Measure and Integration: A Crash Course€¦ · Measure and Integration: A Crash Course M.T.Nair...
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Measure and Integration: A Crash Course * M.T.Nair Department of Mathematics, IIT Madras May 25, 2012 Abstract In these notes we intend to give a crash course on measure and integration motivating the essential concepts of 1. Lebesgue measure on R, 2. Lebesgue measurable sets , 3. σ-algebra M of measurable subsets of R, 4. Lebesgue measure on measurable subsets of R, 5. measure μ on a general σ-algebra A on a none-empty set Ω, 6. measurable function on a general measurable space (X, A), 7. integral of positive, real and complex valued measurable functions on a measure space (X, A,μ), 8. L p -spaces and Sobolev spaces. * Lectures at IIT Delhi under NPDE-TCA, May 17-19, 2012. 1
Transcript of Measure and Integration: A Crash Course€¦ · Measure and Integration: A Crash Course M.T.Nair...
Measure and Integration: A Crash Course∗
M.T.Nair
Department of Mathematics, IIT Madras
May 25, 2012
Abstract
In these notes we intend to give a crash course on measure and integration motivating
the essential concepts of
1. Lebesgue measure on R,
2. Lebesgue measurable sets ,
3. σ-algebra M of measurable subsets of R,
4. Lebesgue measure on measurable subsets of R,
5. measure µ on a general σ-algebra A on a none-empty set Ω,
6. measurable function on a general measurable space (X,A),
7. integral of positive, real and complex valued measurable functions on a measure
space (X,A, µ),
8. Lp-spaces and Sobolev spaces.
∗Lectures at IIT Delhi under NPDE-TCA, May 17-19, 2012.
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1 Lecture 1: Lebesgue measure
1.1 The need for the concept of a measure
We recall the following:
• Every continuous function f : [a, b]→ R is Riemann integrable.
• Every bounded function f : [a, b]→ R which is discontinuous only at a finite number
of points in [a, b] is Riemann integrable.
• Every function f : [a, b]→ R which is either monotonically increasing or monotoni-
cally decreasing is Riemann integrable.
Further, it is also known that
• Every bounded function f : [a, b] → R which is discontinuous only on a set of
measure zero is Riemann integrable.
In the last statement, by a set of measure zero we mean a set E ⊆ R such that for every
ε > 0, there exists a countable collection In of open intervals such that
E ⊆⋃n
In and∑n
`(In) < ε.
However, the concept of Riemann integral has certain draw backs:
• If (fn) is a sequence of Riemann integrable functions on [a, b] such that fn(x)→ f(x)
for every x ∈ [a, b] for some bounded function f , then f need not be Riemann
integrable.
Exercise 1.1. Let
fn(x) =
1 if x ∈ r1, . . . , rn0 if x 6∈ r1, . . . , rn,
Show that each fn is Riemann integrable, and (fn(x)) converges for every x ∈ [0, 1], but
the limit function is not Riemann integrable. ♦
We would like to extend the concept of integrability to a larger class of functions so
that the draw back of Riemann integral as in Example 1.1, does not occur. This is done
by using the concept of a measure of a subset of R which is a generalization of the concept
of length of an interval.
1.2 Lebesgue outer measure on R
Definition 1.2. For E ⊆ R, the Lebesgue outer measure of E is defined as
m∗(E) := infI
∑n
`(In),
where the infimum is taken over the set I of all countable collection In of open intervals
such that E ⊆ ∪nIn. ♦
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Note that 0 ≤ m∗(E) ≤ ∞ for every E ⊆ R. Thus, m∗ can be thought of as a function
from the family of all subsets of R into the set [0,∞].
Exercise 1.3. Let E ⊆ R. Prove that for every ε > 0 there exists a countable family
In of open intervals such that E ⊆∞⋃n
In and∑n
`(In) ≤ m∗(E) + ε. Strict inequality
occurs in the above if m∗(E) <∞. ♦
Exercise 1.4. Prove the following:
(i) m∗(∅) = 0.
(ii) If E is a singleton set, then m∗(E) = 0.
(iii) If E is a countable set, then m∗(E) = 0.
(iv) If E = (a, b) with a < b then m∗(E) ≤ b− a. ♦
Exercise 1.5. Prove the following:
(i) E1 ⊆ E2 ⇒ m∗(E1) ≤ m∗(E2).
(ii) m∗(A ∪B) ≤ m∗(A) +m∗(B).
(iii) If E ⊆ A and m∗(E) = 0, then m∗(A \ E) = m∗(A).
(iv) If E ⊆ R and x ∈ R, then m∗(E + x) = m∗(E). ♦
Remark 1.6. The property (i) in Exercise 1.5 is called the monotonicity property of m∗,
and the property (iv) is called the translation invariance of m∗. ♦
Exercise 1.7. Prove that for every sequence (In) of open intervals,
m∗(⋃In) ≤
∑n
`(In). ♦
Exercise 1.8. If E ⊆ R, then for every ε > 0, there exists an open set G ⊆ R such that
G ⊇ E and m∗(G) ≤ m∗(E) + ε. ♦
THEOREM 1.9. (Sub-additivity) Suppose An ⊆ R for n ∈ N. Then
m∗
( ∞⋃n=1
An
)≤∞∑n=1
m∗(An).
Proof. Use Exercise 1.3.
Exercise 1.10. Derive Exercise 1.4(iii) from Exercise 1.4(ii) by using Theorem 1.9 ♦
THEOREM 1.11. The Lebesgue outer measure of any interval is its length.
Proof. Let I be an interval.
Case 1: Suppose I = [a, b] with −∞ < a < b <∞, and let ε > 0. Let Iε := (a− ε, b+ ε).
Clearly,
m∗(I) ≤ m∗(Iε) ≤ b− a+ 2ε.
This is true for all ε > 0. Hence, m∗(I) ≤ b− a.
Also, there exists a countable collection In of open intervals of finite length such
that I ⊆ ∪nIn and∑n `(In) ≤ m∗(I) + ε. By compactness of I, there exists a finite
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sub-collection In1, , . . . , Ink
⊆ In such that I ⊆ ∪ki=1Ini. Let Ini
:= (ai, bi) for
i ∈ 1, . . . , k. Without loss of generality, assume that Ji ∩ [a, b] 6= ∅ and ai+1 < bi for
i ∈ 1, . . . , k − 1. Then
k∑i=1
`(Ini) =
k∑i=1
(bi − ai)
= bk +
k−1∑i=1
(bi − ai+1)− a1 ≥ bk − a1 ≥ b− a.
Thus,
b− a ≤k∑i=1
`(Ini) ≤∑i
`(Ii) ≤ m∗(I) + ε.
This is true for all ε > 0. Hence, b−a ≤ m∗(I). This completes the proof of m∗(I) = b−a.
Case 2: Suppose I is an interval of finite length with end points a and b with a < b.
Then for sufficiently small ε > 0, we have [a+ ε, b− ε] ⊆ I ⊆ [a− ε, b+ ε] so that
b− a− 2ε = m∗([a+ ε, b− ε]) ≤ m∗(I)
≤ m∗([a− ε, b+ ε]) = b− a+ 2ε.
Since this is true for every ε > 0, it follows that m∗(I) = b− a.
Case 3: Suppose I is of infinite length. Then for every M > 0 there exists a closed interval
IM of length M such that IM ⊆ I. Hence M = m∗(IM ) ≤ m∗(I). Thus, M ≤ m∗(I) for
all M > 0 so that m∗(I) =∞.
COROLLARY 1.12. Every non-degenerate interval is an uncountable set.
1.3 Lebesgue Measurable Sets
Suppose A1 and A2 are disjoint subsets of R. We may expect that
m∗(A1 ∪A2) = m∗(A1) +m∗(A2). (1.1)
Is it true for any two disjoint sets A1 and A2?
Exercise 1.13. Prove that if (1.1) holds for any two disjoint sets A1 and A2, then
m∗( ∞⋃n=1
An
)=
∞∑n=1
m∗(An) (1.2)
holds for any pairwise disjoint denumerable family An∞n=1. ♦
Exercise 1.14. Prove that (1.2) need not hold for every denumerable disjoint family
An∞n=1.
Hint: On [0, 1] consider the relation x ∼ y ⇐⇒ x− y ∈ Q.• Show that ∼ is an equivalence relation on [0, 1].
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Let E be the subset of [0, 1] such that its intersection with each equivalence class is a
singleton set. (Such a set E exists by using the axiom of choice.) Let r1, r2, . . . =
Q ∩ [−1, 1] and En := E + rn for n ∈ N.
• Show that En is a pairwise disjoint family.
Observe: [0, 1] ⊆⋃∞n=1En ⊆ [−1, 2] so that 1 ≤ m∗
(⋃∞n=1En
)≤ 3. Therefore, if (1.2)
is true, we arrive at a contradiction. ♦
Exercise 1.15. There exist disjoint subsets A1 and A2 of R such that
m∗(A1 ∪A2) 6= m∗(A1) +m∗(A2).
Why? ♦
The above discussion motivates us to consider a family of sets in which the relations
(1.2) holds for all possible pairwise disjoint denumerable family of sets involved.
Definition 1.16. A set E ⊆ R is said to be Lebesgue measurable if for every A ⊆ R,
m∗(A) = m∗(A ∩ E) +m∗(A ∩ Ec).
We denote the set of all Lebesgue measurable sets by M. ♦
Exercise 1.17. Prove the following:
(i) E ∈M ⇐⇒ m∗(A) ≥ m∗(A ∩ E) +m∗(A ∩ Ec) ∀A ⊆ R.
(ii) ∅ ∈M.
(iii) E ∈M ⇒ Ec ∈M.
(iv) m∗(E) = 0 ⇒ E ∈M. ♦
THEOREM 1.18. Suppose A1 ∩A2 = ∅. If one of A1, A2 belongs to M, then
m∗(A1 ∪A2) = m∗(A1) +m∗(A2).
Proof. Suppose A1 ∩A2 = ∅. Assume A1 ∈M. Then
m∗(A1 ∪A2) = m∗((A1 ∪A2) ∩A1) +m∗((A1 ∪A2) ∩Ac1).
But, (A1 ∪A2) ∩A1) = A1 and (A1 ∪A2) ∩Ac1) = A2. Hence the result
Exercise 1.19. Prove that if A1, . . . , An is a disjoint family in M, then
m∗( n⋃i=1
Ai
)=
n∑i=1
m∗(Ai).
[Hint: Use Theorem 1.18.] ♦
THEOREM 1.20. If An∞n=1 is a disjoint family in M, then
m∗( ∞⋃n=1
An
)=
∞∑n=1
m∗(An).
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Proof. Let An∞n=1 be a disjoint family in M. Then, we know that by Theorem 1.18,
that
m∗(A1 ∪A2) = m∗(A1) +m∗(A2),
and hence,
m∗( n⋃i=1
Ai
)=
n∑i=1
m∗(Ai) ∀n ∈ N.
Thus, for every n ∈ N,
n∑i=1
m∗(Ai) = m∗( n⋃i=1
Ai
)≤ m∗
( ∞⋃i=1
Ai
)≤∞∑i=1
m∗(Ai)
so thatn∑i=1
m∗(Ai) ≤ m∗( ∞⋃i=1
Ai
)≤∞∑i=1
m∗(Ai).
This is true for every n ∈ N. Hence,
∞∑i=1
m∗(Ai) ≤ m∗( ∞⋃i=1
Ai
)≤∞∑i=1
m∗(Ai).
More importantly, we have the following theorem, whose proof we omit.
THEOREM 1.21. If En ∈M for n ∈ N, then⋃nEn ∈M.
Definition 1.22. 1. The outer measure m∗ restricted to M is called the Lebesgue
measure on R, and it is denoted by m.
2. For E ∈M, m(E) := m∗(E) is called the Lebesgue measure of E.
♦
Now the question is: How large the class M is?
THEOREM 1.23. For any a ∈ R, (a,∞) ∈M.
Proof. For a ∈ R, let E = (a,∞). Let A ⊆ R and ε > 0. By Proposition ??, there exists
a sequence (In) of open intervals such that
A ⊆⋃n
In,∑n
`(In) < m∗(A) + ε.
Let A1 = A ∩ (a,∞) and A2 = A ∩ (−∞, a]. Clearly, A is the disjoint union of A1 and
A2. If we write I ′n = In ∩ (a,∞) and I ′′n = In ∩ (−∞, a], then we have In is the disjoint
union of I ′n and I ′′n , so that `(In) = `(I ′n)+ `(I ′′n). Also, A1 ⊆⋃n I′n, A2 ⊆
⋃n I′′n . Hence,
m∗(A1) +m∗(A2) ≤∑n
`(I ′n) +∑n
`(I ′′n)
=∑n
`(In) ≤ m∗(A) + ε.
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This is true for any ε > 0, so that we get
m∗(A1) +m∗(A2) ≤ m∗(A),
which completes the proof.
Using the Theorem 1.23 and the results which we have already listed, the following
can be deduced.
• For any a ∈ R, [a,∞) ∈M.
• For any a ∈ R, (−∞, a) ∈M.
• For any a ∈ R, (−∞, a] ∈M.
• For any a, b ∈ R with a < b, (a, b) ∈M.
• For any a, b ∈ R with a < b, [a, b] ∈M.
• Every open subset of R belongs to M.
• Every closed subset of R belongs to M.
• Every Gδ subset of R belongs to M.
• Every Fσ subset of R belongs to M.
Is there any subset of R which does not belong to M?
Exercise 1.24. There exists E ⊆ R such that E 6∈ M - Why?. ♦
Exercise 1.25. If E ∈ M with m(E) <∞ then show that for every ε > 0, there exists
an open set G ⊆ R such that G ⊇ E and m(GE) < ε. ♦
Analogues to the definition of outer measure and Lebesgue measure on R, we can
define outer measure and Lebesgue measure on Rk for any k ∈ N.
Definition 1.26. 1. For E ⊆ Rk, the Lebesgue outer measure of E is defined as
m∗(E) := infD
∑n
vol(Dn),
where the infimum is taken over the set D of all countable collection Dn of open
parallelopipeds (or open rectangles if k = 2) such that E ⊆ ∪nDn, and vol(Dn)
denotes the volume (area if k = 2) of Dn .
2. A set E ⊆ Rk is said to be Lebesgue measurable if
m∗(A) = m∗(A ∩ E) +m∗(A ∩ Ec) ∀A ⊆ Rk.
♦
We denote the set of all Lebesgue measurable sets by Mk.
It can be shown that Mk has properties analogous to M.
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2 Lecture 2: Measure and measurable functions
2.1 Measure on an Arbitrary σ-Algebra
Recall:
• R ∈M,
• E ∈M implies Ec ∈M,
• En ∈M for n ∈ N implies⋃nEn ∈M, and
• En is a countable disjoint family in M implies
m(⋃
n
En
)=∑n
m(En).
Definition 2.1. Let X be a set and A be a family of subsets of X. Then A is called a
σ-algebra if
(a) X ∈ A,
(b) A ∈ A implies Ac ∈ A,
(c) An ∈ A for n ∈ N implies⋃nAn ∈M.
The pair (X,A) is called a measurable space, and members of a σ-algebra are called
measurable sets.
♦
Definition 2.2. Let (X,A) be a measurable space. A function µ : A → [0,∞] is called
a positive measure or simply a measure if
(a) µ(∅) <∞, and
(b) for any countable disjoint family An in A,
µ(⋃
n
An
)=∑n
µ(An).
The pair (X,A, µ) is called a measure space. For A ∈ A, µ(A) is called the measure
of A. ♦
According to the above definitions, (R,M) is a measurable space and (R,M,m) is a
measure space.
Let (X,A, µ) be a measure space. Then the following results can be easily verified:
• For A,B ∈ A, if A ⊆ B then µ(A) ≤ µ(B).
• For A,B ∈ A, if A ⊆ B and µ(B) <∞ then
µ(B \A) = µ(B)− µ(A).
Example 2.3. Let X be any set, A = ∅, X and let µ be defined by µ(∅) = 0 and
µ(X) = 1. ♦
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Example 2.4. Let X be any set, A be the power set of X, and µ be defined by
µ(E) =
#(E) if E is a finite set
∞ if E is an infinite set.
Here, #(E) denotes the cardinality of E. The above measure is called the counting
measure on X. ♦
Example 2.5. Let X be any set, A be the power set of X, x0 ∈ X, and µ be defined by
µ(E) =
1 if x0 ∈ E0 if x0 6∈ E.
This measure is called the Dirac measure on X centered at x0.
[Think of a thin wire of negligible weight. Suppose a bead of weight 1 unit kept at some
point x0 on the wire. Then weight of a part, say E, of the wire is 1 if x0 belongs to E,
and the wight is zero if x0 does not belong to E.] ♦
More generally we have the following.
Example 2.6. Let X be a set, A be the power set of X, xi ∈ X and wi ≥ 0 for
i ∈ 1, . . . , k. For E ⊆ X, let ∆E = i : xi ∈ E, and let µ be defined by
µ(E) =∑i∈∆E
wi.
[Think of a thin wire of negligible weight. Suppose a finite number of beads of weights
w1, . . . , wk are kept at points x1, . . . , xk respectively on the wire. Then the weight of a
part, say E, of the wire is∑i∈∆E
wi where ∆E = i : xi ∈ E.] ♦
THEOREM 2.7. Suppose Aα : α ∈ Λ is a family of σ-algebras on a set X. Then⋂α∈ΛAα is also a σ-algebra.
Proof. Exercise.
Definition 2.8. Let X be a set and S be a family of subsets of X. Then the intersection
of all σ-algebras containing S is called the σ-algebra generated by S. ♦
Definition 2.9. Let X be a topological space with topology T . Then the σ-algebra
generated by T is called the Borel σ-algebra or Borel field on X.
♦
Definition 2.10. The Borel σ-algebra on Rk is denoted by Bk. ♦
Since the σ-algebra Mk contains all open sets, it follows that Bk ⊆Mk.
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THEOREM 2.11. Suppose (X,A) is a measurable space and X0 ∈ A. Then
A0 := E ⊆ X0 : E ∈ A
is a σ-algebra on X0, and
A0 = A ∩X0 : A ∈ A.
Proof. Exercise
• The σ-algebra A0 in Theorem 2.11 is called the restriction of A to X0.
• Members of A0 are called measurable subsets of A0.
• If µ is a measure on (X,A), then µ0 := µ|A0is a measure on the measurable space
(X0,A0).
Definition 2.12. A measure µ on a measurable space (X,A) is called a complete
measure if for every A ∈ A with µ(A) = 0, we have E ∈ A for every E ⊆ A. ♦
• Lebesgue measure on M is complete.
• Lebesgue measure on the Borel σ-algebra is not complete.
[Its proof requires the concept of Lebesgue measurable functions which we shall
introduce shortly.
• If #(X) ≥ 2, and A = X,∅, and µ defined by µ(X) = 1, µ(∅) = 0, then µ is not
complete.
THEOREM 2.13. Suppose Ai ∈ A such that
An ⊆ An+1 ∀n ∈ N.
Then
µ(An)→ µ
( ∞⋃i=1
Ai
)as n→∞.
Proof. We write⋃∞i=1Ai as a disjoint union
⋃∞i=1Ei by taking
E1 = A1, Ei = Ai \Ai−1, i = 2, 3, . . . .
Then
µ( ∞⋃i=1
Ai
)= µ
( ∞⋃i=1
Ei
)=
∞∑i=1
µ(Ei)
= limn→∞
n∑i=1
µ(Ei) = limn→∞
µ( n⋃i=1
Ei
).
But,⋃ni=1Ei = An. Hence,
µ( ∞⋃i=1
Ai
)= limn→∞
µ(An).
This completes the proof.
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THEOREM 2.14. Suppose An ∈ A such that
An ⊇ An+1 ∀n ∈ N and µ(A1) <∞.
Then
µ(An)→ µ
( ∞⋂i=1
Ai
)as n→∞.
Proof. We Theorem 2.13: Write Bn = A1 \ An, n ∈ N. Then Bn ⊆ Bn+1 for all n ∈ N,
so that by the above theorem,
µ( ∞⋃i=1
Bi
)= limn→∞
µ(Bn).
But,∞⋃i=1
Bi = A1 \∞⋂i=1
Ai, Bn = A1 \An.
Therefore, since µ(A1) <∞, we have
µ( ∞⋃i=1
Bi
)= µ
(A1 \
∞⋂i=1
Ai
)= µ(A1)− µ
( ∞⋂i=1
Ai
),
µ(Bn) = µ(A1 \An) = µ(A1)− µ(An),
from which we obtain the required relations.
Measurable Functions
Definition 2.15. Let (X,A) be a measurable space. Then a function f : X → R is said
to be a real measurable function if f−1(G) ∈ A for every open set G in R. ♦
Similarly we can define a complex measurable function if the codomain of f is Cinstead of R
Observe:
• If X = E ⊆ R and A is eitherME or BE , then every continuous function f : E → Ris measurable.
More generally:
Definition 2.16. Let (X,A) be a measurable space and (Y, T ) be a topological space.
Then a function f : X → Y is said to be a measurable with respect to the pair (A, T )
if f−1(G) ∈ A for every G ∈ T .
♦
Exercise 2.17. Let (X,A) be a measurable space. Show that a real valued function
f : X → R is measurable if and only if f−1(I) ∈ A for every open interval I. ♦
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Some of the topological spaces that we frequently use in the study of measure and
integration are the following:
• R with usual topology,
• C with usual topology,
• Rk with usual topology.
• R := [−∞,∞] with topology T consisting of unions of intervals of the form (a, b),
(a,∞], [−∞, b).
THEOREM 2.18. Let X be a measurable space, and Y and Z be topological spaces. If
f : X → Y is measurable and g : Y → Z is continuous, then g f : X → Z is measurable.
Proof. Obvious from the definition.
THEOREM 2.19. Let X be a measurable space and Y be a topological space. If u, v
are real measurable functions on X and g : R2 → Y is continuous, then the function
ϕ : X → Y defined by
ϕ(x) = g(u(x), v(x)), x ∈ X,
is a measurable function from X to Y .
Proof. Define f : X → R2 by f(x) = (u(x), v(x)), x ∈ X. Then ϕ = g f . Since g is
continuous, by the last theorem, it is enough to prove that f is a measurable function. For
this it is sufficient to show that for open intervals I1 and I2, f−1(I1 × I2) is measurable.
Note that
x ∈ f−1(I1 × I2) ⇐⇒ f(x) ∈ I1 × I2
⇐⇒ u(x) ∈ I1 & v(x) ∈ I2 ⇐⇒ x ∈ u−1(I1) ∩ v−1(I2).
Thus f−1(I1 × I2) = u−1(I1) ∩ v−1(I2) which is measurable.
The following two theorems can be deduced from the above theorems.
THEOREM 2.20. Let X be a measurable space.
(i) If u, v are real measurable functions on X, then u + iv is a complex measurable
function on X.
(ii) If f is a complex measurable function on X, then Ref , Imf and |f | are real
measurable functions on X.
THEOREM 2.21. Let X be a measurable space. If f and g are real (resp. complex)
measurable functions on X, then f ± g and fg are real (resp. complex) measurable
functions on X.
THEOREM 2.22. Let X be a measurable space.
(i) For E ⊆ X, χE
is measurable if and only if E is measurable.
(ii) If f : X → C is measurable, then there exists a measurable function α : X → Csuch that |α| = 1 and f = α|f |.
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Proof. (i) Let E ⊆ X. For an open set G,
χ−1E
(G) =
E, 1 ∈ G, 0 6∈ G,Ec, 1 6∈ G, 0 ∈ G,X, 1 ∈ G, 0 ∈ G,∅, 1 6∈ G, 0 6∈ G.
Thus, χE
is measurable if and only if E is measurable.
(ii) Let
α(x) :=
f(x)|f(x)| , f(x) 6= 0,
1, f(x) = 0.
That is, if E = x : f(x) = 0, then
α(x) =f(x) + χ
E(x)
|f(x) + χE
(x)|= Φ(f(x) + χ
E(x), x ∈ X,
where Φ(z) := z/|z| for z ∈ C \ 0. Since Φ is continuous and f and χE
are measurable,
α is measurable. Also, f(x) = α(x)|f(x)| and |α(x)| = 1 for all x ∈ X.
In the following (X,A) is a measurable space.
THEOREM 2.23. Let E ⊆ X. Then E ∈ A if and only if χE is a measurable function.
THEOREM 2.24. Let (Y, T ) be a topological space.
(i) S := E ⊆ Y : f−1(E) ∈ A is a σ-algebra on Y .
(i) If f is measurable, then S contains the Borel field on Y .
As a corollary to the above theorem, we have the following.
THEOREM 2.25. Let (Y, T ) be a topological space. Then f : X → Y is measurable if
and only if for every Borel set A in Y , f−1(A) is measurable in X.
Remark 2.26. Suppose (X1,A1) and (X2,A2) are two measurable spaces. Then we may
define a function f : X1 → X2 to be measurable if f−1(A) ∈ A1 for every A ∈ A2. Then,
in view of the above theorem, our definition becomes a particular case by taking X2 a
topological space and A2 the Borel filed on X2. ♦
The following theorem can be verified easily.
THEOREM 2.27. For a function f : X → [−∞,∞], the following are equivalent:
(i) f is measurable.
(ii) x ∈ X : f(x) > a ∈ A ∀a ∈ R.
(iii) x ∈ X : f(x) ≥ a ∈ A ∀a ∈ R.
(iv) x ∈ X : f(x) < a ∈ A ∀a ∈ R.
(v) x ∈ X : f(x) ≤ a ∈ A ∀a ∈ R.
13
THEOREM 2.28. Let fn : X → [−∞,∞] be measurable for each n ∈ N. Then
supnfn, inf
nfn, lim sup
nfn, lim inf
nfn
are measurable functions.
Proof. For α ∈ R, we note that
x ∈ X : supnfn > α =
⋃n
x ∈ X : fn > α,
x ∈ X : infnfn < α =
⋃n
x ∈ X : fn < α.
From this it follows that supn fn and infn fn are measurable functions. Since
lim supn
fn = infk
supn≥k
fn, lim infn
fn = supk
infn≥k
fn
it also follows that lim supn fn and lim infn fn are measurable functions.
COROLLARY 2.29. Let (fn) be a sequence of extended real valued (resp. complex
valued) measurable functions on X. The following hold:
(i) If fn(x)→ f(x) as n→∞ for each x ∈ X, then f is measurable.
(ii) If
∞∑n=1
fn(x) converges for each x ∈ X and f(x) =
∞∑n=1
fn(x), then f is measurable.
THEOREM 2.30. Let f : X → [−∞,∞]. If f is measurable, then f+ and f− are
measurable. In case f is real valued, then f is measurable if and only if f+ and f− are
measurable, and in that case |f | is also measurable.
Proof. Note that f+(x) := maxf(x), 0, f−(x) := −minf(x), 0, and |f | = f+ + f−.
Hence, the proof follows from Theorem 2.28.
Definition 2.31. A function ϕ : X → R is called a simple function if it takes only a
finite number of values.
Suppose ϕ : X → R is a simple function taking distinct values α1, . . . , αn. Then
ϕ =
n∑i=1
αiχEi, where Ei = x ∈ X : ϕ(x) = αi, i = 1, . . . , n. The above representation
of ϕ is called its canonical representation, and X = ∪ni=1Ei. ♦
It is obvious that ϕ : X → R is a simple function if and only if there exist a1, . . . , an
in R and Ai ⊆ X, i = 1, . . . , n, such that
ϕ =
n∑i=1
aiχAi.
14
Exercise 2.32. Let (X,A) be a measurable space and ϕ is a simple function on X with
canonical representation ϕ =
n∑i=1
αiχEi. Then ϕ is measurable if and only if Ei ∈ A for
i = 1, . . . , n – Justify. ♦
THEOREM 2.33. Let (X,A) be a measurable space and f : X → [0,∞] be a measurable
function. For each n ∈ N, let
Ei,n :=x ∈ X :
i− 1
2n≤ f(x) <
i
2n
, i = 1, . . . , n2n,
Fn := x ∈ X : f(x) ≥ n,
ϕn =
n2n∑i=1
i− 1
2nχ
Ei,n+ nχ
Fn.
Then each ϕn is measurable, 0 ≤ ϕn ≤ ϕn+1 for every n ∈ N and ϕn(x) → f(x) as
n→∞ for each x ∈ X.
Proof. Let x ∈ X. If f(x) = ∞, then ϕn(x) = n for all n ∈ N, so that ϕn(x) → f(x) as
n → ∞. In case f(x) ∈ R, then there exists k ∈ N such that f(x) ≤ k, so that x ∈ Ei,nfor all n > k and for some i ∈ 1, 2, . . . , n2n, and in that case |f(x) − ϕn(x)| ≤ 1/2n.
Hence, in this case also, ϕn(x)→ f(x) as n→∞.
Next suppose that x ∈ Ei,n for some n ∈ N and for some i ∈ 1, 2, . . . , n2n. Then
ϕn(x) = (i− 1)/2n, and
ϕn+1(x) ∈i− 1
2n,i− 1
2n+
1
2n+1
.
Thus, ϕn(x) ≤ ϕn+1(x). If x ∈ Fn, then ϕn(x) = n and ϕn+1(x) ∈ n, n+ 12n+1 . Thus,
we get ϕn(x) ≤ ϕn+1(x) for every x ∈ X and for every n ∈ N.
According to the above theorem, every extended real valued non-negative measurable
function f can be approximated pointwise by an increasing sequence (ϕn) of simple
measurable functions. Thus, if (X,A) is endowed with a measure µ, then a natural
procedure of defining the integral of a extended real valued non-negative measurable
function f would be to define the integral∫Xϕdµ of simple measurable functions ϕ, and
then define∫Xfdµ as the limit (if exists) of the sequence
(∫Xϕndµ
)whenever (ϕn) is
an increasing sequence of non-negative simple measurable functions which converges to
f pointwise. But, to have such an integral to be well-defined, we must also show the
following:
• If (ϕn) and (ψn) are increasing sequence of non-negative simple measurable functions
which converge to f pointwise, then the limits limn→∞
∫X
ϕndµ and limn→∞
∫X
ψndµ exist and
must be equal.
We shall do this in the next chapter.
15
3 Lecture 3: Integral of measurable functions
3.1 Integral of a Simple Measurable Function
Definition 3.1. Let (X,A, µ) be a measure space. Let ϕ be a non-negative simple
measurable function with the canonical representation ϕ :=
n∑i=1
αiχAi, i.e., α1, . . . , αn
are distinct values of ϕ and Ai := x ∈ X : ϕ(x) = αi for i ∈ 1, . . . , n. Then we define∫X
ϕdµ :=
n∑i=1
αiµ(Ai)
and it is called the integral of ϕ over X with respect to µ. ♦
Remark 3.2. In the above definition of the integral we assumed ϕ to be non-negative,
because otherwise in the sum
n∑i=1
αiµ(Ai) may not be well-defined since it can happen
that µ(Ai) = µj(Aj) =∞ for some i 6= j and αi < 0 < αj . ♦
It is easy to see that if ϕ =∑ni=1 αiχAi
is the canonical representation of a simple
measurable function ϕ, then for any E ∈ A, χEϕ =
∑ni=1 αiχAi
∩E is the canonical
representation of χEϕ so that∫
X
χEϕdµ =
n∑i=1
αiµ(Ai ∩ E).
Definition 3.3. If ϕ is a non-negative simple measure function on a measure space
(X,A, µ) and if E ∈ A, then we define∫E
ϕdµ :=
∫X
χEϕdµ,
and it called the integral of ϕ over E with respect to µ. ♦
We observe that
• A function ϕ : X → [0,∞) is simple measurable if and only if there exists α1, . . . , αn
in [0,∞) and measurable sets E1, . . . , En such that ϕ =
n∑i=1
αiχEi.
A natural question is whether
∫X
ϕdµ =
n∑i=1
αiµ(Ei). The answer is in affirmative:
THEOREM 3.4. Suppose B1, . . . , Bk are measurable sets, β1, . . . , βk are non-negative
real numbers and ψ :=∑ki=1 βiχBi
. Then
∫X
ψdµ =
k∑i=1
βiµ(Bi).
16
Proof. Case (i): B1, . . . , Bk are disjoint:
In case β1, . . . , βk are distinct, then the given representation of ψ is its canonical
representation and hence the proof follows from the definition.
Next assume that some of β1, . . . , βk are repeated. Suppose α1, . . . , αn are the distinct
numbers among β1, . . . , βk. For each i ∈ 1, . . . , n, let
∆i := j : βj = αi.
Then it is clear that
ψ =
k∑i=1
βiχBi=
n∑i=1
αi∑j∈∆i
χBj
=
n∑i=1
αiχAi,
where Ai =⋃j∈∆i
Bj . Hence,
∫X
ψdµ =
n∑i=1
αiµ(Ai) =
n∑i=1
αi∑j∈∆i
µ(Bj)
=
n∑i=1
∑j∈∆i
βjµ(Bj) =
k∑i=1
βiµ(Bi).
Case (ii): B1, . . . , Bk are not necessarily disjoint:
In this case first we may write⋃ki=1Bi as a disjoint union
⋃kj=1 Ci. Then
Bi = Bi ∩k⋃j=1
Cj =
k⋃j=1
(Bi ∩ Cj)
so that χBi
=∑kj=1 χBi∩Cj
and hence
ψ :=
k∑i=1
βi
k∑j=1
χBi∩Cj
=
k∑i=1
k∑j=1
βiχBi∩Cj.
This representation of ψ fits into the case (i) discussed above.
THEOREM 3.5. If ϕ and ψ are non-negative simple measurable functions and c ∈ R,
then ∫X
(ϕ+ ψ)dµ =
∫X
ϕdµ+
∫X
ψdµ,∫X
c ϕdµ = c
∫X
ϕdµ.
Proof. Suppose ϕ :=∑ni=1 αiχAi
and ψ :=∑mj=1 βjχBj
are the canonical representations
of ϕ and ψ respectively. Then ϕ+ ψ and c ϕ have the representations
ϕ+ ψ =
n∑i=1
αiχAi+
m∑j=1
βjχBj, c ϕ =
n∑i=1
c αiχAi
respectively. Hence, the proof follows by applying last theorem.
17
THEOREM 3.6. If ϕ and ψ are non-negative simple measurable functions such that
ϕ ≤ ψ, then ∫X
ϕdµ ≤∫X
ψdµ.
Proof. Since ψ = ϕ + (ψ − ϕ) and ψ − ϕ ≥ 0 and since∫X
(ψ − ϕ)dµ ≥ 0, by the last
theorem, ∫X
ψdµ =
∫X
ϕdµ+
∫X
(ψ − ϕ)dµ ≥∫X
ϕdµ.
This competes the proof.
THEOREM 3.7. Suppose ϕ : X → [0,∞) is a simple measurable function. For E ∈ A,
let
ν(E) :=
∫E
ϕdµ.
Then ν is a measure on X.
Proof. Clearly ν(∅) = 0. Suppose ϕ = χA for some A ∈ A. Then for E ∈ A,
ν(E) =
∫E
ϕdµ = µ(A ∩ E).
If En is a countable disjoint family in A, then
ν(∪∞n=1En) = µ(A ∩ (∪∞n=1En)) = µ(∪∞n=1(A ∩ En))
=
∞∑n=1
µ(A ∩ En) =
∞∑n=1
ν(En).
Since a simple function is a finite linear combination of characteristic functions, the result
follows by using Theorem 3.5.
Exercise 3.8. Suppose ϕ is a non-negative simple measurable function and E is a mea-
surable set. Let µE
be the restriction of µ to the restricted σ-algebra AE := A ⊆ E :
A ∈ A. Show that∫Eϕdµ =
∫Eϕdµ
E. ♦
3.2 Integral of Positive Measurable Functions
Let (X,A, µ) be a measure space and f : X → [0,∞] be a measurable function. We know
that there exists a sequence (ϕn) of non-negative simple measurable functions such that
• 0 ≤ ϕn ≤ ϕn+1 ≤ f ∀n ∈ N,
• ϕn → f pointwise.
Also we know that
• 0 ≤∫Xϕndµ ≤
∫Xϕn+1dµ ∀n ∈ N
18
so that ∫Xϕndµ converges. Hence, we expect that our definition of
∫Xfdµ should be
such that ∫X
fdµ = limn→∞
∫X
ϕndµ.
THEOREM 3.9. Let (X,A, µ) be a measure space and f : X → [0,∞] be a measurable
function. Let (ϕn) be an increasing sequence of non-negative simple measurable functions
which converge to f pointwise. Then the limit limn→∞
∫X
ϕndµ exists and
limn→∞
∫X
ϕndµ = supϕ∈Φf
∫E
ϕdµ,
where Φf is the set of all simple measurable functions ϕ such that 0 ≤ ϕ ≤ f .
Proof. By Theorem 3.6,∫Xϕn ≤
∫ϕn+1 for all n ∈ N. Hence
(∫Xϕndµ
)converges in
[0,∞]. Let
α := limn→∞
∫X
ϕndµ and β := supϕ∈Φf
∫E
ϕdµ.
Since ϕn(x)→ ϕ(x) as n→∞ for every x ∈ X we also have ϕn ≤ f for all n ∈ N. Hence,∫Xϕndµ ≤ β for all n ∈ N so that α ≤ β.
Now, it remains to prove that β ≤ α. That is to show that∫Xϕ ≤ α for every ϕ ∈ Φf .
So, let ϕ ∈ Φf . We are going to show that∫X
r ϕ ≤ α ∀ r ∈ (0, 1), (∗)
so that by taking limit as r → 1, the result will follow. Hence, let r ∈ (0, 1). We claim
that
X =
∞⋃n=1
x ∈ X : rϕ(x) ≤ ϕn(x) =
∞⋃n=1
En,
where En = x ∈ X : rϕ(x) ≤ ϕn(x), n ∈ N. To see this, note that for x ∈ X, if f(x) =
0, then we have ϕ(x) = 0 so that x ∈ E1. If f(x) > 0, then r ϕ(x) ≤ rf(x) < f(x) so
that, since ϕn(x)→ f(x) as n→∞, there exists k ∈ N such that r ϕ(x) ≤ ϕk(x) ≤ f(x).
Hence, x ∈ Ek. Also, we have En ⊆ En+1 for all n ∈ N. Now,∫X
ϕn ≥∫En
ϕn ≥∫En
r ϕ = r ν(En),
where ν : X → [0,∞] is defined by ν(E) =∫Eϕdµ for E ∈ A. We know, by Theorem
3.7, ν is a measure. Hence, by Theorem 2.13, that
ν( ∞⋃n=1
En
)= limn→∞
ν(En).
Thus,
r
∫X
ϕ = r ν(X) = r ν( ∞⋃n=1
En
)= limn→∞
r ν(En) ≤ limn→∞
∫X
ϕn = α.
Thus (∗) is proved, which completes the proof.
19
Motivated by the above theorem we have the following definition.
Definition 3.10. Let (X,A, µ) be a measure space and f : X → [0,∞] be a measurable
function. Then we define the integral of f over E ∈ A as∫E
fdµ := supϕ∈Φf
∫E
ϕdµ,
where Φf is the set of all simple measurable functions ϕ satisfying 0 ≤ ϕ ≤ f . ♦
Once the measure µ under consideration is understood, we may write the integral∫Efdµ as
∫Ef for every E ∈ A, and we write
∫f for
∫Xfdµ.
Now we list some of the properties of the integral in the following theorem.
THEOREM 3.11. Suppose that f, g are extended real valued non-negative measurable
functions defined on a measure space X, c ∈ R and E is a measurable set. Then we have
the following.
(i) f ≤ g ⇒∫Ef ≤
∫Eg (ii) A,B ∈ A, A ⊆ B ⇒
∫Af ≤
∫Bf .
(iii)∫Ec f = c
∫Ef , (iv) f(x) = 0 ∀x ∈ E ⇒
∫Ef = 0.
(v) µ(E) = 0 ⇒∫Ef = 0, (vi)
∫Ef =
∫Xχ
Ef .
Proof. (i) Suppose ϕ ∈ Φf . Then ϕ ∈ Φg, so that∫Eϕ ≤
∫Eg. This implies that∫
E
f := supϕ∈Φf
∫E
ϕ ≤∫E
g.
(ii) Suppose A,B ∈ A such that A ⊆ B, and ϕ ∈ Φf . Then, clearly,∫Aϕ ≤
∫Bϕ ≤∫
Bf . Hence, ∫
A
f := supϕ∈Φf
∫A
ϕ ≤∫E
g.
(iii) Suppose ϕ ∈ Φf . We know that∫Ec ϕ = c
∫Eϕ. Taking supremum over all
ϕ ∈ Φf , we get ∫E
c f = c
∫E
f.
(iv) Suppose ϕ ∈ Φf . If f(x) = 0 for all x ∈ E, then ϕ(x) = 0 for all x ∈ E so that∫Eϕ = 0. Hence, Taking supremum over all ϕ ∈ Φf , we get
∫Ef = 0.
(v) Suppose ϕ ∈ Φf . If µ(E) = 0, then it is clear that∫Eϕ = 0 for all ϕ ∈ Φf . Hence,∫
Ef = 0.
(vi) Suppose ϕ ∈ Φf . Then χEϕ ≤ χ
Ef so that by (i) we have
∫Eϕ =
∫Xχ
Eϕ ≤∫
Xχ
Ef . Hence,
∫Eϕ ≤
∫Xχ
Ef . This is true for all ϕ ∈ Φf . Therefore∫
E
f ≤∫X
χEf.
Next, suppose ϕ ∈ ΦχEf . Then ϕ ≤ χ
Ef ≤ f and
∫Eϕ =
∫Xϕ. Hence,
∫Xϕ =∫
Eϕ ≤
∫Ef This is true for all ϕ ∈ Φχ
Ef . Therefore,∫
X
χEf ≤
∫E
f.
This completes the proof.
20
3.3 Convergence Theorems
THEOREM 3.12. (Monotone convergence theorem) Suppose (fn) is a sequence of
extended real valued non-negative measurable functions on a measure space (X,A, µ) such
that fn ≤ fn+1 for all n ∈ N and (fn(x)) converges for every x ∈ X. Then f : X → [0,∞]
defined by f(x) = limn→∞ fn(x), x ∈ X, is a measurable function on X and
limn→∞
∫X
fn =
∫f.
Proof. By by Corollary 2.29, f is measurable, and by Theorem 3.11(i),∫X
fn ≤∫fn+1 ≤
∫X
f ∀n ∈ N.
Also, there exists α ∈ [0,∞] such that∫Xfn → α as n→∞ so that α ≤
∫Xf .
Now, it remains to prove that∫Xf ≤ α. This part of the proof is exactly same as in
Theorem 3.9 with fn in place of ϕn.
COROLLARY 3.13. Suppose f and g are extended real valued non-negative measurable
functions on a measure space (X,A, µ) such that f ≤ g. Then∫E
(f + g) =
∫E
f +
∫E
g ∀E ∈ A.
Proof. Use Theorem 2.33, Theorem 3.5 and Theorem 3.12.
COROLLARY 3.14. Suppose (fn) is a sequence of extended real valued non-negative
measurable functions on a measure space (X,A, µ) and f :=∑∞n=1 fn. Then∫
X
f =
∞∑n=1
∫fn.
Proof. Use Corollary 3.13 and Theorem 3.12,
COROLLARY 3.15. Suppose f : X → [0,∞) is a measurable function. For E ∈ A, let
ν(E) :=
∫E
fdµ.
Then ν is a measure on X. More over, for any measurable function g : X → [0,∞),∫X
gdν =
∫X
gfdµ.
Proof. Clearly, ν(∅) = 0. Suppose En is a disjoint countable family in A. Then,
ν(⋃i
Ei) :=
∫⋃
i Ei
fdµ =
∫X
χ⋃i Ei
fdµ.
Now, we observe that
χ⋃i Ei
f = limn→∞
χ⋃ni=1
Eif = lim
n→∞
n∑i=1
χEif,
21
where(∑n
i=1 χEif)
is an increasing sequence of non-negative measurable functions. Hence
by Monotone Convergence Theorem,∫X
χ⋃i Ei
fdµ = limn→∞
∫X
n∑i=1
χEif = lim
n→∞
n∑i=1
∫X
χEif =
∞∑i=1
∫X
χEif.
Thus,
ν(⋃i
Ei) =
∫X
χ⋃i Ei
fdµ =
∞∑i=1
∫X
χEif =
∞∑i=1
ν(Ei).
This completes the first part of the theorem. For the second part, first we observe that
the relation∫Xgdν =
∫gfdµ holds if g is a characteristic function of a measurable
set, and hence it holds for all simple non-negative measurable functions as well. Since
any measurable function g : X → [0,∞) is a point-wise limit of an increasing sequence
of simple non-negative measurable functions, the proof can be completed by invoking
Monotone Convergence Theorem.
Remark 3.16. The relation∫Xgdν =
∫gfdµ usually written as
dν = fdµ ordν
dµ= f,
and f is called the (Radon-Nikodym) derivative of ν with respect to µ.
Note that if µ and ν are as in Corollary 3.15, then ν(E) = 0 whenever µ(E) = 0. A
question naturally arises would be the following:
If µ and ν are measures such that ν(E) = 0 whenever µ(E) = 0, then does
there exists a non-negative measurable function f such that ν(E) =∫Efdµ
for all E ∈ A?
The answer is in affirmative when µ is a σ-finte measure , that is, if there exists a countable
family Xn of measurable sets such that X = ∪Xn and µ(Xn) < ∞ for every n ∈ N.
This result is known as the Radon-Nikodym theorem. ♦
COROLLARY 3.17. (Fatou’s lemma) Suppose (fn) is a sequence of extended real
valued non-negative measurable functions on a measure space (X,A, µ). Then∫X
(lim inf
nfn
)≤ lim inf
n
∫f.
Proof. For each k ∈ N, let gk = infn≥k fn. Then gk ≤ gk+1 for all k ∈ N, and
limk→∞
gk = f := lim infn
fn. Hence by Monotone Convergence Theorem, limk→∞
∫X
gk =
∫X
f .
But, since gk ≤ fk for all k ∈ N, we have∫Xgk ≤
∫Xfk so that∫
X
f = limk→∞
∫X
gk = lim infk
∫X
gk ≤ lim infk
∫X
fk.
This completes the proof.
22
Example 3.18. Let X = x1, . . . , xk with the counting measure µ on the σ-algebra 2X .
Let f : X → [0,∞). Then
f(x) =
k∑i=1
f(xi)χxi(x), x ∈ X
and ∫X
fdµ =
k∑i=1
f(xi)µ(xi) =
k∑i=1
f(xi).
♦
Exercise 3.19. Suppose X = x1, x2, . . . with the counting measure µ. If f is an
extended real valued non-negative measurable function on X, then show that∫X
fdµ =
∞∑i=1
f(xi).
♦
Exercise 3.20. Suppose aij ≥ 0 for all i, j ∈ N. Then show that
∞∑i=1
∞∑j=1
aij =
∞∑j=1
∞∑i=1
aij .
♦
Exercise 3.21. Suppose X = x1, . . . , xn, and w1, . . . , wn are non-negative reals. For
x ∈ X, define w(x) = wj whenever x = xj , and µ(E) =∑x∈E w(x) for E ⊆ X. Show that
µ is a measure on (X, 2X), and for every extended real valued non-negative measurable
function f on X, ∫X
fdµ =
n∑i=1
f(xi)wi.
♦
Exercise 3.22. Suppose X = x1, x2, . . ., and w1, w2, . . . are non-negative reals. For
x ∈ X, define w(x) = wj whenever x = xj , and µ(E) =∑x∈E w(x) for E ⊆ X. Show that
µ is a measure on (X, 2X), and for every extended real valued non-negative measurable
function f on X, ∫X
fdµ =
n∑i=1
f(xi)wi.
♦
3.4 Integral of Complex Measurable Functions
Recall that if α is a real number, then α = α+ − α− and |α| = α+ + α−, where α+ :=
maxα, 0 and α− := max−α, 0. Similarly, if f is a real valued function defined on a
set X, then
f = f+ − f−, |f | = f+ + f−,
23
where f+ and f− are defined as
f+ := maxf, 0, f− := max−f, 0.
Now, let (X,A, µ) be a measure space and f be a real measurable function on X.
Then we know that both f+ and f− are positive measurable functions. Hence, we can
define∫Xf+dµ and
∫Xf−dµ.
In view of this we have the following definition.
Definition 3.23. Suppose f is a real measurable function on X. If at least one of∫Xf+dµ and
∫Xf−dµ is finite, then we define∫
X
fdµ =
∫X
f+dµ−∫X
f−dµ.
♦
Since |f | = f+ +f−, it follows that if∫X|f |dµ <∞, then both
∫Xf+dµ and
∫Xf−dµ
are finite, and in that case∫Xfdµ =
∫Xf+dµ−
∫Xf−dµ. This observation prompts the
following definition.
Definition 3.24. A real measurable function f on X is said to be integrable (over X)
if
∫X
|f |dµ <∞. ♦
Nothing prevents us from having the following definition as well.
Definition 3.25. A complex valued measurable function f on X is said to be integrable
(over X) if
∫X
|f |dµ <∞. ♦
We denote the set of all complex valued integrable functions by L(X,A, µ) or simply
by L(µ) if the set X and the sigma algebra are understood.
We may recall that if f is any complex valued function on X, then f = Ref + iImf .
We may observe that if f ∈ L(µ), then Ref and Imf are integrable. In view of this
we have the following definition.
Definition 3.26. If f ∈ L(µ), then we define the integral of f (over X) with respect to
µ as ∫X
f dµ :=
∫X
Ref dµ+ i
∫X
Imf dµ.
♦
Notation. We shall denote the integral
∫X
f dµ by
∫X
f or by
∫f .
Exercise 3.27. If f ∈ L(µ) such that∫Xf ≥ 0, then show that∫
X
f =
∫X
Ref ≤∫X
|f |.
♦
24
THEOREM 3.28. If f and g are in L(µ) and α ∈ C, then f + g and αf are in L(µ),
and ∫X
(f + g) =
∫X
f +
∫X
g,
∫X
α f = α
∫X
f.
Proof. Suppose f and g are in L(µ) and α ∈ C. Since |f +g| ≤ |f |+ |g| and |αf | = |α| |f |it follows that f + g and αf are in L(µ).
Now to prove the equalities of the integrals, we consider two cases.
Case 1: f and g are real valued.
In this case, we have
(f + g)+ − (f + g)− = f + g = (f+ − f−) + (g+ − g−).
Thus,
(f + g)+ + f− + g− = (f + g)− + f+ + g+.
Hence, ∫X
(f + g)+ +
∫X
f− +
∫X
g− =
∫X
(f + g)− +
∫X
f+ +
∫X
g+
so that ∫X
(f + g) =
∫X
(f + g)+ −∫X
(f + g)−
=
∫X
f+ −∫X
f− +
∫X
g+ −∫X
g−
=
∫X
f +
∫X
g.
Case 2: f and g are complex valued.
Let h := f+g, and let f1 = Ref, g1 = Reg, h1 = Reh, f2 = Imf, g2 = Img, h2 = Imh.
Then
h = h1 + ih2 = (f1 + g1) + i(f2 + g2)
so that ∫X
h =
∫X
(f1 + g1) + i
∫X
(f2 + g2).
By case 1, ∫X
(f1 + g1) =
∫X
f1 +
∫X
g1,
∫X
(f2 + g2) =
∫X
f2 +
∫X
g2.
Hence, ∫X
(f + g) =
(∫X
f1 +
∫X
g1
)+ i
(∫X
f2 +
∫X
g2
)=
(∫X
f1 + i
∫X
f2
)+
(∫X
g1 + i
∫X
g2
)=
∫X
f +
∫X
g.
Similarly we can reduce the case of complex valued f and complex number α to the case
of real valued f and real number α, and then prove∫Xα f = α
∫Xf .
25
Exercise 3.29. Show that L(µ) is a vector space over C, and f 7→∫Xf is a linear
functional on L(µ). ♦
THEOREM 3.30. If f ∈ L(µ), then∣∣∣∣∫X
f
∣∣∣∣ ≤ ∫X
|f |.
Proof. Let f ∈ L(µ), and let θ ∈ R be such that∫Xf =
∣∣∫Xf∣∣ eiθ. Then (cf. Exercise
3.27) ∣∣∣∣∫X
f
∣∣∣∣ = e−iθ∫X
f =
∫X
e−iθf =
∫X
Re(e−iθf) ≤∫X
|f |.
This completes the proof.
Exercise 3.31. Show that f 7→∫X|f | is a semi-norm on the vector space L(µ). ♦
Exercise 3.32. Show that N := f ∈ L(µ) :∫X|f | = 0 is subspace of L(µ), and
[f ] 7→∫X|f | is a norm on the quotient space L(µ)/N . ♦
CONVENTION. The quotient space L(µ)/N considered in Exercise 3.32 is denoted
by L(µ). Abusing the notation, we shall denote the element [f ] in L(µ) by f .
Exercise 3.33. Prove that, if f ∈ L(µ), then E ∈ A, µ(E) = 0 imply∫Ef dµ = 0. ♦
In view of the above exercise, a set of measure zero has no effect as far as integral of
a function is concerned.
Definition 3.34. Let (X,A, µ) be a measure space and E ∈ A. We say that a property
P concerning points in X holds almost everywhere on E, or P holds for almost all
x ∈ E, if the set x ∈ E : P does not hold at x is measurable and is of measure zero,
i.e.,
µ (x ∈ E : P (x) does not hold atx) = 0.
In this case, we write P holds a.e. on E. ♦
Using the above definition, we can state that if f and g are in L(µ) such that f = g
a.e. on X, then∫Xf =
∫Xg.
PROPOSITION 3.35. Suppose f : X → [0,∞] and E ∈ A. If∫Ef = 0, then f = 0
a.e. on E.
Proof. Suppose
∫E
f = 0. Note that F := x ∈ E : f(x) 6= 0 = ∪∞n=1En where
En := x : f(x) > 1/n. Hence
0 =
∫E
f ≥∫F
f ≥∫En
f ≥ 1
nµ(En).
Hence µ(En) = 0 for all n ∈ N so that µ(F ) = 0. Thus, f = 0 a.e. on E.
26
Exercise 3.36. Prove that, if f and g are complex measurable functions such that f = 0
a.e. on X and g = 0 a.e. on X, then f + g = 0 a.e. on X. ♦
Exercise 3.37. Prove that, if f ∈ L(µ) is such that∫Ef = 0 for all E ∈ A, then f = 0
a.e.
Hint: It is enough to prove for the case of real valued f . Take E = x ∈ X : f(x) ≥ 0and show that
∫Xf+ = 0. Similarly show that
∫Xf− = 0. ♦
THEOREM 3.38. (Dominated convergence theorem) Suppose (fn) is a sequence
of complex measurable functions such that (fn(x)) converges for every x ∈ X, and let
f(x) := limn→∞
fn(x), x ∈ X. Suppose there exists g ∈ L(µ) such that |fn| ≤ |g| for all
n ∈ N. Then fn and f are integrable, and
limn→∞
∫X
|fn − f | = 0.
In particular,
limn→∞
∫X
fn =
∫X
f.
Proof. Since |fn| ≤ |g| for some g ∈ L(µ) and (fn(x)) converges for every x ∈ X, it
follows that |f | ≤ |g| and fn ∈ L(µ), f ∈ L(µ). Also, we note that 2|g| − |fn − f | ≥ 0 for
all n ∈ N, and 2|g| − |fn − f | → 2|g| as n→∞. Hence, by Fatou’s lemma,∫X
2|g| =
∫X
lim infn
(2|g| − |fn − f |)
≤ lim infn
∫X
(2|g| − |fn − f |)
=
∫X
2|g| − lim supn
∫X
|fn − f |.
Thus,
0 ≤ lim infn
∫X
|fn − f | ≤ lim supn
∫X
|fn − f | ≤ 0,
consequently, limn→∞
∫X
|fn − f | exists and it is equal to 0.
Definition 3.39. Let (X,A, µ) be a measure space, E ∈ A and (fn) is a sequence of
complex valued measurable functions defined on E. Then we say (fn) converges to f
almost everywhere on E, denoted by fn → f a.e. on E, if and only if µ(x ∈ E : fn(x) 6→f(x)) = 0. ♦
Remark 3.40. The conclusions in Theorem 3.38 holds good if we replace pointwise
convergence by convergence almost everywhere. ♦
LEMMA 3.41. Suppose f : X → [0,∞] is such that
∫X
f < ∞. Then f(x) ∈ R for
almost all x ∈ X, i.e., µ(x : f(x) =∞) = 0.
27
Proof. We note that F := x : f(x) = ∞ = ∩∞n=1An, where
An := x : f(x) > n. Then F ⊆ An for all n ∈ N and we have µ(F ) ≤ µ(An) for
all n ∈ N. Thus,∫X
f ≥∫An
f =
∫X
fχAn≥∫X
nχAn= nµ(An) ≥ nµ(F ).
Thus
µ(F ) ≤ 1
n
∫X
f ∀n ∈ N,
so that µ(F ) = 0.
THEOREM 3.42. Suppose (fn) is a sequence of complex measurable functions such
that
∞∑n=1
∫X
|fn| converges. Then
∞∑n=1
fn converges for almost all x ∈ X, and
∫X
( ∞∑n=1
fn
)=
∞∑n=1
∫X
fn.
Proof. Let gn :=∑nj=1 fj . Then |gn| ≤
∑nj=1 |fj(x)| ≤
∑∞j=1 |fj(x)|. Let h :=
∑∞n=1 |fn|.
Since ∫X
h =
∞∑n=1
∫X
|fn| <∞,
by Lemma 3.41, h =∑∞n=1 |fn| is real valued almost everywhere. Let S := x ∈ X :
h(x) < ∞. Then we have µ(Sc) = 0, and g(x) :=
∞∑n=1
fn(x) converges for every x ∈ S.
Then we have |gn| ≤ h on S and h ∈ L(µ), and gn(x)→ g(x) for every x ∈ S. Hence, by
Dominated convergence Theorem, g ∈ L(µ) and
∫S
gn →∫S
g as n→∞. But,
∫S
gn =
∫X
n∑j=1
fj
=
n∑j=1
∫X
fj ,
∫S
g =
∫X
g =
∫X
( ∞∑n=1
fn
).
Thus,∞∑j=1
∫X
fj = limn→∞
n∑j=1
∫X
fj =
∫X
( ∞∑n=1
fn
).
Thus the proof is completed.
28
4 Lecture 4: Lp-spaces
For 1 ≤ p <∞ and for a complex valued measurable function f , let
‖f‖p :=
(∫X
|f |pdµ)1/p
.
THEOREM 4.1. (Holder’s inequality) Let (X,A, µ) be a measure space and f, g be
complex measurable functions on X. For 1 < p < ∞, let q > 1 be such that p+ q = pq.
Then ∫X
|fg|dµ ≤ ‖f‖p ‖g‖q.
Proof. First we observe that if one of ‖f‖p and ‖g‖q is zero, then the inequality holds.
Hence, assume that both ‖f‖p and ‖g‖q are strictly positive. Also, if one of ‖f‖p and
‖g‖q is infinity, then the inequality holds. Thus, it is enough to prove by assuming that
0 < ‖f‖p < ∞ and 0 < ‖g‖q < ∞. For this purpose, first we may observe that for any
non-negative real numbers a and b,
ab ≤ ap
p+bq
q.
Thus, for x ∈ X, taking a = |f(x)|/‖f‖p and b = |g(x)|/‖g‖q, we have
|f(x)g(x)|‖f‖p‖g‖q
≤ |f(x)|p
p‖f‖pp+|g(x)|q
q‖g‖qq.
Now, taking integrals over X, we get
1
‖f‖p‖g‖q
∫X
|fg| ≤∫X|f |p
p‖f‖pp+
∫X|g|q
q‖g‖qq= 1.
Hence,
∫X
|fg| ≤ ‖f‖p‖g‖q, which completes the proof.
THEOREM 4.2. (Minkowski inequality) Let (X,A, µ) be a measure space and f, g be
complex measurable functions on X. Then for 1 ≤ p <∞,
‖f + g‖p ≤ ‖f‖p + ‖g‖p.
Proof. First we observe that if one of ‖f‖p and ‖g‖p is infinity, then the inequality holds.
Hence, it is enough to prove for the case when ‖f‖p < ∞ and ‖g‖p < ∞. We note that
if p = 1, then the proof is trivial, as the result follows by observing |f + g| ≤ |f | + |g|.Thus assume that 1 < p <∞. Also, we can assume that ‖f + g‖p 6= 0. Note that∫
X
|f + g|pdµ =
∫X
|f + g|p−1|f + g|dµ ≤∫X
|f + g|p−1|f |dµ+
∫X
|f + g|p−1|g|dµ.
Hence, with q > 1 satisfying p+ q = pq and applying Holder’s inequality, we have∫X
|f + g|p−1|f | ≤ ‖f‖p(∫
X
|f + g|(p−1)q
)1/q
= ‖f‖p‖f + g‖q/qp ,
29
∫X
|f + g|p−1|f | ≤ ‖g‖p(∫
X
|f + g|(p−1)q
)1/q
= ‖f‖p‖f + g‖q/qp .
Thus,
‖f + g‖pp =
∫X
|f + g|pdµ ≤ (‖f‖p + ‖g‖p) ‖f + g‖q/qp .
Now canceling out ‖f + g‖q/qp , as ‖f + g‖p 6= 0, we get ‖f + g‖p ≤ ‖f‖p + ‖g‖q.
For 1 ≤ p <∞, let
Lp := Lp(X,A, µ) := f : X → C : f measrable and ‖f‖p <∞.
and
Zp := f ∈ Lp : f = 0 a.e. .
Definition 4.3. A measurable function f : X → C is said to be an essentially bounded
function if there exists Mf > 0 such that |f | ≤Mf a.e. on X. ♦
The set of all essentially bounded functions is denoted by L∞ := L∞(X,A, µ), and
we denote
‖f‖∞ := infMf > 0 : |f | ≤Mf a.e.,
and
Z∞ := f ∈ L∞ : f = 0 a.e..
With the above definitions and the Minkowski’s inequality, it follows that for any p
satisfying 1 ≤ p ≤ ∞,
‖f + g‖p ≤ ‖f‖p + ‖g‖q ∀ f, g ∈ Lp(µ).
Now, the proof of the following theorem is immediate.
THEOREM 4.4. For 1 ≤ p ≤ ∞, Lp is a complex vector space, and the map f 7→ ‖f‖pis a seminorm on Lp. Also, the map [f ] 7→ ‖f‖p is a norm on Lp/Zp.
CONVENTION. The quotient space Lp(µ)/Zp considered in the above theorem is
denoted by Lp(X,A, µ) or simply by Lp(µ) or Lp(X) or by Lp. Abusing the notation, we
shall denote the element [f ] in L(µ) by f .
Observe that, if X = N, A = 2N and µ is the counting measure, then for any p ∈ [1,∞],
Zp = 0 and
Lp = Lp = `p(N).
Also, if X = 1, . . . , k for some k ∈ N, A = 2N and µ is the counting measure, then for
any p ∈ [1,∞], Zp = 0 and
Lp = Lp = Ck.
30
Exercise 4.5. Show that if µ(X) <∞, then for 1 ≤ p ≤ r ≤ ∞,
L∞ ⊆ Lr ⊆ Lp ⊆ L1.
♦
Exercise 4.6. Show that if for 1 ≤ p ≤ r ≤ ∞,
`∞(N) ⊇ `r(N) ⊇ `p(N) ⊇ `(N)1.
♦
Next we prove the completeness of Lp(µ). Before that, let us observe the following.
LEMMA 4.7. Let 1 ≤ p < ∞ and let (fn) be a Cauchy sequence in Lp(µ) having a
subsequence which converges almost everywhere to a function f . Then f ∈ Lp(µ) and
(fn) converges to f in Lp(µ).
Proof. Suppose (fn) is a Cauchy sequence in Lp(µ) with 1 ≤ p <∞ having a subsequence
(fnk) which converges a.e. to a function f . Then for each n ∈ N, |fnk
− fn| → |f − fn|a.e. on X. Therefore, by Fatou’s lemma,∫
X
|f − fn|p ≤ lim infk
∫X
|fnk− fn|p.
Now, let ε > 0 and let N ∈ N be such that ‖fk − fn‖p < ε for all n, k ≥ N . Then
‖fnk− fn‖p < ε for all n, k ≥ N so that∫
X
|f − fn|p ≤ lim infn
∫X
|fnk− fn|p < εp ∀n ≥ N,
showing that f ∈ Lp and ‖f − fn‖p → 0 as n→∞.
We shall also make use of the following result.
THEOREM 4.8. Let Ω be any nonempty set. Then `∞(Ω), the space of all bounded
complex valued functions on Ω, is a Banach space with respect to the supremum norm,
‖f‖∞ := supt∈Ω|f(t)|, f ∈ `∞(Ω).
Proof. Let (fn) be a Cauchy sequence in `∞(Ω). Then for every t ∈ Ω, the sequence
(fn(t)) is a Cauchy sequence of complex numbers. Hence, (fn(t)) converges for every
t ∈ Ω. Let
f(t) := limn→∞
fn(t), t ∈ Ω.
Now, let k ∈ N such that ‖fn − fk‖∞ ≤ 1 for all n ≥ k. Then we have
|fn(t)| ≤ ‖fn − fk‖∞ + ‖fk‖∞ < 1 + ‖fk‖∞
for all n ≥ N and for all t ∈ Ω. Hence,
|f(t)| ≤ max1 + ‖fk‖∞, ‖fk‖1, ‖fk‖2, . . . , ‖fk‖k ∀ t ∈ Ω
31
showing that f ∈ `∞(Ω). Next we show that ‖f − fn‖∞ → 0 as n → ∞. For this let
ε > 0 be given. Let N ∈ N be such that ‖fn− fm‖∞ ≤ ε for all n,m ≥ N . Let t inΩ and
let m ≥ N be such that ‖f(t)− fm(t)| < ε. Then we have
|f(t)− fn(t)| ≤ |f(t)− fm(t)|+ |fm(t)− fn(t)| < 2ε
for all n ≥ N . Note that this N is independent of x. Hence we have proved that
‖f − fn‖∞ < 2ε for all n ≥ N . Thus, the Cauchy sequence (fn) converges in `∞(Ω).
THEOREM 4.9. For 1 ≤ p ≤ ∞, Lp is a Banach space with respect to the norm
f 7→ ‖f‖p.
Proof. First we consider the case 1 ≤ p <∞: Let (fn) be a Cauchy sequence in Lp. By
Lemma 4.7, it is enough to show that (fn) has a subsequence which converges almost
every where.
Since (fn) is a Cauchy sequence, for each i ∈ N, there exists ni ∈ N such that
‖fn − fni‖p < 1/2i for all n ≥ ni. Without loss of generality we may assume that
ni ≤ ni+1 for all i ∈ N. Then we have ‖fni+1 − fni‖p < 1/2i for all i ∈ N. Note that
fnk=
k−1∑i=1
(fni+1− fni
).
Thus, it is enough to show that∑∞i=1(fni+1
− fni) converges a.e. on X. Now, let
gk :=
k∑i=1
|fni+1 − fni |, g :=
∞∑i=1
|fni+1 − fni |.
Then gk(x)→ g(x) as k →∞ for every x ∈ X, and by Monotone Convergence Theorem,
limk→∞
∫X
gpk =
∫X
gp. But,
(∫X
gpk
)1/p
= ‖gk‖p ≤k∑i=1
‖fni− fni−1
‖p ≤ 1.
Hence,∫Xgp <∞. Therefore, by Lemma 3.41, g(x) <∞ a.e. on X. Thus
∑∞i=1(fni+1 −
fni) converges a.e. on X, completing the proof for the case 1 ≤ p <∞.
Next we consider the case p = ∞. Let (fn) be a Cauchy sequence in L∞. Then the
sets
Ak := x ∈ X : |fk(x)| > ‖fk‖∞,
Bm,n := x ∈ X : |fn(x)− fm(x)| > ‖fn − fm‖∞
are of measure zero for every k,m, n ∈ N. Hence, the set E := (∪kAk) ∪ (∪m,nBm,n) is
of measure zero, and for each x ∈ Ω := Ec,
|fk(x)| ≤ ‖fk‖∞ ∀ k ∈ N, |fn(x)− fm(x)| ≤ ‖fn − fm‖∞ ∀m,n ∈ N.
32
Since (fn) is cauchy in L∞, it follows that (fn) with fn = fn|Ω is a Cauchy sequence in
`∞(Ω). Since `∞(Ω) is a Banach space, it follows that there exists f ∈ `∞(Ω) such that
‖fn − f‖`∞(Ω) → 0 as n → ∞. Defining f(x) = f(x) for x ∈ Ω and f(x) = 0 for x 6∈ Ω,
it follows that f ∈ L∞, and ‖fn − f‖∞ → 0 as n→∞.
Note that as part of the proof of the above theorem, we have proved the following.
PROPOSITION 4.10. If 1 ≤ p < ∞, then every Cauchy sequence in Lp(µ) has a
subsequence which converges almost everywhere.
In particular, if (fn) is a sequence in Lp(µ) which converges to f ∈ Lp(µ), then (fn)
has a subsequence which converges a.e. to f ∈ Lp(µ).
4.1 Sobolev spaces
One of the most important measure spaces which is important in applications is (Ω,A, µ),
where Ω is an open subset of Rk, A is the σ-algebra of all Borel subsets or Lebesgue
measurable subsets of Ω and µ is the Lebesgue measure. For further discussion, let us
introduce the following:
• C(Ω): The set of all continuous (C-valued) functions on Ω.
• For α := (α1, . . . , αk) with αj ∈ N0 := N ∪ 0,
|α| = α1 + . . .+ αk, Dα :=( ∂
∂x1
)α1( ∂
∂x2
)α2
· · ·( ∂
∂xk
)αk
.
• For m ∈ N, Cm(Ω): The set of all f ∈ C(Ω) such that Dαf exists and Dαf ∈ C(Ω)
for all α with |α| ≤ m.
• C∞(Ω) =
∞⋂m=1
Cm(Ω).
• For f : Ω→ C, the support of f is the set
supp(f) := clx ∈ Ω : f(x) 6= 0.
• A function A f : Ω→ C is said to be of compact support if supp(f) is compact.
• Cc(Ω): The set of all functions in C(Ω) with compact support.
• For m ∈ N, Cmc (Ω) := Cm(Ω) ∩ Cc(Ω).
• C∞c (Ω) := C∞(Ω) ∩ Cc(Ω).
We observe that
C∞(Ω) ⊆ Cm(Ω) ⊆ C(Ω) ∀m ∈ N,
C∞c (Ω) ⊆ Cmc (Ω) ⊆ Cc(Ω) ⊆ Lp(Ω) ∀m ∈ N.
For m ∈ N0 and f ∈ Cm(Ω), let
‖f‖m,p =( ∑|α|≤m
‖Dαf‖pp)1/p
, 1 ≤ p <∞,
‖f‖m,∞ = max‖Dαf‖∞ : |α| ≤ m.
33
THEOREM 4.11. For m ∈ N0 and 1 ≤ p ≤ ∞,
Wm,p(Ω) := f ∈ Cm(Ω) : ‖f‖m,p <∞
is a linear space and
f 7→ ‖f‖m,p
is a norm on Wm,p(Ω). Further,
〈f, g〉 :=∑|α|≤m
〈Dαf,Dαg〉L2(Ω) =∑|α|≤m
∫Ω
DαfDαgdµ
defines an inner product on Wm,2(Ω).
Definition 4.12. The completion of Wm,p(Ω) is called a Sobolev space, and it is
denoted by Wm,p(Ω). The Hilbert space Wm,2(Ω) is usually denoted by Hm(Ω). ♦
We note that Cmc (Ω) is subspace of Wm,p(Ω).
• The closure of Cmc (Ω) in Wm,p(Ω) is denoted by Wm,p0 (Ω), and the closure of Cmc (Ω)
in Hm(Ω) is denoted by Hm0 (Ω).
Definition 4.13. Let f ∈ Lp(Ω) for 1 ≤ p < ∞. Then f is said to have generalized
derivative of order α ∈ Nk0 if there exists fα ∈ L1loc(Ω) such that
〈fα, ϕ〉L2(Ω) = (−1)|α|〈f,Dαϕ〉 ∀ϕ ∈ C∞c (Ω),
and in that case fα is called the generalized derivative of f of order α, and we write
this fα by f (α). ♦
It is known that
Wm,p(Ω) = f ∈ Lp(Ω) : f (α) exists for |α| ≤ m and f (α) ∈ Lp(Ω)
and
‖f‖m,p =( ∑|α|≤m
‖f (α)‖pp)1/p
.
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5 Problems
1. Let E ⊆ R. Prove that for every ε > 0 there exists a countable family In of open
intervals such that E ⊆∞⋃n
In and∑n
`(In) ≤ m∗(E) + ε. Strict inequality occurs
in the above if m∗(E) <∞.
2. Prove the following:
(i) m∗(∅) = 0.
(ii) If E is a singleton set, then m∗(E) = 0.
(iii) If E is a countable set, then m∗(E) = 0.
(iv) If E = (a, b) with a < b then m∗(E) ≤ b− a.
3. Prove the following:
(i) E1 ⊆ E2 ⇒ m∗(E1) ≤ m∗(E2).
(ii) m∗(A ∪B) ≤ m∗(A) +m∗(B).
(iii) If E ⊆ A and m∗(E) = 0, then m∗(A \ E) = m∗(A).
(iv) If E ⊆ R and x ∈ R, then m∗(E + x) = m∗(E).
4. Prove that for every sequence (In) of open intervals,
m∗(⋃In) ≤
∑n
`(In).
5. If E ⊆ R, then for every ε > 0, there exists an open set G ⊆ R such that G ⊇ E
and m∗(G) ≤ m∗(E) + ε.
6. Derive Problem 1.4(iii) from Problem1.4(ii) by using sub-additivity of m∗.
7. Prove that if
m∗(A1 ∪A2) = m∗(A1) +m∗(A2). (5.3)
holds for any two disjoint sets A1 and A2, then
m∗( ∞⋃n=1
An
)=
∞∑n=1
m∗(An) (5.4)
holds for any pairwise disjoint denumerable family An∞n=1.
Prove that (5.4) need not hold for every denumerable disjoint family An∞n=1.
Hint: On [0, 1] consider the relation x ∼ y ⇐⇒ x− y ∈ Q.• Show that ∼ is an equivalence relation on [0, 1].
Let E be the subset of [0, 1] such that its intersection with each equivalence class is a
singleton set. (Such a set E exists by using the axiom of choice.) Let r1, r2, . . . =
Q ∩ [−1, 1] and En := E + rn for n ∈ N.
• Show that En is a pairwise disjoint family.
Observe: [0, 1] ⊆⋃∞n=1En ⊆ [−1, 2] so that 1 ≤ m∗
(⋃∞n=1En
)≤ 3. Therefore, if
(1.2) is true, we arrive at a contradiction.
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8. There exist disjoint subsets A1 and A2 of R such that
m∗(A1 ∪A2) 6= m∗(A1) +m∗(A2).
Why?
9. Prove the following:
(i) E ∈M ⇐⇒ m∗(A) ≥ m∗(A ∩ E) +m∗(A ∩ Ec) ∀A ⊆ R.
(ii) ∅ ∈M.
(iii) E ∈M ⇒ Ec ∈M.
(iv) m∗(E) = 0 ⇒ E ∈M.
10. Prove that if A1, . . . , An is a disjoint family in M, then
m∗( n⋃i=1
Ai
)=
n∑i=1
m∗(Ai).
[Hint: Use Theorem 1.18.]
11. Prove that if An∞n=1 is a disjoint family in M, then
m∗( ∞⋃n=1
An
)=
∞∑n=1
m∗(An).
[Hint: Use Problem 10 and Problem 7.]
12. There exists E ⊆ R such that E 6∈ M - Why?.
13. If E ∈M with m(E) <∞ then show that for every ε > 0, there exists an open set
G ⊆ R such that G ⊇ E and m(GE) < ε.
14. Let (X,A, µ) be a measure space and E ∈ A. Prove that AE := A ⊂ E : A ∈ Ais a σ-algebra on E and µE(A) := µ(A), A ∈ AE , defines a measure on AE .
15. Let µ∗ : 2X → [0,∞] be an outer measure on X. Prove that
(i) A := E ⊆ X : ∀A ⊆ X, µ∗(A) = µ∗(A ∩ E) + µ∗(A ∩ Ec) is a σ-algebra on
X, and
(ii) µ := µ∗|A is a measure on A.
16. Let (X,A) be a measurable space and µ be a measure on it. For A ⊆ X, let
µ∗(A) = inf∞∑i=1
µ(Ei) : Ei∞i=1 ⊆ A such that A ⊆ ∪∞i=1Ei.
Prove that µ∗ is an outer measure and µ = µ∗|A.
17. Let (X,A) be a measurable space. Show that a real valued function f : X → R is
measurable if and only if f−1(I) ∈ A for every open interval I.
18. Let (X,A) be a measurable space and E ⊆ X. Prove that χE
is measurable if and
only if E ∈ A.
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19. Let (X,A) be a measurable space and ϕ is a simple function on X with canonical
representation ϕ =
n∑i=1
αiχEi. Then ϕ is measurable if and only if Ei ∈ A for
i = 1, . . . , n – Justify.
20. Suppose ϕ is a non-negative simple measurable function and E is a measurable set.
Let µE
be the restriction of µ to the restricted σ-algebra AE := A ⊆ E : A ∈ A.Show that
∫Eϕdµ =
∫Eϕdµ
E.
21. Suppose X = x1, x2, . . . with the counting measure µ. If f is an extended real
valued non-negative measurable function on X, then show that∫X
fdµ =
∞∑i=1
f(xi).
22. Suppose aij ≥ 0 for all i, j ∈ N. Then show that
∞∑i=1
∞∑j=1
aij =
∞∑j=1
∞∑i=1
aij .
23. Suppose aij ∈ R for all i, j ∈ N such that∑∞i=1
∑∞j=1 |aij | <∞. Show that
∞∑i=1
∞∑j=1
aij =
∞∑j=1
∞∑i=1
aij .
24. Suppose X = x1, . . . , xn, and w1, . . . , wn are non-negative reals. For x ∈ X,
define w(x) = wj whenever x = xj , and µ(E) =∑x∈E w(x) for E ⊆ X. Show
that µ is a measure on (X, 2X), and for every extended real valued non-negative
measurable function f on X, ∫X
fdµ =
n∑i=1
f(xi)wi.
25. Suppose X = x1, x2, . . ., and w1, w2, . . . are non-negative reals. For x ∈ X, define
w(x) = wj whenever x = xj , and µ(E) =∑x∈E w(x) for E ⊆ X. Show that µ is
a measure on (X, 2X), and for every extended real valued non-negative measurable
function f on X, ∫X
fdµ =
∞∑i=1
f(xi)wi.
26. If f ∈ L(µ) such that∫Xf ≥ 0, then show that∫
X
f =
∫X
Ref ≤∫X
|f |.
27. Show that L(µ) is a vector space over C, and f 7→∫Xf is a linear functional.
28. Show that f 7→∫X|f | is a semi-norm on the vector space L(µ).
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29. Show that N := f ∈ L(µ) :∫X|f | = 0 is subspace of L(µ), and [f ] 7→
∫X|f | is a
norm on the quotient space L(µ)/N .
30. Prove that, if f and g are complex measurable functions such that f = 0 a.e. on X
and g = 0 a.e. on X, then f + g = 0 a.e. on X.
31. Prove that, if f ∈ L(µ) is such that∫Ef = 0 for all E ∈ A, then f = 0 a.e.
Hint: It is enough to prove for the case of real valued f . Take E = x ∈ X : f(x) ≥0 and show that
∫Xf+ = 0. Similarly show that
∫Xf− = 0.
32. Show that if µ(X) <∞, then for 1 ≤ p ≤ r ≤ ∞,
L∞ ⊆ Lr ⊆ Lp ⊆ L1.
33. Show that if for 1 ≤ p ≤ r ≤ ∞,
`∞(N) ⊇ `r(N) ⊇ `p(N) ⊇ `1(N).
34. If X = N, A = 2N and µ is the counting measure, then prove that Lp = Lp = `p(N).
35. If X = 1, . . . , k for some k ∈ N, A = 2N and µ is the counting measure, then
prove that Lp = Lp = Ck.
38
