-one mapping on L^ onto f-. Moreover, the mapping nearly preserves the norm in the sense that || f\\^ — T[\^a\ + Y.i<^\ + b^\


if {Uk}-, {bk} are the Fourier coefficients of f. The calculations of the following problem give us most of the information we need.

Problem 7. Let f e L^[~7t, TC] and let {^ j, {bk} be the Fourier coefficients of f. Let

1 "" Pn{x) = -Qfo + ^(Oik c o s kx + ^k s iu kx)

be any nth-degree trigonometric polynomial. Show that

f-Vnt^Wft-n 4 + E ^4 + bl) k=i

(11)

Hint: This is nothing more than a moderately unpleasant exercise in algebra. The simplest way may be to verify the case w = 1, which involves completing a square, and then use induction. ""HI

Now we milk the identity (11). One obvious but significant consequence of (11) is that the nth partial sum of the Fourier series for f is the trigonometric polyomial of degree n or less which best approximates f in the L^ norm. Any other nth-degree trigonometric polynomial P«(x) gives a strictly larger value for || /" - P„||. We also see from (11) that if P„ is the nth partial sum of the Fourier series for f (i.e., ak = ak, ^k = bk for ^ = 0, 1 , . . . , n) then, since || f - P„||^ > 0,

7t 4 + 224 + bl k=i

< II fr (12)

This is called BessePs inequality. It follows immediately that if f e L^, the sequence ^UQ, ai, bi, ai, ^2. • • • of its Fourier coefficients is in i^. Moreover, (12) shows that the mapping from functions in L^ to their Fourier coefficients in i^ is continuous. Our final result shows that Bessel's inequality (12) is an equality, and that the mapping from L^ to i^ is onto. That is, every i^ sequence makes a Fourier series which represents an

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L^-function, and the norms are preserved from L^ to (} with a constant factor, so

n -al + Y. (4 + bl) = \\ft. (13)

Proposition 5. {Sometimes called the Riesz-Fischer Theorem because the completeness of L^ is the essential ingredient in this proof.) If {an}, {bn} are sequences in i^, then

1 OO

: < o + '^^k cos kx + bk sin kx (14)

is the Fourier series of a function f e L^. If Sn(x) is the nth partial sum of the series (14), then Sn—>f in L^; i.e., W^n — f\\ —> 0. The equality (14) holds between the l^ norm of the coefficients and the L^ norm of f.

Proof We show first that if Sn(x) is the nth partial sum of (14), then {s„} is a Cauchy sequence in L^, and hence has a hmit f. Then we show that (14) is the Fourier series of this limit f. A calculation like that of Problem 5 shows that

\Sfi Sf Jl^ = / X] akCOskx +bks'mkx] ^ \k=m+l J

all the cross product integrals being zero. Since {^„}, {bn} e £^, the identity above shows that {s„} is a Cauchy sequence in Since L^ is complete, there is f e L^ so \\Sn — f\\ —> 0. Since II f — SnW —> 0 it follows that the inner product (f — Sn, g) —> 0 for every g e L^.In particular, if g(x) = cos kx^ we have

0 = lim if — Sn, cos kx) n—>0Q

= l i m [( f c o s kx) — (Sn, COS kx)] n—>OQ

= (f cos^x) — Ttak.


Therefore Uk is the ^th Fourier cosine coefficient, and a similar calculation shows that bk is the ^th sine coefficient. That is, the general trigonometric series (14) formed from i^ sequences {^„}, {bn]/is the Fourier series of the L^ function which is the L^ limit of its partial sums. We know from Bessel's inequality that

\\Sn\\ = ^ <\\ft

Since ||s„ - f\ —> 0, it follows that ||s^|| f II; i .e. ,

n ff, ill

To sum up: Each £ sequence defines a trigonometric series whose partial sums converge in 1} to an iJ function f, The Fourier coefficients of this f are the given sequence in €^. The 1} functions all determine sequences of Fourier coefficients in f}^ and the above result says this mapping from L^ to t^ is onto. The mapping is obviously linear, since integrals are linear. The mapping is also one-to-one. This is our one article of faith. There is, therefore, a one-to-one linear mapping ip on 1} onto t^ such that7r| |(^(/ ') | |^=||/ ' | |^forall /" G L^.

INDEX

absolutely continuous 129 absolutely convergent 18, 154 absolute value of a vector 150 admissible function 63, 118 admissible partition 63, 118 algebra of sets 136 algebra generated by a family 138 almost everywhere (a.e.) 49 Axiom of Choice 37

g basket of numbers 17 Bessel's inequality 158 Bounded Convergence Theorem 58, 123

Cantor set 39, 60 Caratheodory criterion 30 Cauchy net 18 Cauchy's inequality 151 characteristic function 46 choice function 12, 53 compact set 22 complete measure 108, 110 complete normed space 154 convergent net 10 convex function 153 countable additivity 31 countable subadditivity 24, 87 —for signed measures 111 countable partition 62 counting measure 105, 108 covering of a set 22 cross section 98

161


J ) dyadic interval 86 dyadic square 86 differentiation under the integral sign 71 Dini derivates 77 directed set 10 distance in R"" 151 Dirichlet problem 81 dominated convergence theorem 68, 123 double integral 135

Egoroff's Theorem 58, 122

Fatou's Lemma 68, 123, 124 Fourier coefficients 157 Fourier series 157 Fubini's Theorem 101, 145 Fundamental Theorem of Calculus 73

f^ Hahn decomposition 114 harmonic function 81 Heine-Borel Theorem 22

improper integral 6 inner measure (m*) 41 inner product in L^ 152 integrable function, Riemann 2 integrable function, Lebesgue 45, 118 integral, Riemann 2 —, improper Riemann 6 —, Lebesgue 45 iterated integral 135

T Jordan decomposition 115

Uf,P)2 V- 152 i^ 149, 152 Lebesgue Dominated Convergence Theorem 68, 123 Lebesgue integrable function 45 Lebesgue (outer) measure 21, 22, 85

INDEX 163

—, in the plane 85 Lebesgue singular function 75 linear space 152 little boy 17 lower sum, upper sum 2

J\4 m%m*28,41 measure 21, 22 —, complete 109 measurable function 46, 117 measurable set 27, 88, 140 measure space 107 metric 152 Minkowski's inequality 151 Monotone Convergence Theorem 69, 124 monotonicity of measures 23, 167

] ^ negative set 111 net 10 nonmeasurable set 37 norm 19, 153, 156 null set 109, 112

Q outer measure {nf) 28

p partition 1, 43 point mass 108 Poisson kernel 82 positive set 111

Q Q (the rationals) 21 , 25

R R(/-,P,c:)13 Ra. RaS 141 Radon-Nikodym Theorem 130 Radon-Nikodym derivative 132 rational numbers 21, 25 rectangles 85, 136, 137 refinement 3, 43 regular measure 91 Riemann integrable 2, 13, 60 Riemann integral 1 Riemann sum 12, 53 Riesz-Fischer Theorem 155, 159


§ scalar multiplication 149 scalar product 150 section of a set 141 or-algebra 35, 107 a-finitelOl, 110 signed measure 111 simple function 46, 119 singular measures 114, 132 step function 5 subadditive 24 summable 17 symmetric difference 39

X Tonelli's Theorem 101, 147 total length of a covering 22 total variation of a measure 115 trigonometric series 156

U U(/-,P)2 unordered sum 17 upper and lower sums 2 —, for Lebesgue integral 43

^y vector space 149 Vitali covering 75 Vitali's Theorem 75

7 Zorn's Lemma 37

Bear - A Primer of Lebesgue Integration

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