[Emmanuele DiBenedetto (Auth.)] Real Analysis(BookZa.org) Pages155,156,157,Etc

5/20/2018 [Emmanuele DiBenedetto (Auth.)] Real Analysis(BookZa.org) Pages155,156,157,Etc

1/17

6. Integral of simple functions

135

Integral of simple functions

Let

{X

A, L} be a measure space, and let E

E

A.

For a measurable set

A

and

x E JR, define

if x

=I= 0

if x = O.

Since /L E nA) E

JR ,

the first

of

these is well defined

as an

element of

JR

for

all

x

E JR - {O}.

Let f : E --7 JR be a nonnegative simple function, with canonical representation

n

f = 2:. fi XEi

i=1

6.1)

where {E1

E2,

,

En}

is a finite collection of mutually disjoint measurable sets

exhausting E and {f l 12, ... , f;,} is a finite collection of mutually distinct, non

negative numbers. The Lebesgue integral

of

a nonnegative simple function f is

defined by

Ie

f x)d/L

=

Ie

fiXEid/L

= fi/L E

i

).

6.2)

This could be finite or infinite.

I f

it is finite, then f is said

to

be integrable in E.

Remark 6.1. I f f : E --7 JR is nonnegative. simple. and integrable. the set

[ f > 0] has finite measure.

Now let {E

1

E2,

Em}

be a finite collection of measurable disjoint sets

exhausting

E

and consider the nonnegative. simple function

m

f =

2:. fj XEj '

j=1

6.1 )

where

/ j

are nonnegative numbers. This is not, in general, in canonical form. Since

the sets E are mutually disjoint and exhaust

E

this would occur if the numbers

f j are mutually distinct. We put it into its canonical form by setting

Ei = U{Ej l f j =

fil,

i = 1,2,

..

n,

6.3)

and then by writing the representation 6.1) by means of the sets E

i

.

From the definitions 6.2) and 6.3), it follows that

lef

d

/L=tfi 2:..

/L Ej) =

t / j /L (E j ) .

1=1

{ j : j= t l J=1

Thus the integral of a nonnegative simple function is independent of the represen

tation of f.


2/17

136 HI. The Lebesgue Integral

Let

f.

g :

E -

lit be nonnegative simple functions. If

I

g a.e. in

E,

then

L

x)df l

L

i x)dfl

6.4)

If both I and g are nonnegative, simple, and integrable,

L

al

+

f3g)dfl =

aL

df l

+ 3L

dfl

6.5)

for all a, f3 E

lit.

7 The Lebesgue integral of nonnegative functions

Let

I : E - lIt*

be measurable and nonnegative, and let Sf denote the collection

of all nonnegative simple functions : E - lit such that

: S I

Since == 0 is

one such function, the class Sf is nonempty.

The Lebesgue integral of lover E is defined by

I

fd f l = sup r d f l

E ~ S r J

7.1 )

This could be finite or infinite. The elements

E

Sf are

not

required

to

vanish

outside a set of finite measure. For example, if I is a positive constant on a

measurable set of infinite measure, its integral is well defined by (7.1) and is

infinity.

The key new idea of this notion of integral is that the range of a nonnegative

function

I

is partioned, as opposed to its domain as in the notion

of

the Riemann

integra1.

6

If

I

:

E - lIt* is

measurable and nonpositive, we define

(7.1)-

A nonnegative measurable function

I : E - lIt* is

said

to

be

integrable if

the

number defined by (7.1) is finite. For example, if

fl is

the counting measure on N,

a nonnegative function I : N -

lit

is integrable if and only if L I n)

0, there exists an index

nO

x) such that

fn x) 2: t x) - E for all n 2: n(x).

By the version

of

Egorov s theorem as stated in Proposition 2.4, having fixed

IJ > 0, there exists a set Fry C F such that JJ-(F - Fry) ::: IJ and this inequality

holds uniformly in

Fry;

i.e., for every fixed

E

>

0, there exists

n

such that

fn x) 2: t x) - E for all n 2: n and for all x E Fry.

7P. J l atou, Series trigonometriques et series de Taylor, Acta Math., 30 (1906), 335-400.


4/17

138 III. The Lebesgue Integral

From this, for

n ::::: n

e

,

r

dj1.:::::

r

j ~ d f l : : : : :

r

S- -

E)df.l.

J

hi

JF'I

i -dj1. -i-F,; ~ d f l - Ej1.(F)

::::: [ S-df.l. - 1] sup S -

Ef.I.(F).

Since

j1.(F)

is finite, this implies

liminf

[j;,dj1.

:::::

[S-dj1.

for all integrable

S

E

Sf

(8.2)

If S

is not integrable, it equals some positive number 8 on a measurable set

F e

of

infinite measure. Having fixed

E

(0, 8), set

Fn

= {x E ElfJ x) :::::

8 -

for all

j

n}.

From the definition

of

lower limit Fn C

Fn+l

and F e U

Fn.

Therefore,8

From this,

lim inf r

n

dj1.:::::

lim inf r t;,dj1.::::: 8 - E) lim inf f.I.(Fn).

J JF

Thus in either case, (S.2) holds for all ESt

D

In the conclusion 8.1)

of

Fatou s lemma, equality does not hold in general. For

example, in

lR

with the Lebesgue measure, the sequence

{

I

forxE[n, n+l)J,

f,, x)

= .

o

otherwIse

satisfies

S.l)

with strict inequality. This raises the issue

of

when

S.I)

holds with

equality or, equivalently, when one can pass to the limit under the integral.

Theorem

8.2 (monotone convergence). Let {fn} be a monotone increasing se-

quence

of

measurable, nonnegative functions in

E;

i.e.,

O.:s

fn x).:s fn+l(X) for all

x E and foralln E N.

Then

l m [

f dj1.

=

lim fndf.l..

H

Sec (3.3) of Proposition 3.1 of Chapter II.


5/17

9.

Basic properties of

the

Lebesgue integral 139

Remark 8 1 The integrals are meant in the sense

of

(7.1). In particular, both sides

could be infinite.

Proof

o f Theorem 8.2. The sequence

Un}

converges for all x E E to a measurable

function f E --* JR . Therefore, by Fatou s lemma,

9 Basic properties o the Lebesgue integral

Proposition

9 1

Let

f,

g : E

--*

R

be

integrable. Then

for

all

a,

f

E

R

f

f g a.e. in

E

then

l fdJ-L ~

19dJ-L.

l dJ-L1 ::: l l f ldJ-L.

If E

is a

measurable subset of

E then

D

(9.1)

(9.2)

(9.3)

(9.4)

Proof

For

a

0, denote by

a s

the collection

of

functions

of

the form

al;,

where

l;

E

Sf f

a 0 and f 0, then aSf =

Sexf

Therefore,

{ afdJ-L=

sup (

rydJ-L=a

sup

, l ;dJ-L=a , fdJ-L.

JE

I} Saf

JE

~ S f

E E

Similarly,

if a

0 and f is integrable and

of

variable sign, then (9.5) continues to hold in

view of (7.2) and the decomposition

A similar argument applies if a < 0, and we conclude that (9.5) holds true for

every integrable function and every

a

E

lR.

Therefore, it suffices to prove (9.1) for

a = f 3 = l .


6/17

140 III The Lebesgue Integral

Assume first that both

f

and g are nonnegative. There exist monotone increasing

sequences

of

simple functions

{ ~ }

and

{ ~ }

converging pointwise in

E

to

f

and

g respectively. By the monotone convergence theorem,

e f

g dfJ- =

lim e

~ n

n ) d f J -

= lim e ~ n d f lim e ~ n d f -

= e fdfJ- e gdfJ-.

Next, we assume that

f

0 and

g

:s

O.

First, we observe that f

g

is integrable

since I f gl :s If I

gl

From the decomposition

f g +

-

g

=

f g)-

f

and 9.1) proven for the sum of two nonnegative functions,

This and the definition 7.2)

proves 9.1) for f 0 and g

:s

O.

f

f

and

g

are integrable with no further sign restriction,

To prove 9.2), observe that from f

- g

0 and 9.1),

Inequality 9.3) follows from 9.2) and

If :s f

:s

I f

I

Finally, 9.4) follows from 9.1) upon writing

f

= fXE fXE-E

D

orollary

9.2.

Let f

: E

JR

be integrable and let

E

be

of

inite measure.

Then

fJ- E) inf f (x):s [ fdp,:s

p,(E)

sup f(x).

XEE

JE

XEE


7/17

10

Convergence theorems 141

1

Convergence theorems

The properties

of

the Lebesgue integral permit one to formulate various versions

of

Fatou s lemma and

of

the monotone convergence theorem.

For example, the conclusion

of

Fatou s lemma continues to hold

ifthe

functions

fn are of variable sign, provided they are uniformly bounded below by some

integrable function

g

Proposition 10.1. Let

g

:

E

~

be integrable

and

assume that fn

:::

g a.e. in

E

for

all n E

N

Then

liminf

e

fnd/-L :::

lel iminf

fnd/-L.

Proof

Since

fn - g)

:::

0,

the sequence

Un - g}

satisfies the assumptions

of

Fatou s lemma. Thus

liminf

e

fnd/-L :::

e

gd/-L

e

liminf

fn

- g)d/-L.

o

Proposition 10.2. Let Un} be a sequence of nonnegative, measurable functions

on E Then

Proof The sequence {L:7= fi} is a monotone sequence

of

nonnegative, measur

able functions.

0

Remark 10.1.

t s not required that the fn be integrable or that

L:

fn be integrable.

The integral ofmeasurable, nonnegative functions, finite or infinite, is well defined

by (7.1).

Theorem 10.3 dominated convergence).

Let Un} be a dominated

and

conver-

gent sequence

of

ntegrable functions in E, i.e.,

lim fn x)

= f x)

for all x E

E,

and there exists an integrable function g

:

E ~ such that

Ifni :::: g a.e. in E for all n EN.

Then the limit function f

:

E ~ is integrable and

lim

e

fnd/-L

=

e

lim

fnd/-L.

Proof

The limit function f is measurable, and by Fatou s lemma,


8/17

142

III. The Lebesgue Integral

Thus I is integrable. Next,

g-./;,)::::O

and

f,,+g)::::O

for all n EN.

Therefore, by Fatou s lemma,

L

df l

:::

lim inf

L/;,

lim sup

L

dfl :::

Ldfl

D

Absolute continuity o the integral

Theorem 11 1 (Vitali

9

). Let E be measurable, and let

I :

E

-- ]]{

be integrable.

For every

E

> O.

there exists

8

>

0

such that

lor

eve }' measurable subset [;

c

E

olmeasure less than

8.

Proof We may assume that I O For = 1. 2 consider the functions

1

-

{/ X)

1

n

Since {f,,} is increasing,

if I x) < n.

if

I (x) :::: n.

Having fixed E > 0, there exists some index n

E

such that

{ In[dfl > { fdf l ~ E

J J 2

Choose 8

=

- 2 .

Then for every measurable set [; C

E

of

measure less than

8,

1[

2 Product o measures

D

Let {X.

A.

fl}

and

{Yo

B.

v}

be two measure spaces. Any pair

of

sets A C X and

eY

generates a subset A x B of the Cartesian product X x Y called a

generalized rectangle.

YG.

Vitali. SuI Ie funzioni integrali. Alii Rend. Accad. Sci. Torino.

40

(1905). 1021-1034.


9/17

12 Product of measures 143

There are subsets o

X

x Y) that are not rectangles. The intersection o any two

rectangles is a rectangle by the formula

The mutual complement o any two rectangles, while not a rectangle, can be written

as the disjoint union o two rectangles by the decomposition

A2

x

B2)

AI

x BI) =

{ A2

AI) x

B2}

U

AIn 2)

x

B2

BI)}

.

Thus the collection

R

o

all rectangles

is

a semialgebra.

O

f A E

A

and B E B the rectangle A x B is called a measurable rectangle. The

collection o all measurable rectangles

is

denoted by Ro. By the previous remarks,

Ro

is a semialgebra.

Since X

x

Y E

Ro

such a collection forms a sequential covering o X

x

Y).11

The semialgebra Ro can be endowed with a nonnegative set function by setting

A A x

B)

=

fL A)v B)

12.1 )

for all measurable rectangles A x B.

Proposition 12 1

Let

{ n

X

Bn}

be a countable collection

of

disjoint measurable

rectangles whose union is a measurable rectangle A x B. Then

Proof

For each

x

E

A

B

=

U{B j l x ,

Y

E A j

x

Bi;

Y

E B}.

Since, for each

x

E A fixed, this is a disjoint union,

Integrating in dfL over A and using Proposition 10.2 now gives

o

Thus A is unambiguously defined since the measure o a measurable rectangle

does not depend on its partitions into countably many pairwise-disjoint measurable

rectangles.

Proposition 12 1 also implies that A is a measure on the semia1gebra

Ro.

There

fore, Acan be extended to a complete measure fL x v) on X x y), which coincides

with A on

Ro.12

IOSec

Section 9 o Chapter 11

II See Section 4 o Chapter II.

12See Theorem 1

J . J o

Chapter II.


10/17

44

Ill. The Lebesgue Integral

Theorem 12 2 Every pair {X, A, Id and {f, E, v} ofmeasure spaces generates

a complete product measure space

{ X

x

Y),

A x

3 ,

IL x

v)},

where (A x 3 is a

5

-algebra containing Ro and

IL

x v is a measure on (A x 3

that coincides with 12.1) on measurable rectangles.

3 On the structure

o

A x B)

Denote by A x

3)0

the smallest 5-algebra generated by the collection of all

measurable rectangles. Also set

Ra =

{countable unions of elements of

R

o

},

RaJ

= {countable intersections of elements of Ra

}.

By construction,

For each

E c X

x y , the two sets

E,

=

{vl x,y)

E

E)

fora

fixed

x

EX

X

E, =

xl x,y) E

E) forafixedy

E

Y

are, respectively, the X -section and the Y -section of

E.

Proposition 13 1

Let E

E (A x 3)0

Then

for

every y

E

Y, the Y -section E

y

is

in A, and for every x

E

X, the X -section E, is in E.

Proof The collection F of all sets E E

(A

x E such that Ex E E for all x E X

is a 5-algebra. Since F contains all the measurable rectangles, it must contain the

smallest 5-algebra generated by the measurable rectangles. D

Remark 13 1 The converse is false as there exist nonmeasurable sets E

c

X x

Y) such that all the

x

and v sections are measurable. An example is in Section 13.5

of the Problems and Complements.

Remark 13 2

There exist

IL

x v)-measurable rectangles

A

x

B

that are not

measurable rectangles. To construct an example, let Ao

C

X not be IL-measurable

but be included into a measurable set of finite IL-measure. Also let Bo

E

E be of

zero v-measure. The rectangle Ao x

Bo

is IL x v)-measurable and has measure

zero.

For

each c

>

0, there exists a measurable rectangle

Rc

containing Ao x

Bo

and

of

measure less than c. Therefore, Ao x

Bn

is

IL

x v)-measurable by the

criterion of measurability of Proposition 10.2 of Chapter II.

Remark 13 3

This example implies that Proposition 13.1 does not hold

if A

x

3)0 is replaced by

A x

3 . In particular, the inclusion

A x

3 0

c

A

x

3

is strict.


11/17

13. On the structure of A x B 145

Proposition 13 2

Let E E

Ra/i

be

of

finite measure. Then the function x ---?

v

Ex)

is J-t-measurable

and

the function

y

---?

J-t(E

y) is v-measurable. Moreover,

]

XEd(J-t

x v)

= [

v Ex)dJ-t

= [

J-t Ey)dv.

Xxy )

}x

}y

Proof

The statement

is

obvious

if

E is a measurable rectangle.

I f

E

E R

a

, it

can

be decomposed into the countable union

of

disjoint measurable rectangles En. The

functions

and

are measurable, and by monotone convergence,

]

XEd(J-t x

v)

] XE d(J-t x

v)

Xxy )

Xxy)

L x v En,x)dJ-t

=

L i -t En,y)dv

x

Lv E n,x)dJ-t

=

i

LJ-t En,y)dv

x v E

x

)dJ-t

i

-t Ey)dv.

I f E E

Ra/i,

there exists a countable collection {En} of elements

of Ra

such that

En+l

C

En and

E = n

En. The functions

x

---?

v Ex)

lim

v En,x)

and

Y

---?

J-t(E),)

=

limJ-t En.y),

are measurable. Since

J-t x v) E) lE be measurable and nonnegative. Then the measurability

statements

in 14.1)

and the double-integral formula

14.2)

hold. The integrals

in 14.2)

could be either finite or infinite.

Proof

The integrability requirement in the Fubini theorem was used

to

insure the

existence

of

a sequence

U; }

of

integrable functions each vanishing outside a set

of

finite measure and converging to f. The positivity

of f

and the a-finiteness in

the Tonelli theorem provide similar information. 0

If

f

is integrable in ~ x v . then Fubini s theorem holds and equality occurs

in 14.2). f

f

is not integrable, then the left-hand side of

04.2)

is infinite. Tonelli s

theorem asserts that in such a case, the right-hand side also is well defined and is

infinity, provided x \I) is

a

-finite.

In particular. Tonelli s theorem could be used to establish whether a nonnegative.

measurable function

f :

(X x

Y)

c>

JR

is integrable through the equality

of

the

two right-hand sides of

4.2).

The requirement that x \I) be a-finite cannot be removed. as shown by the

example in 14.5 of the Problems and Complements.

5 Some applications o the Fubini Tonelli

theorem

15.1 Integrals in

terms

of distribution functions. Let

f

: E

c>

JR be mea

surable and nonnegative. The distribution function of f relative to E is defined

as

Jl{+ 0) t ----+ ~ [ f > tl .

15.1)

This is a nonincreasing function of

t

and if f is finite a.e. in E, then

lim

~ [ f > t]

=

0

unless

~ [ f > t] 00.

1- 00

15.2)

f

f

is integrable, such a limit can be given a quantitative

fom1.

Indeed,

15.3)

15L. Tonelli. Sull integrazione per parti, Atti Accad. No;. Lincei (5). 18-2 1909).246-253.


15/17

15.

Some applications of the Fubini-Tonelli theorem 149

Proposition 15 1 Let

{X, A, I.t} be a-finite and let f : E

-- jR*

be measurable

and nonnegative. Also let v be a a-finite measure on

jR+

such that v [O, t)) =

v [O,

tl). Then

l

v [O,

f])dJJ- =

fooo

JJ- Lf

>

t])dv.

15.4)

In

particular, ifv [O,

t]) =

t

P

for some p

> 0,

then

l5.4)p

where

dt

is the Lebesgue measure on jR+.

Proof

The function

f

: E -- jR*, when regarded as a function from E x jR+ into

jR* , is measurable in the product measure JJ- xv). Likewise, the function g t) =

t

from

jR+

into jR*, when regarded as a function from

E

x

jR

into jR*, is measurable

in the product measure JJ- x v). Therefore, the difference

f

t is measurable in

the product measure JJ- x v). This implies that the set

Lf - t

> 0] =

Lf

>

t]

is measurable in the product measure JJ- x v). Therefore, by the Tonelli theorem,

foOO

JJ Cff

> tl)dv =

lC O

l

X[f>tjdJJ-) dv

=

l

foOO Xlf>t]dV)

dJJ

=

l

fofdV) dJJ

=

l

v [O,

f])dJJ-.

o

Both sides of 15.4) could be infinity and the formula could be used to verify

whether v [O, f]) is t-t-integrable over

E.

In the next two applications in Sections 15.2 and 15.3, the measure space

{X A,

JJ-}

is jRN with the Lebesgue measure.

15 2 Convolution integrals A measurable function

f

from

jRN

into jR*, when

regarded as a function defined on

jR2N,

is measurable. Indeed, for every c

E R

the set

Lf

> c]

X

jRN is a measurable rectangle in the product space. This implies

that the function

jR2N

:3

x,

y)

----+ f x -

y)

is

also measurable with respect to the product measure. Indeed, the set

{ x, y)

E jR2N

x y)

> c)

coincides with the measurable rectangle

Lf

> c]

X

jRN in the rotated coordinates

=

x -

y.


16/17

ISO

III.

The Lebesgue Integral

Given any two nonnegative measurable functions f, g :

m;,N

--+ m;, , their

con-

volution

is defined

as

x

- f g) x)

= f .

g y) f x

-

y)dy.

llR

15.5)

Since

f

and

g

are both nonnegative, the right-hand side, finite or infinite, is well

defined for all x E

m;

N .

Proposition 15 2 Let f and g

be nonnegative

and integrable in m;,N. Then

f

g) x) isfinitefora.e.

x

E m;,N,

thefunction

f g

is integrable in

m;,N

and

f

f

g)dx

::s

f

f

dX

)

f

gdX) .

lR I

llR.v

IIA

v

15.6)

Proof The function

(x.

Y --+ K(y) f (x - y) is nonnegative and measurable with

respect to the product measure. Therefore, by the Tonelli theorem,

j

f,. g :v)f x

-

y)dxdy

=

f .

f , g y) f x -

Y)dX) dy

llR-\

llR 1;R\

= lv f

dX

) lv gdX) . 0

The convolution of any two integrable functions

f

and g is defined as in 15.5).

Since

Ig y)f x -

y)1

::s

Ig y)ll f x

-

y l.

the convolution f g) is well defined as an integrable function over

m;,2N.

15 3 The Marcinkiewicz integral

Let E be a nonvoid set in

m;

N and let 0

(x)

denote the distance from x to E; i.e ..

OE X)

= inf Ix

-

zl.

~

By definition, 0E

(x)

=

or all x E E.

Lemma 15 3

Let

E be a nonemp v

set

in m;,N. Then

the distance

function x

--+

o x) is

Lipschitz continuous

with

Lipschitz

constant 1.

P r o ~ f Fix x and

y in m;,N

and assume that

(5 E

x) ::::

(5E

v). From the definition

of

o (Y), having fixed E

>

0, there exists z E E such that

Then estimate

OEC.V) :::: Iy - z l - E.

o :s

0E X)

- OEc V) ::s inf Ix - zl -

Iy

- z l

+

E

~

::s Ix

- z l - Iy - z l

+ E ::s Ix

- ) 1

+ E.

The conclusion follows since E

>

0 is arbitrary.

15.7)

o


17/17

16. Signed measures and the Hahn decomposition 151

Let

E

be a bounded, closed set in ][tN. Fix a positive number A and a cube Q

containing

E.

The Marcinkiewicz integral relative to

E

and A is the function

1

8A

v)

][tN

3 X ME.A X)

=

E .

NH dy.

Q x

- yl

15.8)

The right-hand side is well defined as the integral

of

a measurable, nonnegative

function.

Proposition 15 4 Marcinkiewicz

I6

). The Marcinkiewicz integral ME.A X) isfi-

nitefora.e.

x E

E

Moreover, thefunction x

-+

ME.A X) is integrable

in E

and

where

WN

is the measure

of

the unit sphere

in ][tN.

Proof Since 8E Y) = 0 for all y E E, by the Tonelli theorem,

1

E.A X)dx

=

1 8 ~ ( Y ) 1 dx N+A) dy

E Q E x

- yl

1

A

1 dx

)

DE Y) N dv.

Q E t. x - yl

A

.

For each fixed y E Q - E) and x E E, since E is closed, we estimate

Therefore, for each fixed

y

E Q - E ,

1

dx

[Emmanuele DiBenedetto (Auth.)] Real Analysis(BookZa.org) Pages155,156,157,Etc

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