Measure and Integration: Concepts, Examples and Exercises · Measure and Integration: Concepts, Examples and Exercises INDER K. RANA Indian Institute of Technology Bombay India Department

Measure and Integration:

Concepts, Examples and Exercises

INDER K. RANA

Indian Institute of Technology Bombay

India

Department of Mathematics, Indian Institute of Technol-

ogy, Bombay, Powai, Mumbai 400076, India

Current address: Department of Mathematics, Indian Institute of Tech-nology, Powai, Mumbai 400076, India

E-mail address: [email protected]

These notes were specially prepared for the participants of the first AnnualFoundation School, May 2004, held at IIT Bombay, India.

Abstract. These notes present a quick overview of the theory of Mea-

sure and Integration. For a more detailed and motivated text, the reader

may refer author’s book:

An Introduction to Measure and Integration,

Narosa Publishers, Delhi, 1997

or,

An Introduction to Measure and Integration,Second Edition,

Graduate Text in Mathematics, Volume 45,

American Mathematical Society, 2002.

May, 2004 Mumbai 400076 Inder K. Rana

Contents

Chapter 1. Classes of sets 1

§1.1. Semi-algebra and algebra of sets 1

§1.2. Sigma algebra and monotone class 5

Chapter 2. Measure 9

§2.1. Set functions 9

§2.2. Countably additive set functions on intervals 14

§2.3. Set functions on algebras 15

§2.4. Uniqueness problem for measures 17

Chapter 3. Construction of measures 19

§3.1. Extension from semi-algebra to the generated algebra 19

§3.2. Extension from algebra to the generated σ-algebra 20

§3.3. Choosing nice sets: Measurable sets 22

§3.4. Completion of a measure space 24

§3.5. The Lebesgue measure 27

Chapter 4. Integration 35

§4.1. Integral of nonnegative simple measurable functions 35

§4.2. Integral of nonnegative measurable functions 38

§4.3. Intrinsic characterization of nonnegative measurable functions 42

§4.4. Integrable functions 50

§4.5. The Lebesgue integral and its relation with the Riemannintegral 55

vii

viii Contents

§4.6. L1[a, b] as the completion of R[a, b] 59

Chapter 5. Measure and integration on product spaces 63

§5.1. Introduction 63

§5.2. Product of measure spaces 65

§5.3. Integration on product spaces: Fubini’s theorems 69

§5.4. Lebesgue measure on R2 and its properties 75

Chapter 6. Lp-spaces 79

§6.1. Integration of complex-valued functions 79

§6.2. Lp-spaces 82

§6.3. L∞(X,S, µ) 86

§6.4. L2(X,S, µ) 87

§6.5. L2-convergence of Fourier series 93

Appendix A. Extended real numbers 97

Appendix B. Axiom of choice 101

Appendix C. Continuum hypothesis 103

Appendix. References 105

Appendix. Index 107

Chapter 1

Classes of sets

1.1. Semi-algebra and algebra of sets

Concepts and examples:

1.1.1. Definition:Let X be a nonempty set and let C be a collection of subsets of X. We sayC is a semi-algebra of subsets of X if it has the following properties:

(i) ∅, X ∈ C.

(ii) A ∩B ∈ C for every A,B ∈ C.

(iii) For every A ∈ C there exist n ∈ N and sets C1, C2, . . . , Cn ∈ C suchthat Ci ∩ Cj = ∅ for i 6= j and Ac =

⋃ni=1Ci.

1.1.2. Definition:Let X be a nonempty set and F a collection of subsets of X. The collectionF is called an algebra of subsets of X if F has the following properties:

(i) ∅, X ∈ F .

(ii) A ∩B ∈ F , whenever A,B ∈ F .

(iii) Ac ∈ F , whenever A ∈ F .

1.1.3. Examples:

(i) LetX be any nonempty set. The collections {∅, X} and P(X) := {E |E ⊆X} are trivial examples of algebras of subsets of X. The collection P(X)is called the power set of X.

1

2 1. Classes of sets

(ii) The collection I of all intervals forms a semi-algebra of subsets of R. For

a, b ∈ R with a < b, consider the collection I of all intervals of the form(a, b], (−∞, b], (a,∞), (−∞,+∞). We call I the collection of all left-open,

right-closed intervals of R. It is easy to check that I is also a semi-algebraof subsets of R.

(iii) The collection

F(I) :=

{

E ⊆ R E =n⋃

k=1

Ik, Ik ∈ I, Ik ∩ Iℓ = ∅ for k 6= ℓ, n ∈ N

}

.

is an algebra of subsets of R. So is the class

F(I) :=

{

E ⊆ R |E =

n⋃

k=1

Ik, Ik ∈ I, Ik ∩ Iℓ = ∅ for k 6= ℓ, n ∈ N

}

.

(iv) Let X be any nonempty set. Let

C := {E ⊆ X | either E or Ec is finite}.

Then C is an algebra of subsets of X.

(v) LetX and Y be two nonempty sets, and F and G semi-algebras of subsetsof X and Y , respectively. Let

F×G = {F×G | F ∈ F , G ∈ G}.

Then, F×G is a semi-algebra of subsets of X×Y .

Exercises:

(1.1) Let F be any collection of subsets of a set X. Show that F is analgebra if and only if the following hold:(i) φ,X ∈ F .(ii) Ac ∈ F whenever A ∈ F .(iii) A ∪B ∈ F whenever A,B ∈ F .

(1.2) Let F be an algebra of subsets of X. Show that(i) If A,B ∈ F then A△B := (A \B) ∪ (B \A) ∈ F .(ii) If E1, E2, . . . , En ∈ F then ∃ F1, F2, . . . , Fn ∈ F such that

Fi ⊆ Ei for each i, Fi∩Fj = ∅ for i 6= j and⋃n

i=1Ei =⋃n

j=1 Fj .

The next set of exercise describes some methods of constructing algebrasand semi-algebras.

1.1. Semi-algebra and algebra of sets 3

(1.3) Let X be a nonempty set. Let ∅ 6= E ⊆ X and let C be a semi-algebra (algebra) of subsets of X. Let

C ∩E := {A ∩ E |A ∈ C}.

Show that C ∩ E is a semi-algebra (algebra) of subsets of E. Notethat C∩E is the collection of those subsets of E which are elementsof C when E ∈ C.

(1.4) Let X,Y be two nonempty sets and f : X −→ Y be any map. ForE ⊆ Y, we write f−1(E) := {x ∈ X | f(x) ∈ E}. Let C be anysemi-algebra (algebra) of subsets of Y. Show that

f−1(C) := {f−1(E) | E ∈ C}

is a semi-algebra (algebra) of subsets of X.

(1.5) Give examples of two nonempty sets X,Y and algebras F , G ofsubsets of X and Y, respectively such that F × G := {A× B |A ∈F , B ∈ G} is not an algebra. (It will of course be a semi-algebra,as shown in example 1.1.3(v).)

(1.6) Let {Fα}α∈I be a family of algebras of subsets of a set X. LetF :=

⋂

α∈I Fα. Show that F is also an algebra of subsets of X.

(1.7) Let {Fn}n≥1 be a sequence of algebras of subsets of a set X. Underwhat conditions on Fn can you conclude that F :=

⋃∞n=1 Fn is also

an algebra?

(1.8) Let C be a semi-algebra of subsets of a set X. A set A ⊆ X is calleda σ-set if there exist sets Ci ∈ C, i = 1, 2, . . . , such that Ci∩Cj = ∅for i 6= j and

⋃∞i=1Ci = A. Prove the following:

(i) For any finite number of sets C,C1, C2, . . . , Cn in C, C\(⋃n

i=1Ci)is a finite union of pairwise disjoint sets from C and hence is aσ-set.

(ii) For any sequence {Cn}n≥1 of sets in C,⋃∞

n=1Cn is a σ-set.(iii) A finite intersection and a countable union of σ-sets is a σ-set.

(1.9) Let C be any collection of subsets of a set X. Then there exists aunique algebra F of subsets of X such that C ⊆ F and if A is anyother algebra such that C ⊆ A, then F ⊆ A. This unique algebragiven is called the algebra generated by C and is denoted byF(C).

(1.10) Show that he algebra generated by I, the class of all intervals, is{E ⊆ R | E =

⋃nk=1 Ik, Ik ∈ I, Ik ∩ Iℓ = ∅ if 1 ≤ k 6= ℓ ≤ n}.

(1.11) Let C be any semi-algebra of subsets of a setX. Show that F(C), thealgebra generated by C, is given by {E ⊆ X | E =

⋃ni=1Ci, Ci ∈ C

and Ci ∩ Cj = ∅ for i 6= j, n ∈ N}.


Remark:Exercise 1.11 gives a description of F(C), the algebra generated bya semi-algebra C. In general, no description is possible for F(C)when C is not a semi-algebra.

(1.12) Let X be any nonempty set and C = {{x} | x ∈ X}⋃

{∅, X}. Is Ca semi-algebra of subsets of X? What is the algebra generated byC? Does your answer depend upon whether X is finite or not?

(1.13) Let Y be any nonempty set and let X be the set of all sequenceswith elements from Y, i.e.,

X = {x = {xn}n≥1 | xn ∈ Y, n = 1, 2, . . .}.

For any positive integer k let A ⊆ Y k, the k-fold Cartesian productof Y with itself, and let i1 < i2 < · · · < ik be positive integers. Let

C(i1, i2, . . . , ik;A) := {x = (xn)n≥1 ∈ X | (xi1 , . . . , xik) ∈ A}.

We call C(i1, i2, . . . , ik;A) a k-dimensional cylinder set in Xwith base A. Prove the following assertions:(a) Every k-dimensional cylinder can be regarded as a n-dimensional

cylinder also for n ≥ k.(b) Let

A = {E ⊂ X | E is an n-dimensional cylinder set for some n}.

Then, A ∪ {∅, X} is an algebra of subsets of X.

(1.14) Let C be any collection of subsets of a set X and let E ⊆ X. Let

C ∩ E := {C ∩ E | C ∈ C}.

Then the following hold:(a)

C ∩E ⊆ F(C) ∩ E := {A ∩ E | A ∈ F(C)}.

Deduce that

F(C ∩ E) ⊆ F(C) ∩ E.

(b) Let

A = {A ⊆ X | A ∩ E ∈ F(C ∩E)}.

Then, A is an algebra of subsets of X, C ⊆ A and

A ∩ E = F(C ∩ E).

(c) Using (a) and (b), deduce that F(C) ∩ E = F(C ∩E).

1.2. Sigma algebra and monotone class 5

1.2. Sigma algebra and monotone class


1.2.1. Definition:Let X be any nonempty set and let S be a class of subsets of X with thefollowing properties:

(i) ∅ and X ∈ S.

(ii) Ac ∈ S whenever A ∈ S.

(iii)⋃∞

i=1Ai ∈ S whenever Ai ∈ S, i = 1, 2, . . . .

Such a class S is called a sigma algebra (written as σ-algebra) of subsetsof X.

1.2.2. Examples:

(i) Let X be any set. Then {∅, X} and P(X) are obvious examples of σ-algebras of subsets of X.

(ii) Let X be any uncountable set and let

S = {A ⊆ X | A orAc is countable}.

Then S is a σ-algebra of subsets of X.

(iii) Let X be any set and let C be any class of subsets of X. Let S(C) :=⋂

S,where the intersection is taken over all σ-algebras S of subsets of X suchthat S ⊇ C (note that P(X) is one such σ-algebra). It is easy to see thatS(C) is also a σ-algebra of subsets of X and S(C) ⊇ C. In fact, if S is anyσ-algebra of subsets of X such that S ⊇ C, then clearly S ⊇ S(C). ThusS(C) is the smallest σ-algebra of subsets of X containing C, and is calledthe σ-algebra generated by C. In general it is not possible to representan element of S(C) explicitly in terms of elements of C.

1.2.3. Definition:Let X be a nonempty set and M be a class of subsets of X. We say M is amonotone class if

(i)⋃∞

n=1An ∈ M, whenever An ∈ M and An ⊆ An+1 for n = 1, 2, . . . ,

(ii)⋂∞

n=1An ∈ M, whenever An ∈ M and An ⊇ An+1 for n = 1, 2, . . . .

1.2.4. Examples:

(i) Clearly, every σ-algebra is also a monotone class.


(ii) Let X be any uncountable set. Let M := {A ⊆ X | A is countable}.Then M is a monotone class but not a σ-algebra.

(iii) Let X be any nonempty set and let C be any collection of subsets of X.Clearly P(X) is a monotone class of subsets of X such that C ⊆ P(X).Let M(C) :=

⋂

M, where the intersection is over all those monotoneclasses M of subsets of X such that C ⊆ M. Clearly, M(C) is itself amonotone class, and if M is any monotone class such that C ⊆ M, thenM(C) ⊆ M. Thus M(C) is the smallest monotone class of subsets ofX such that C ⊆ M(C). The class M(C) is called the monotone classgenerated by C.

Exercises:

(1.15) Let S be a σ-algebra of subsets of X and let Y ⊆ X. Show thatS ∩ Y := {E ∩ Y | E ∈ S} is a σ-algebra of subsets of Y.

(1.16) Let f : X → Y be a function and C a nonempty family of subsetsof Y . Let f−1(C) := {f−1(C) | C ∈ C}. Show that S(f−1(C)) =f−1(S(C)).

(1.17) Let X be an uncountable set and C = {{x} | x ∈ X}. Identify theσ-algebra generated by C.

(1.18) Let C be any class of subsets of a set X and let Y ⊆ X. Let A(C)be the algebra generated by C.(i) Show that S(C) = S(A(C)).(ii) Let C ∩Y := {E∩Y | E ∈ C}. Show that S(C ∩Y ) ⊆ S(C)∩Y.(iii) Let

S := {E ∪ (B ∩ Y c) | E ∈ S(C ∩ Y ), B ∈ C}.

Show that S is a σ-algebra of subsets of X such that C ⊆ Sand S ∩ Y = S(C ∩ Y ).

(iv) Using (i), (ii) and (iii), conclude that S(C ∩ Y ) = S(C) ∩ Y.

(1.19) Let C be a class of subsets of a set X such that ∅ ∈ C. Then E ∈ S(C)iff ∃ sets C1, C2, . . . in C such that E ∈ S({C1, C2, . . .}).

Hint:The technique used to prove of exercise 1.19 is very useful, andis often used to prove various properties of σ-algebras under con-sideration, is as follows: The sets satisfying the required propertyare collected together. One shows that this collection itself is aσ-algebra and includes a subfamily of the original σ-algebra which

1.2. Sigma algebra and monotone class 7

generates it. The claim then follows by the definition of the gener-ated σ-algebra. We call this the σ-algebra technique.

(1.20) Let X be any topological space. Let U denote the class of all opensubsets of X and C denote the class of the all closed subsets of X.(i) Show that

S(U) = S(C).

This is called the σ-algebra of Borel subsets of X and isdenoted by BX .

(ii) Let X = R. Let I be the class of all intervals and I the class ofall left-open right-closed intervals. Show that I ⊂ S(U), I ⊂S(I), I ⊂ S(I) and hence deduce that

S(I) = S(I) = BR.

(1.21) Prove the following statements:(i) Let Ir denote the class of all open intervals of R with rational

endpoints. Show that S(Ir) = BR.(ii) Let Id denote the class of all subintervals of [0, 1] with dyadic

endpoints (i.e., points of the form m/2n for some integers m andn). Show that S(Id) = BR ∩ [0, 1].

(1.22) Let C be any class of subsets of X. Prove the following:(i) If C is an algebra which is also a monotone class, show that C is

a σ-algebra.(ii) C ⊆ M(C) ⊆ S(C).

(1.23) (σ-algebra monotone class theorem ) Let A be an algebra ofsubsets of a set X. Then , S(A) = M(A).

Prove the above statement by proving the following:(i) M(A) ⊆ S(A).(ii) Show that M(A) is closed under complements by proving that

for

B := {E ⊆ X | Ec ∈ M(A)},

A ⊆ B, and B is a monotone class. Hence deuce that M(A) ⊆B.

(iii) For F ∈ M(A), let

L(F ) := {A ⊆ X | A ∪ F ∈ M(A)}.

Show that E ∈ L(F ) iff F ∈ L(E), L(F ) is a monotone class,and A ⊆ L(F ) whenever F ∈ A. Hence, M(A) ⊆ L(F ), forF ∈ A.


(iv) Using (iii), deduce that M(A) ⊆ L(E) for everyE ∈ M(A),i.e.,M(A) is closed under unions also. Now use exercise (1.22) todeduce that S(A) ⊆ M(A).

Chapter 2

Measure

2.1. Set functions


2.1.1. Definition:Let C be a class of subsets of a set X. A function µ : C −→ [0,+∞] is calleda set function. Further,

(i) µ is said to be monotone if µ(A) ≤ µ(B) whenever A,B ∈ C andA ⊆ B.

(ii) µ is said to be finitely additive if

µ

(

n⋃

i=1

Ai

)

=n∑

i=1

µ(Ai).

whenever A1, A2, . . . , An ∈ C are such that Ai ∩ Aj = ∅ for i 6= jand

⋃ni=1Ai ∈ C.

(iii) µ is said to be countably additive if

µ

(

∞⋃

n=1

An

)

=∞∑

n=1

µ(An)

wheneverA1, A2, . . . in C withAi∩Aj = ∅ for i 6= j and⋃∞

n=1An ∈ C.

9

10 2. Measure

(iv) µ is said to be countably subadditive if

µ(A) ≤∞∑

n=1

µ(An).

whenever A ∈ C, A =⋃∞

n=1An with An ∈ C for every n.

(v) µ is called a measure on C if ∅ ∈ C with µ(∅) = 0 and µ is countablyadditive on C.

Here are some more examples of finitely/countably additive set func-tions:

2.1.2.Example:Let X be any infinite set and let xn ∈ X,n = 1, 2, . . . . Let {pn}n≥1 be asequence of nonnegative real numbers. For any A ⊆ X, define

µ(A) :=∑

{i|xi∈A}

pi.

It is easy to show that µ is a countably additive set function on the algebraP(X). We say µ is a discrete measure with ‘mass’ pi at xi. The measureµ is finite (i.e., µ(X) < +∞) iff

∑∞i=1 pi < +∞. If

∑∞i=1 pi = 1, the measure

µ is called a discrete probability measure/distribution. Note thatµ({xi}) = pi ∀ i and µ({x}) = 0 if x 6= xi. So, one can regard µ as a setfunction defined on the subsets of the set Y := {xn : n ≥ 1}. Some of thespecial cases when X = {0, 1, 2, . . .} are:

(a) Binomial distribution: Y := {0, 1, 2, . . . , n} and, for 0 < p < 1,

pk =

(

n

k

)

pk(1 − p)n−k, 0 ≤ k ≤ n.

(b) Poisson distribution: Y := {0, 1, 2, . . .} and

pk := λk e−λ/k!

for k = 0, 1, 2, . . . , where λ > 0.

(c) Uniform distribution: Y := {1, 2, . . . , n},

pk := 1/k ∀ k.

An important example of a set function on the collection of intervals inR.

2.1.3 Example:We denote the set of real numbers by R. Let R∗ denote the set of extendedreal numbers.

Let I denote the collection of all intervals of R. If an interval I ∈ I hasend points a and b we write it as I(a, b). By convention, the open interval

2.1. Set functions 11

(a, a) = ∅ ∀ a ∈ R. Let [0,+∞] := {x ∈ R∗|x ≥ 0} = [0,+∞) ∪ {+∞}.Define the function λ : I −→ [0,∞] by

λ(I(a, b)) :=

{

|b− a| if a, b ∈ R,+∞ if either a = −∞ or b = +∞ or both.

The function λ, as defined above, is called the length function and hasthe following properties:

Property (1): λ(∅) = 0.

Property (2): λ(I) ≤ λ(J) if I ⊆ J.

This is called the monotonicity property of λ (or one says that λ ismonotone).

Property (3): Let I ∈ I be such that I =⋃n

i=1 Ji, where Ji ∩ Jj = ∅ fori 6= j. Then

λ(I) =

n∑

i=1

λ(Ji).

This property of λ is called the finite additivity of λ, or one says thatλ is finitely additive.

Property (4): Let I ∈ I be a finite interval such that I ⊆⋃∞

i=1 Ii, whereIi ∈ I. Then

λ(I) ≤∞∑

i=1

λ(Ii).

Property (5): Let I ∈ I be a finite interval such that I =⋃∞

n=1 In, whereIn ∈ I and In ∩ Im = ∅ for n 6= m. Then

λ(I) =∞∑

n=1

λ(In).

Property (6): Let I ∈ I be any interval. Then

λ(I) =∞∑

n=−∞

λ(I ∩ [n, n+ 1)).

Property (7): Let I ∈ I be any interval such that I =⋃∞

n=1 In, In ∈ Iand In ∩ Im = ∅ for n 6= m. Then

λ(I) =∞∑

n=1

λ(In).

12 2. Measure

This property of λ is called the countable additivity of λ, or one saysthat λ is countably additive.

Property (8): Let I ∈ I and I ⊆⋃∞

n=1 In, In ∈ I. Then

λ(I) ≤∞∑

n=1

λ(In).

This property of λ is called the countable subadditivity of λ, or onesays that λ is countably subadditive.

Property (9): λ(I) = λ(I + x), for every I ∈ I and x ∈ R, whereI + x := {y + x | y ∈ I}.

This property of the length function is called translation invariance, orone says that λ is translation invariant.

Exercises:

(2.1) Let X be any countably infinite set and let

C = {{x} | x ∈ X}.

Show that the algebra generated by C is

F(C) := {A ⊆ X | A or Ac is finite}.

Let µ : F(C) −→ [0,∞) be defined by

µ(A) :=

{

0 if A is finite,1 if Acis finite.

Show that µ is finitely additive but not countably additive. If X isan uncountable set, show that µ is also countably additive.

(2.2) Let X = N, the set of natural numbers. For every finite set A ⊆ X,let #A denote the number of elements in A. Define for A ⊆ X,

µn(A) :=#{m : 1 ≤ m ≤ n,m ∈ A}

n.

Show that µn is countably additive for every n on P(X). In a sense,µn is the proportion of integers between 1 to n which are in A. Let

C = {A ⊆ X | limn→∞

µn(A) exists}.

Show that C is closed under taking complements, finite disjointunions and proper differences. Is it an algebra?

2.1. Set functions 13

(2.3) Let µ : I ∩ (0, 1] −→ [0,∞] be defined by

µ(a, b] :=

{

b− a if a 6= 0, 0 < a < b ≤ 1,+∞ otherwise.

(Recall that I ∩ (0, 1] is the class of all left-open right-closed inter-vals in (0, 1].)Show that µ is finitely additive. Is µ countably additive also?

(2.4) Let X be a nonempty set.(a) Let µ : P(X) −→ [0,∞) be a finitely additive set function

such that µ(A) = 0 or 1 for every A ∈ P(X). Let

U = {A ∈ P(X) | µ(A) = 1}.

Show that U has the following properties:(i) ∅ 6∈ U .(ii) If A ∈ X and B ⊇ A, then B ∈ U .(iii) If A,B ∈ U , then A ∩B ∈ U .(iv) For every A ∈ P(X), either A ∈ U or Ac ∈ U .

(Any U ⊆ P(X) satisfying (i) to (iv) is called an ultrafilterin X.)

(b) Let U be any ultrafilter in X. Define µ : P(X) −→ [0,∞) by

µ(A) :=

{

1 if A ∈ U ,0 if A 6∈ U .

Show that µ is finitely additive.

(2.5) Let A be an algebra of subsets of a set X.(i) Let µ1, µ2 be measures on A, and let α and β be nonnegative

real numbers. Show that αµ1 + βµ2 is also a measure on A.(ii) For any two measures µ1, µ2 on A, we say

µ1 ≤ µ2 if µ1(E) ≤ µ2(E), ∀ E ∈ A.

Let {µn}n≥1 be a sequence of measures on A such that

µn ≤ µn+1, ∀ n ≥ 1.

Define ∀ E ∈ A,

µ(E) := limn→∞

µn(E).

Show that µ is also a measure on A and ∀ E ∈ B,

µ(E) = sup {µn(E) | n ≥ 1}.

(2.6) Let X be a compact topological space and A be the collection ofall those subsets of X which are both open and closed. Show thatA is an algebra of subsets of X. Further, every finitely additive setfunction on A is also countably additive.

14 2. Measure

2.2. Countably additive set functions on

intervals


We saw in the previous section that the length function is a countably ad-ditive set function on the class of all intervals. One can ask the question:do there exist countably additive set functions on intervals, other than thelength function? The answer is given by the following:

2.2.1. Proposition:Let F : R −→ R be a monotonically increasing function. Let µF : I −→[0,∞] be defined by

µF (a, b ] := F (b) − F (a),

µF (−∞, b ] := limx→∞

[F (b) − F (−x)],

µF (a,∞) := limx→∞

[F (x) − F (a)],

µF (−∞,∞) := limx→∞

[F (x) − F (−x)].

Then, µF is a well-defined finitely additive set function on I. Further, µF

is countably additive if F is right continuous.

One calls µF the set function induced by F .

The converse of proposition 2.2.1 is also true.

2.2.2. Proposition:Let µ : I −→ [0,∞] be a finitely additive set function such that µ(a, b] < +∞for every a, b ∈ R. Then there exists a monotonically increasing functionF : R −→ R such that µ(a, b] = F (b) − F (a) ∀ a, b ∈ R. If µ is alsocountably additive, then F is right-continuous.

(Hint: Define F as follows:

F (x) :=

µ(0, x] if x > 0,0 if x = 0,

−µ(x, 0] if x < 0.)

2.2.3. Remarks:

(i) In case µ(R) < +∞, a more canonical choice for the required function Fin proposition 2.2.2 is given by F (x) := µ(−∞, x], x ∈ R.

2.3. Set functions on algebras 15

(ii) Propositions 2.2.1 and 2.2.2 completely characterize the non-trivial count-ably additive set functions on intervals in terms of functions F : R −→ R

which are monotonically increasing and right continuous. Such functionsare called distribution functions on R. The set function µF induced bythe distribution function F is non-trivial in the sense that it assigns finitenon-zero values to bounded intervals.

Exercises:

(2.7) Let F (x) = [x], the integral part of x, x ∈ R. Describe the setfunction µF .

(2.8) α ∈ R. Show that F1 := F + α is also a distribution function andµF = µF1

. Is the converse true?

(2.9) (i) Let C be a collection of subsets of a set X and µ : C → [0,∞]be a set function. If µ is a measure on C, show that µ is finitelyadditive. Is µ monotone? Countably subadditive?

(ii) If C be a semi-algebra, then µ is countably subadditive iff ∀ A ∈C with A ⊆

⋃∞i=1Ai, Ai ∈ C implies

µ(A) ≤∞∑

i=1

µ(Ai).

2.3. Set functions on algebras


In this section, we give some general properties of a set function µ definedon an algebra A of subsets of an arbitrary set X.

2.3.1. Theorem:Let A be an algebra of subsets of a set X and let µ : A −→ [0,∞] be a setfunction. Then the following hold:

(i) If µ is finitely additive and µ(B) < +∞, then µ(B − A) = µ(B) − µ(A)for every A,B ∈ A with A ⊆ B. In particular, µ(∅) = 0 if µ is finitelyadditive and µ(B) < +∞ for some B ∈ A.

(ii) If µ is finitely additive, then µ is also monotone.

(iii) Let µ(∅) = 0. Then µ is countably additive iff µ is both finitely additiveand countably subadditive.

16 2. Measure

Another characterization of countable additivity of set functions definedon algebras is given in the next theorem.

2.3.2. Theorem:Let A be an algebra of subsets of a set X and let µ : A −→ [0,∞] be suchthat µ(∅) = 0.

(a) If µ is countably additive then the following hold:(i) For any A ∈ A, if A =

⋃∞n=1An, where An ∈ A and An ⊆

An+1 ∀ n, then

µ(A) = limn→∞

µ(An).

This is called the continuity from below of µ at A.(ii) For any A ∈ A, if A =

⋂∞n=1An, where An ∈ A with An ⊇

An+1 ∀ n and µ(An) < +∞ for some n, then

limn→∞

µ(An) = µ(A).

This is called the continuity from above of µ at A.

Conversely,

(b) If µ is finitely additive and (i) holds, then µ is countably additive.

(c) If µ(X) < +∞, µ is finitely additive and (ii) holds, then µ is count-ably additive.

Exercises:

(2.10 ) (i) In the proofs of part (ii) and part (c) of theorem 2.3.2, where doyou think we used the hypothesis that µ(X) < +∞? Do you thinkthis condition is necessary?

(ii) Let A be an algebra of subsets of a set X and µ : A → [0,∞] bea finitely additive set function such that µ(X) < +∞. Show thatthe following statements are equivalent:(a) lim

k→∞µ(Ak) = 0, whenever {Ak}k≥1 is a sequence in A with

Ak ⊇ Ak+1 ∀ k, and⋂∞

k=1Ak = ∅.(b) µ is countably additive.

(2.11 ) Extend the claim of theorem 2.3.2 when A is only a semi-algebraof subsets of X.(Hint: Use exercise 1.8)

(2.12 ) Let A be a σ-algebra and µ : A → [0,∞] be a measure. For anysequence {En}n≥1 in A, show that

2.4. Uniqueness problem for measures 17

(i) µ (lim infn→∞En) ≤ lim infn→∞ µ(En).(ii) µ (lim supn→∞En) ≥ lim supn→∞ µ(En).

(Hint: For a sequence {En}n≥1 of subsets of a set X,

lim infn→∞

En :=∞⋃

n=1

∞⋂

k=n

Ek ⊆ lim supn→∞

En :=∞⋂

n=1

∞⋃

k=n

Ek. )

2.4. Uniqueness problem for measures


The problem that we want to analyze is the following: Let µ be a σ-finitemeasure on an algebra A of subsets of X. Let µ1 and µ2 be two measureson S(A), the σ-algebra generated by the algebra A, such that µ1(A) =µ2(A) ∀ A ∈ A. Is µ1 = µ2?

2.4.1. Definition:Let C be a collection of subsets ofX and let µ : C −→ [0,∞] be a set function.We say µ is totally finite (or just finite) if µ(A) < +∞ ∀ A ∈ C. Theset function µ is said to be sigma finite (written as σ-finite) if there existpairwise disjoint sets Xn ∈ C, n = 1, 2, . . . , such that µ(Xn) < +∞ for everyn and X =

⋃∞n=1Xn.

2.4.2. Examples:

(i) The length function λ on the class of intervals is σ-finite.

(ii) Let A = A(I), the algebra generated by left-open right-closed intervalsin R. For A ∈ A, let µ(A) = +∞ if A 6= ∅ and µ(∅) = 0. Then µ is ameasure on A and it is not σ-finite. Let ξ ∈ R be chosen arbitrarily andfixed. Let Aξ denote the algebra of subsets of R generated by A and {ξ}.Define for A ∈ Aξ,

µξ(A) :=

{

+∞ if A \ {ξ} 6= ∅,0 if either A = ∅ or A = {ξ}.

It is easy to check that µξ is also a measure on Aξ and is not σ-finite.

(iii) Let X denote the set of rationals in (0,1] and let A be as in (ii) above.Show that the σ-algebraX∩S(A) = P(X) and that every nonempty set inthe algebra X∩A has an infinite number of points. For any E ∈ X∩S(A)and c > 0, define µc(E) = c times the number of points in E. Show thatµc is a measure on X ∩ S(A) = S(X ∩ A) and is not σ-finite.

18 2. Measure

2.4.3. Proposition:Let µ1 and µ2 be totally-finite measures on a σ-algebra S. Then the classM = {E ∈ S | µ1(E) = µ2(E)} has the following properties:

(i) M is a monotone class.

(ii) If S = S(A), and µ1(A) = µ2(A) ∀ A ∈ A, then µ1(A) = µ2(A) ∀A ∈S(A).

Exercises:

(2.13 ) Let A be an algebra of subsets of a set X. Let µ1 and µ2 be σ-finitemeasures on a σ-algebra S(A) such that µ1(A) = µ2(A) ∀ A ∈ A.Then, µ1(A) = µ2(A) ∀A ∈ S(A).

(2.14) Show that a measure µ defined on an algebra A of subsets of a setX is finite if and only if µ(X) < +∞.

Chapter 3

Construction of

measures


3.1. Extension from semi-algebra to the

generated algebra

3.1.1. Definition:Let Ci, i = 1, 2 be classes of subsets of a set X, with C1 ⊆ C2. Let functionµ1 : C1 −→ [0,+∞] and µ2 : C2 −→ [0,+∞] be set functions. The setfunction µ2 is called an extension of µ1 if µ1(E) = µ2(E) for every E ∈ ⌋1.

3.1.2. Examples:In example 2.4.2(ii), each µξ is an extension of the measure µ. Similarly, inexample 2.4.2(iii) each µc is an extension of µ.

Above examples show that in general a measure µ on an algebra A canhave more than one extension to S(A), the σ-algebra generated by A.

Our next theorem describes a method of uniquely extending a measurefrom a semi-algebra to the algebra generated by it.

3.1.3. Theorem:Given a measure µ on a semi-algebra C, there exists a unique measure µ onF(C) such that µ(E) = µ(E) for every E ∈ C.

The measure µ is called the extension of µ.

19

20 3. Construction of measures

[Hint: For E ∈ F(C), with E =⋃n

i=1Ei for pairwise disjoint setsE1, . . . , En ∈C, define

µ(E) :=∞∑

i=1

µ(Ei).]

Exercises:

(3.1) Let C be a collection of subsets of a set X and µ : C → [0,∞] be aset function. If µ is a measure on C, show that µ is finitely additive.Is µ monotone? Countably subadditive?

(3.2) If C be a semi-algebra, then µ is countably subadditive iff ∀ A ∈ Cwith A ⊆

⋃∞i=1Ai, Ai ∈ C implies

µ(A) ≤∞∑

i=1

µ(Ai).

(3.3) Using theorem 3.1.3, show that length function, which is initiallydefined on the semi-algebra I of all intervals, can be uniquely ex-tended to a set function on F(I), the algebra generated. It isworth mentioning a result due to S.M. Ulam (1930) which, underthe assumption of the “continuum hypothesis”, implies that it isnot possible to extend the length function to all subsets of R.Theorem (Ulam): Let µ be a measure defined on all subsets ofR such that µ((n, n + 1]) < ∞ ∀ n ∈ Z and µ({x}) = 0 for everyx ∈ R. Then µ(E) = 0 for every E ⊆ R.

3.1.3. Remark:Ulam’s theorem shows the impossibility of extending the length functionfrom intervals to all subsets of R, assuming the continuum hypothesis. Weshall see later that similar results can be proved if one assumes the ‘axiomof choice’. Ulam’s result uses the property of λ that λ({x}) = 0 ∀ x ∈ R,and the fact that λ([n, n + 1]) < +∞ for every n ∈ Z. In the later resultswe shall use the translation invariance property of the length function λ.

3.2. Extension from algebra to the generated

σ-algebra

Concepts and Examples:

Given an arbitrary measure µ on an algebra A of subsets of a set X, ouraim is to try to extend µ to a class of subsets of X which is larger than

3.2. Extension from algebra to the generated σ-algebra 21

A. Intuitively, sets A in A are those whose size µ(A) can be measuredaccurately. The approximate size of any set E ⊆ X is given by the outermeasure as defined next. Recall that, for any nonempty set A ⊆ [0,+∞],we write inf(A) := inf A ∩ [0,+∞) if A ∩ [0,+∞) 6= ∅, and inf(A) := +∞otherwise.

3.2.1. Definition:Let A be an algebra of subsets of a set X and µ : A −→ [0,∞] be a measureon A. For E ⊆ X, define

µ∗(E) := inf

{

∞∑

i=1

µ(Ai) Ai ∈ A,∞⋃

i=1

Ai ⊇ E

}

.

The set function µ∗ is called the outer measure induced by µ.

3.2.2. Proposition (Properties of outer measure):The set function µ∗ : P(X) −→ [0,∞] has the following properties:

(i) µ∗(∅) = 0 and µ∗(A) ≥ 0 ∀ A ⊆ X.

(ii) µ∗ is monotone, i.e.,

µ∗(A) ≤ µ∗(B) whenever A ⊆ B ⊆ X.

(iii) µ∗ is countably subadditive, i.e.,

µ∗(A) ≤∞∑

i=1

µ∗(Ai) whenever A =∞⋃

i=1

Ai.

(iv) µ∗ is an extension of µ , i.e.,

µ∗(A) = µ(A) if A ∈ A.

3.2.3. Remarks:(i) A set function ν defined on all subsets of a set X is called an outermeasure if ν has properties (i), (ii) and (iii) in proposition 3.2.2. The outermeasure µ∗ induced by µ is characterized by the property that if ν is anyouter measure on X such that ν(A) = µ(A) ∀ A ∈ A, then µ∗(A) ≥ ν(A).In other words, µ∗ is the largest of all the outer measures which agree withµ on A.

(ii) In the definition of µ∗(E) the infimum is taken over the all possiblecountable coverings of E. To see that finite coverings will not suffice, con-sider E := Q ∩ (0, 1), the set of all rationals in (0, 1), and let I1, I2, . . . , Inbe any finite collection of open intervals such that E ⊆

⋃ni=1 Ii. Then it is

easy to see that∑n

i=1 λ(Ii) ≥ 1. This will imply λ∗(E) ≥ 1 if only finitecoverings are considered in the definition of λ∗, which contradicts the factthat λ∗(E) = 0, E being a countable set.


3.2.4. Example:Let

A := {A ⊆ R | Either A or Ac is countable}.

It is easy to see that A is a σ-algebra. For A ∈ A, let µ(A) = 0 if A iscountable and µ(A) = 1 if Ac is countable. Then, µ is a measure on A. Letµ∗ be the outer measure induced by µ on P(R). It follows from proposition3.2.2 that µ∗ is countably subadditive on P(R).

If A ⊂ R is countable, then clearly A ∈ A, and hence µ∗(A) = µ(A) = 0.Further, µ∗(A) = 1 iff A is uncountable. Since, R = (−∞, 0] ∪ (0,∞) and

µ∗(R) = 1 < 2 = µ(−∞, 0] + µ(0,∞).

This shows that µ∗ need not be even finitely additive on all subsets.

Exercises:

(3.4) Show that µ∗(E), as in definition 3.2.1, is well-defined.

(3.5) The set function µ∗(E) can take the value +∞ for some sets E.

(3.6) Show that

µ∗(E) = inf

{

∞∑

i=1

µ(Ai) Ai ∈ A, Ai ∩Aj = ∅ for i 6= j and

∞⋃

i=1

Ai ⊇ E

}

.

(3.7) Let X be any nonempty set and let A be any algebra of subsets ofX. Let x0 ∈ X be fixed. For A ∈ A, define

µ(A) :=

{

0 if x0 6∈ A,1 if x0 ∈ A.

Show that µ is countably additive. Let µ∗ be the outer measureinduced by µ. Show that µ∗(A) is either 0 or 1 for every A ⊆ X,and µ∗(A) = 1 if x0 ∈ A. Can you conclude that µ∗(A) = 1 impliesx0 ∈ A? Show that this is possible if {x0} ∈ A.

3.3. Choosing nice sets: Measurable sets


In the previous section we defined the notion of µ∗, the outer measure in-duced by µ on all subsets of X. We saw that µ∗(A) = µ(A), ∀ A ∈ A, butin general µ∗ need not be even finitely additive on P(X). Let us try to iden-tify some subclass S of P(X) such that µ∗ restricted to S will be countableadditive. This is the class S which we call the class of ‘nice’ subsets of X.

3.3. Choosing nice sets: Measurable sets 23

But the problem is how to pick these ‘nice’ sets? This motivates our nextdefinition.

3.3.1. Definition:A subset E ⊆ X is said to be µ∗-measurable if for every Y ⊆ X,

µ∗(Y ) = µ∗(Y ∩ E) + µ∗(Y ∩ Ec). (3.1)

We denote by S∗ the class of all µ∗-measurable subsets of X. Note thatE ∈ S∗ iff Ec ∈ S∗, due to the symmetry in equation (3.1).

Thus, a set E ⊆ X is a ‘nice’ set if we use it as a knife to cut any subsetY of X into two parts, Y ∩E and Y ∩Ec, so that their sizes µ∗(Y ∩E) andµ∗(Y ∩Ec) add up to give the size µ∗(Y ) of Y. Thus a ‘nice’ set is in a sensea ‘sharp’ knife.

3.3.2. Theorem:Let E ⊆ X. Show that the following statements are equivalent:

(i) E ∈ S∗.

(ii) For every Y ⊆ X,

µ∗(Y ) ≥ µ∗(Y ∩ E) + µ∗(Y ∩ Ec).

(iii) For every Y ⊆ X, with µ∗(Y ) < +∞,

µ∗(Y ) ≥ µ∗(Y ∩ E) + µ∗(Y ∩ Ec).

(iv) For every A ∈ A,

µ(A) ≥ µ∗(A ∩ E) + µ∗(A ∩ Ec).

We give an equivalent definition of measurable sets when µ(X) < +∞.

3.3.3. Theorem:Let µ(X) < +∞. Then E ⊆ X is µ∗-measurable iff

µ(X) = µ(E) + µ(Ec)

Next, we check that S∗ is indeed the required collection of ‘nice’ sets.

3.3.4. Proposition:The collection S∗ has the following properties:

(i) A ⊆ S∗.

(ii) S∗ is an algebra of subsets of X, S(A) ⊆ S∗, and µ∗ restricted toS∗ is finitely additive.


(iii) If An ∈ S∗, n = 1, 2, . . . , then⋃∞

n=1An ∈ S∗ and µ∗ restricted toS∗ is countably additive.

(iv) Let N := {E ⊆ X | µ∗(E) = 0}. Then N ⊆ S∗.

This together with proposition 2.4.3 gives us the following:

3.3.5. Theorem:Let µ be a measure on an algebra A of subsets of a set A. If µ is σ-finite, thenthere exists a unique extension of µ to a measure µ on S(A), the σ-algebragenerated by A.

3.3.6. Remark:Theorems 3.2.2.(iv) and 3.3.4 together give us a method of constructing anextension of a measure µ defined on an algebra A to a class S∗ ⊇ S(A) ⊃ A.

Exercises:

(3.8) Identify the collection of µ∗-measurable sets for µ as in example3.2.4.

(3.9) Let X = [a, b] and let S be the σ-algebra of subsets of X generatedby all subintervals of [a, b]. Let µ, ν be finite measures on S such thatµ([a, c]) = ν([a, c]), ∀ c ∈ [a, b]. Show that µ(E) = ν(E) ∀ E ∈ S.

(3.10) Let µF be the measure on the algebra A(I) as given in proposition2.2.1. Let µF itself denote the unique extension of µF to LF , theσ-algebra of µ∗F -measurable sets, as given by theorem 3.3.4. Showthat(i) BR ⊆ LF .(ii) µF ({x}) = F (x) − lim

y↑xF (y). Deduce that the function F is

continuous at x iff µF ({x}) = 0.(iii) Let F be differentiable with bounded derivative. If A ⊆ R is

a null set, then µ∗F (A) = 0.The measure µF is called the Lebesgue-Stieltjes measure in-duced by the distribution function F.

3.4. Completion of a measure space


Theorem 3.3.4 showed that, given a σ-finite measure µ on an algebra A ofsubsets of a set X, µ can be extended to a unique measure µ∗ on the σ-algebra S∗ of µ∗-measurable subsets of X, and S∗ ⊇ S(A). In this sectionwe describe the relation between S∗ and the sets in S(A).

3.4. Completion of a measure space 25

We first give an equivalent ways of describing µ∗(E) for any set E ⊆X,µ∗ being the outer measure induced by µ. Let Aσ denote the collectionof sets of the form

⋃∞i=1Ai, Ai ∈ A.

3.4.1. Proposition:For every set E ⊆ X,

µ∗(E) = inf {µ∗(A) | A ∈ Aσ, E ⊆ A}

= inf {µ∗(A) | A ∈ S(A), E ⊆ A}

= inf {µ∗(A) | A ∈ S∗, E ⊆ A}.

3.4.2. Proposition:For every E ⊆ X, there exists a set F ∈ S(A) such that E ⊆ F, µ∗(E) =µ∗(F ) and µ∗(F \ E) = 0.

The set F is called a measurable cover of E.

3.4.3. Corollary:Let E ⊆ X. Then there exists a set K ⊆ E,K ∈ S(A), such that µ∗(A) = 0for every set A ⊆ E \K.

The set K is called a measurable kernel of E.

3.4.4. Definition:Let X be a nonempty set, S a σ-algebra of subsets of X and µ a measure onS. The pair (X,S) is called a measurable space and the triple (X,S, µ) iscalled a measure space. Elements of S are normally called measurablesets.

Till now what we have done is that, given a measure on an algebra A ofsubsets of a set X, we have constructed the measure spaces (X,S(A), µ∗),(X,S∗, µ∗) and exhibited the relations between them. The measure space(X,S∗, µ∗) has the property that if E ⊆ X and µ∗(E) = 0, then E ∈ S∗. Thisproperty is called the completeness of the measure space (X,S∗, µ∗).The measure space (X,S(A), µ∗) need not be complete in general. However,S∗ is obtainable from S(A) and N := {E ⊆ X | µ∗(E) = 0} by

S∗ = S(A) ∪N := {E ∪N | E ∈ S(A), N ∈ N}.

One calls (X,S∗, µ∗) the completion of (X,S(A), µ). This construction canbe put in a general context as follows.

3.4.5. Definition:Let (X,S, µ) be a measure space and let N := {E ⊆ X | E ⊆ N for some


N ∈ S with µ(N) = 0}. One says (X,S, µ) is complete if N ⊆ S. Elementsof N are called the µ-null subsets of X.

The abstraction of the relation between the measure spaces (X,S(A), µ∗)and (X,S∗, µ∗) is described in the next theorem.

3.4.6. Theorem:Let (X,S, µ) be a measure space and let N be the class of µ-null sets (as indefinition 3.4.5). Let S △ N := {E △ N | E ∈ S, N ∈ N} and S ∪ N :={E ∪N |E ∈ S, N ∈ N}. Then S △N = S ∪N is a σ-algebra of subsets ofX. Let µ(E△N) = µ(E), ∀ E ∈ S, N ∈ N . Then µ is a measure on S ∪Nand (X,S∪N , µ) is a complete measure space, called the completion of themeasure space (X,S, µ). (The measure space (X,S ∪ N , µ) is also denotedby (X,S, µ).

Finally we describe the relation between µ∗ on P(X) and µ on A.

3.4.7. Proposition:Let µ be a measure on an algebra A of subsets of a set X and let µ∗ be theinduced outer measure. Let E ∈ S∗ be such that µ∗(E) < +∞ and let ǫ > 0be arbitrary. Then there exists a set Fǫ ∈ A such that µ∗(E △ Fǫ) < ǫ.

3.4.8. Note:Whenever (X,S, µ) is a finite measure space with µ(X) = 1, it is called aprobability space and the measure µ is called a probability. The reasonfor this terminology is that the triple (X,S, µ) plays a fundamental rolein the axiomatic theory of probability. It gives a mathematical model foranalyzing statistical experiments. The setX represents the set of all possibleoutcomes of the experiment, the σ -algebra S represents the collection ofevents of interest in that experiment, and for every E ∈ S, the nonnegativenumber µ(E) is the probability that the event E occurs. For more detailssee Kolmogorov [9] and Parthasarathy [10].

Exercises:

(3.11 ) Let E ⊆ X, and let G1, G2 be two measurable covers of E. Showthat µ∗(G1∆G2) = 0.

(3.12) Let E1 ⊆ E2 ⊆ E3 ⊆ . . . be subsets of X. Then

µ∗

(

∞⋃

n=1

En

)

= limn→∞

µ∗(En).

(3.13 ) Let E ⊆ X, and let K1,K2 be two measurable kernels of E. Showthat µ∗(K1∆K2) = 0.

3.5. The Lebesgue measure 27

(3.14 ) Let N := {E ⊆ X | µ∗(E) = 0}. Show that N is closed undercountable unions and

S∗ = S(A) ∪N := {E ∪N | E ∈ S(A), N ∈ N},

where S∗ is the σ-algebra of µ∗-measurable sets. Further, ∀ A ∈ S∗

µ∗(A) = µ∗(E), if A = E ∪N, with E ∈ S(A) and N ∈ N .

3.5. The Lebesgue measure


We now apply the extension theory of measures, developed in previous sec-tions, to the particular case when X = R,A = A(I), the algebra generatedby all intervals, and µ on A is the length function λ as described in section

3.1. The outer measure λ∗, induced by the length function λ, on all sub-

sets of R is called the Lebesgue outer measure and can be described asfollows: for E ⊆ R,

λ∗(E) := inf

{

∞∑

i=1

λ(Ii) Ii ∈ I ∀ i, Ii ∩ Ij = ∅ for i 6= j and E ⊆∞⋃

i=1

Ii

}

.

The σ-algebra of λ∗-measurable sets, as obtained in section 3.4, is called the

σ-algebra of Lebesgue measurable sets and is denoted by LR, or simplyby L. The σ-algebra S(I) = S(A) := BR, generated by all intervals, is called

the σ-algebra of Borel subsets of R. We denote the restriction of λ∗

to L orBR by λ itself. The measure space (R,L, λ) is called the Lebesgue mea-sure space and λ is called the Lebesgue measure. We note that sinceλ on I is σ-finite (e.g., R =

⋃+∞n=−∞(n, n + 1]), the extension of λ to BR is

unique. It is natural to ask the question:What is the relation between the classes BR,L and P(R)?

As a special case of theorem 3.4.6, we have L = BR ∪N , where

N := {N ⊆ R | N ⊆ E ∈ BR, λ(E) = 0}.

Thus,

BR ⊆ L ⊆ P(R).

The question arises:

Is BR a proper subset of L?

That is, are there sets in L which are not Borel sets?. First of all, we note

that Cantor’s ternary set C ∈ N ⊂ L. Further, E ⊆ C, then λ∗(E) = 0

and hence E ∈ L. In other words, P(C) ⊆ L. Thus, the cardinality of L


is at least 2c (here c denotes the cardinality of the real line, also called thecardinality of the continuum). Since L ⊆ P(R), we get the cardinalityof L to be 2c. On the other hand, BR is the σ-algebra generated by allopen intervals of R with rational endpoints. One can show that the σ-algebra BR of Borel subsets of R has cardinality c, that of the continuum.Thus, there exist sets which are Lebesgue measurable but are not Borel sets.The actual construction of such sets is not easy. One such class of sets iscalled analytic sets. An analytic set is a set which can be represented as acontinuous image of a Borel set. For a detailed discussion on analytic sets,see Srivastava [38], Parthasarathy [29].

Since L, the class of all Lebesgue measurable subsets of R, has 2c ele-ments, i.e., same as that of P(R), the natural question arises:

Is L = P(R)?

We stated earlier that, if we assume the continuum hypothesis, it is notpossible to define a countably additive set function µ on P(R) such thatµ({x}) = 0 ∀ x ∈ R. In particular, if we assume the continuum hypothesis,we cannot extend λ to all subsets of R. Hence L 6= P(R). What can be saidif one does not assume the continuum hypothesis? To answer this question,one can either try to construct a set E ⊂ R such that E 6∈ L ,or, assumingthat such a set exists, try to see whether one can reach a contradiction.G. Vitali (1905), F. Bernstein (1908), H. Rademacher (1916) and othersconstructed such sets assuming the ‘axiom of choice’ (see appendix B). Theexample of Vitali used the translation invariance property of the Lebesguemeasure, and that of Bernstein used the regularity properties of the Lebesguemeasure. Rademacher proved that every set of positive outer Lebesguemeasure includes a Lebesgue nonmeasurable set. Even today, more andmore nonmeasurable sets with additional properties are being constructed.For example, one can construct nonmeasurable subsets A of R such that

λ∗(A∩ I) = λ

∗(I) for every interval I ⊂ R. Of course, all these constructions

are under the assumption of the ‘axiom of choice’. Lebesgue himself didnot accept such constructions. In 1970, R. Solovay [37] proved that if oneincludes the statement “all subsets of R are Lebesgue measurable” as anaxiom in set theory, then it is consistent with the other axioms of set theoryif the axiom of choice is not assumed. Construction of a nonmeasurableset(due to Vitali), assuming the axiom of choice, is given in exercise (3.26).

We recall that the σ-algebra BR includes all topologically ‘nice’ subsetsof R, such as open sets, closed sets and compact sets. Also, for E ∈ BR, ifwe transform E with respect to the group operation on R, e.g., for x ∈ R,consider E + x := {y+ x | y ∈ E}, then E + x ∈ BR. For this, note that themap y 7−→ x + y is a homeomorphism of R onto R, and hence E + x ∈ Bfor every open set E. We leave it for the reader to verify (using σ-algebra


techniques) that this is true for all sets E ∈ BR. The relation of λ on Lwith λ on topologically nice subsets of R and the question as to whetherE + x ∈ L for E ∈ L, x ∈ R, i.e., do the group operations on R preserve theclass of Lebesgue measurable sets, will be analyzed in this section. We give

below some properties of λ∗

which are also of interest.

3.5.1. Theorem:Let E ⊆ R and λ

∗(E) < +∞. Then, given ǫ > 0, there exists a set Fǫ which

is a finite disjoint union of open intervals and is such that

λ∗(E △ Fǫ) < ǫ.

We give next some more characterizations of Lebesgue measurable sets.

3.5.2. Theorem:For any set E ⊆ R the following statements are equivalent:

(i) E ∈ L, i.e., E is Lebesgue measurable.

(ii) For every ǫ > 0, there exists an open set Gǫ such that

E ⊆ Gǫ and λ∗(Gǫ \ E) < ǫ.

(iii) For every ǫ > 0, there exists a closed set Fǫ such that

Fǫ ⊆ E and λ∗(E \ Fǫ) < ǫ.

(iv) There exists a Gδ-set G such that

E ⊆ G and λ∗(G \ E) = 0.

(v) There exists an Fσ-set F such that

F ⊆ E and λ∗(E \ F ) = 0.

[Hint: Prove the following implications:

(i) =⇒ (ii) =⇒ (iv) =⇒ (i)

and

(i) =⇒ (iii) =⇒ (v) =⇒ (i).]

3.5.3. Note:Theorem 3.5.2 tells us the relation between L, the class of Lebesgue mea-surable sets, and the topologically nice sets, e.g., open sets and closed sets.The property that for E ∈ L and ǫ > 0, there exists an open set G ⊇ Ewith λ(G \ E) < ǫ can be stated equivalently as:

λ(E) = inf{λ(U) | U open, U ⊇ E}.


This is called the outer regularity of λ. Other examples of outer regularmeasures on R (in fact any metric space) are given in the exercise 3.25.

Another topologically nice class of subsets of R is that of compact subsetsof R. It is natural to ask the question: does there exist a relation betweenL and the class of compact subsets of R? Let K be any compact subset ofR. Since K is closed (and bounded), clearly K ∈ BR ⊂ L and λ(K) < +∞.It is natural to ask the question: can one obtain λ(E) for a set E ∈ BR, ifλ(E) < +∞, from the knowledge of λ(K),K compact in R? The answer isgiven by the next proposition.

3.5.4. Proposition:Let E ∈ L with 0 < λ(E) < +∞ and let ǫ > 0 be given. Then there exists acompact set K ⊆ E such that λ(E \K) < ǫ.

On the set R, we have the group structure given by the binary operation ofthe addition of two real numbers. We analyze the behavior of λ on L underthe map y 7−→ y + x, y ∈ R and x ∈ R fixed. We saw that A + x is aLebesgue measurable set whenever A is Lebesgue measurable and x ∈ R. Itis natural to ask the question: for A ∈ L and x ∈ R, is λ(A + x) = λ(A)?The answer is given by the following:

3.5.5. Theorem (Translation invariance property):Let E ∈ L. Then E + x ∈ L for every x ∈ R, and λ(E + x) = λ(E).

We saw that Lebesgue measure is the unique extension of the length functionfrom the class I of intervals to BR, the σ-algebra of Borel subsets of R. Thisgave us a measure λ on BR with the following properties:

(i) For every nonempty open set U, λ(U) > 0.

(ii) For every compact set K, λ(K) < +∞.

(iii) For every E ∈ BR,

λ(E) = inf{λ(U) | U open, U ⊇ E},= sup{λ(C) | C ⊆ E,C closed}.

If λ(E) < +∞, then we also have

λ(E) = sup{λ(K) | K ⊆ E,K compact}.

(iv) For every E ∈ BR and x ∈ R, E + x ∈ BR and λ(E + x) = λ(E).

Thus the Lebesgue measure is a translation invariant σ-finite regularmeasure on BR. The question arises: are there other σ-finite measures onBR with these properties? Obviously, if c > 0 then cλ defined by (cλ)(E) :=cλ(E), E ∈ BR, is also a σ-finite measure and is translation invariant. Infact the following hold:


3.5.6. Theorem:Let µ be a measure on BR such that

(i) µ(U) > 0 for every nonempty open set U ⊆ R.

(ii) µ(K) < +∞ for every compact set K ⊂ R.

(iii) µ(E + x) = µ(E), ∀ E ∈ BR and ∀ x ∈ R.

Then there exists a positive real number c such that µ(E) = cλ(E)∀ E ∈ BR.

3.5.7. Note:In fact the above theorem has a far-reaching generalization to abstract ‘topo-logical groups’. Let us recall that the set of real numbers R is a group underthe binary operation +, the addition of real numbers. Also, there is a topol-ogy on R which ‘respects’ the group structure, i.e., the maps (t, s) 7−→ t+ sand t −→ −t from R × R −→ R and R −→ R, respectively, are continuouswhen R × R is given the product topology. In an abstract setting, if G is aset with a binary operation ‘·’ and a topology T such that (G, ·) is a groupand the maps G × G −→ G, (g, h) 7−→ g.h and G −→ G, g 7−→ g−1 arecontinuous with respect to the product topology on G × G, one calls G atopological group. Given a topological group, let BG denote the σ-algebragenerated by open subsets of G, called the σ-algebra of Borel subsets of G.The question arises: does there exist a σ-finite measure µ on G such that ithas the properties as given in theorem 3.5.5? A celebrated theorem due to A.Haar states that such a measure exists and is unique up to a multiplicative(positive) constant if G is locally-compact. Such a measure is called a (right)Haar measure on G. Theorem 3.5.5 then states that for the topologicalgroup R, the Lebesgue measure λ is a Haar measure. Consider the group(R \ {0}, ·), where R \ {0} = {t ∈ R|t 6= 0} and ‘·’ is the usual multiplicationof real numbers. Let R \ {0} be given the subspace topology from R. It iseasy to show that R \ {0} is a topological group and, ∀ E ∈ BR\{0},

µ(E) :=

∫

E

1

|x|dλ(x) (3.2)

is a Haar measure on R \ {0}.

3.5.8. Note:In the previous sections we have seen how the general extension theory,as developed earlier, can be applied to the particular situation when thesemi-algebra is that of intervals and the set function is the length function.More generally, if we consider the semi-algebra I of left-open right-closedintervals in R and consider F : R −→ R as a monotonically increasingright continuous function, then we can construct a countably additive setfunction µF on the semi-algebra I, as in example 1.1.2. Using theorem


3.4.6, we can construct a complete measure µF on a σ-algebra of subsetsof R which includes BR. This measure µF is called the Lebesgue-Stieltjesmeasure induced by the function F. Note that µF has the property thatµF (a, b] < +∞ ∀ a, b ∈ R, a < b. Conversely, given a measure µ on BR

such that µ(a, b] < +∞ ∀ a, b ∈ R, a < b, we can restrict it to I and,using proposition 2.2.2, define a monotonically increasing right continuousfunction F : R −→ R such that the unique Lebesgue-Stieltjes measure µF

induced by F is nothing but µ (by the uniqueness of the extension). Thusmeasures µ on BR which have the property that µ(a, b] < +∞ ∀ a < bcan be looked upon as a Lebesgue-Stieltjes measure µF for some F. Wepoint out that it is possible to find different F1, F2 : R −→ R such thatboth are monotonically increasing and right continuous and µF1

= µF2. If

µ is finite measure, i.e., µ(R) < +∞, then it is easy to see that F (x) :=µ(−∞, x], x ∈ R, is a monotonically increasing right continuous functionsuch that µ = µF . This F is called the distribution function of µ. Whenµ(R) = 1, µ is called a probability and its distribution function F, which ismonotonically increasing and is right continuous with lim

x→∞[F (x) − F (−x)]

= µF (R) = 1, is called a probability distribution function.

Exercises:

(3.15) Let I0 denote the collection of all open intervals of R. For E ⊆ X,show that

λ∗(E) = inf

{

∞∑

i=1

λ(Ii) Ii ∈ I0 ∀ i, for i 6= j and E ⊆∞⋃

i=1

Ii

}

.

(3.16) Let E ⊆ R and let ǫ > 0 be arbitrary. Show that there exists an

open set Uǫ ⊇ E such that λ(Uǫ) ≤ λ∗(E)+ǫ. Can you also conclude

that λ(Uǫ \ E) ≤ ǫ?

(3.17) For E ⊆ R, let

diameter(E) := sup{|x− y| | x, y ∈ E}.

Show that λ∗(E) ≤ diameter(E).

(3.18) Show that for E ⊆ R, λ∗(E) = 0 if and only if for every ǫ > 0,

there exist a sequence {In}n≥1 of intervals such that E ⊆ ∪∞n=1 and

λ∗(∪∞

n=1 \ E) < ǫ. Such sets are called Lebesgue null sets.Provethe following:(i) Every singleton set {x}, x ∈ R, is a null set. Also every finite

set is a null set.

(ii) Any countably infinite set S = {x1, x2, x3, . . .} is a null set.(iii) Q, the set of rational numbers, is a null subset of R.


(iv) Every subset of a null set is also a null set.(iv) Let A1, A2, . . . , An, . . . be null sets. Then

⋃∞n=1An is a null

set.(v) Let E ⊆ [a, b] be any set which has only a finite number of

limit points. Can E be uncountable? Can you say E is a nullset?

(vi) Let E be a null subset of R and x ∈ R.What can you say aboutthe sets E + x := {y + x | y ∈ E} and xE := {xy | y ∈ E}?

(vii) Let I be an interval having at least two distinct points. Showthat I is not a null set.

(viii) If E contains an interval of positive length, show that it is nota null set. Is the converse true, i.e., if E ⊆ R is not a null set,then does E contain an interval of positive length?

(ix) Show that Cantor’s ternary set is an uncountable null set.

(3.19) Let E ⊆ [0, 1] be such that λ∗([0, 1] \E) = 0. Show that E is dense

in [0, 1]

(3.20) Let E ⊆ R be such that λ∗(E) = 0. Show that E has empty interior.

(3.21) Let {En}n≥1 be any increasing sequence of subsets (not necessarilymeasurable) of R. Then,

λ∗

(

∞⋃

n=1

En

)

= limn→∞

λ∗(En).

(3.22) Let E ⊆ R. Show that the following statements are equivalent:(i) E ∈ L.

(ii) λ∗(I) = λ

∗(E ∩ I) + λ

∗(Ec ∩ I) for every interval I.

(iii) E ∩ [n, n+ 1) ∈ L for every n ∈ Z.

(iv) λ∗(E ∩ [n, n+ 1)) + λ

∗(Ec ∩ [n, n+ 1)) = 1 for every n ∈ Z.

(3.23) Let A ∈ L and x ∈ R. Using theorem 4.2.2, show that(i) A+ x ∈ L, where A+ x := {y + x | y ∈ A}.(ii) −A ∈ L, where −A := {−y | y ∈ A}.

(3.24) Let (X, d) be any metric space and let µ be a measure on BX , theσ-algebra generated by open subsets of X, called the σ-algebra ofBorel subsets of X. The measure µ is called outer regular if∀ E ∈ BX ,

µ(E) = inf{µ(U) | U open, U ⊇ E},

= sup{µ(C) | C closed, C ⊆ E}. (3.3)

(i) If µ(X) < +∞, show that µ is outer regular iff for every E ∈BX and ǫ > 0 given, there exist an open set Uǫ and a closed


set Cǫ such that

Uǫ ⊇ E ⊇ Cǫ and µ(Uǫ − Cǫ) < ǫ.

(ii) For A ⊆ X, let

d(x,A) := inf{d(x, y) | y ∈ A}.

Show that for every A ⊆ X,x 7−→ d(x,A) is a uniformly con-tinuous function.(Hint: |d(x,A) − d(y,A)| ≤ d(x, y) ∀ x, y.)

(iii) Let µ(X) < +∞ and

S := {E ∈ BX | (3.3) holds for E}.

Show that S is a σ-algebra of subsets of X.(iv) Let C be any closed set in X. Show that C ∈ S.

(Hint: C =⋂∞

n=1{x ∈ X : d(x,C) < 1/n}.)(v) Show that µ is outer regular on BX .

(3.25) Let E ∈ BR. Show that E + x ∈ BR for every x ∈ R.

(3.26) Let E ∈ L and x ∈ R. Let

xE := {xy | y ∈ E} and − E := {−x | x ∈ E}.

Show that −E, xE ∈ L for every x ∈ E. Compute λ(xE) andλ(−E) in terms of λ(E).

(3.27) Example (Vitali)(Existence of nonmeasurable sets:Define a relation on [0,1] as follows: for x, y ∈ [0, 1], we say x is

related to y, written as x ∼ y, if x − y is a rational. Prove thefollowing:

(i) Show that ∼ is an equivalence relation on [0, 1].(ii) Let {Eα}α∈I denote the set of equivalence classes of elements

of [0, 1]. Using the axiom of choice, choose exactly one elementxα ∈ Eα for every α ∈ I and construct the set E := {xα|α ∈ I}.Let r1, r2, . . . , rn, . . . denote an enumeration of the rationals in[−1, 1]. Let

En := rn + E, n = 1, 2, . . . .

Show that En ∩Em = ∅ for n 6= m and En ⊆ [−1, 2] for every n.Deuce that

[0, 1] ⊆∞⋃

n=1

En ⊆ [−1, 2].

(iii) Show that E is not Lebesgue measurable.

Chapter 4

Integration

Unless stated otherwise, we shall work on a fixed σ-finite measure space(X,S, µ).

4.1. Integral of nonnegative simple measurable

functions


4.1.1. Definition:Let s : X −→ [0,∞] be defined by

s(x) =n∑

i=1

aiχAi(x), x ∈ X,

where n is some positive integer; a1, a2, . . . , an are nonnegative extended realnumbers; Ai ∈ S for every i;Ai ∩Aj = ∅ for i 6= j; and

⋃ni=1Ai = X. Such a

function s is called a nonnegative simple measurable function on (X,S)and

∑ni=1 aiχAi

(x) is called a representation of s. We say∑n

i=1 aiχAiis

the standard representation of s if a1, a2, . . . , an are all distinct.

We denote by L+0 the class of all nonnegative simple measurable func-

tions on (X,S).

Note that s ∈ L+0 iff s takes only a finite number of distinct values, say

a1, a2, . . . , an, the value ai being taken on the set Ai ∈ S, i = 1, 2, . . . , n.And in that case its standard representation is

∑ni=1 aiχAi

. Also note that

the class L+0 depends only upon the set X and the σ-algebra S; the measure

µ plays no part in the definition of functions in L+0 .

35

36 4. Integration

4.1.2. Examples:

(i) Clearly, if s(x) ≡ c for some c ∈ [0,+∞], then s ∈ L+0 .

(ii) For A ⊆ X, consider χA : X −→ [0,+∞], the indicator function of the

set A, i.e., χA(x) = 1 if x ∈ A and χA(x) = 0 if x 6∈ A. Then χA ∈ L+0 iff

A ∈ S, for χA = aχA + bχA

c with a = 1 and b = 0.

(iii) Let A,B ∈ S. Then s = χAχB ∈ L+0 , since s = χA∩B .

(iv) Let A,B ∈ S. If A ∩B = ∅, then clearly χA + χB = χA∪B ∈ L+0 .

4.1.3. Definition:For s ∈ L+

0 with a representation s =∑n

i=1 aiχAi, we define

∫

s(x)dµ(x),the integral of s with respect to µ, by

∫

s(x)dµ(x) :=n∑

i=1

aiµ(Ai).

The integral∫

s(x)dµ(x) is also denoted by∫

sdµ. It is well-defined (seeexercise (4.1)).

4.1.4. Proposition:For s, s1, s2 ∈ L+

0 and α ∈ R with α ≥ 0, the following hold:

(i) 0 ≤∫

sdµ ≤ +∞.

(ii) αs ∈ L+0 and

∫

(αs)dµ = α

∫

sdµ.

(iii) s1 + s2 ∈ L+0 and

∫

(s1 + s2)dµ =

∫

s1dµ+

∫

s2dµ.

(iv) For E ∈ S we have sχE ∈ L+0 , and the set function

E 7−→ ν(E) :=

∫

sχEdµ

is a measure on S. Further, ν(E) = 0 whenever µ(E) = 0, E ∈ S.

Before we proceed further, we observe that∫

s(x)dµ(x) is well-defined,(see

Exercise 4.1). The properties of∫

sdµ, for s ∈ L+0 , are given by the next

proposition.

4.1. Integral of nonnegative simple measurable functions 37

4.1.5. Proposition:Let s ∈ L+

0 . Then the following hold:

(i) If {sn}n≥1 is any increasing sequence in L+0 such that limn→∞ sn(x)

= s(x), x ∈ X, then∫

sdµ = limn→∞

∫

sndµ.

(ii)∫

sdµ = sup{

∫

s′

dµ | 0 ≤ s′

≤ s, s′

∈ L+0

}

.

Exercises:

(4.1) Show that for s ∈ L+0 ,∫

s(x)dµ(x) is well-defined by proving thefollowing: let

s =n∑

i=1

aiχAi=

m∑

j=1

bjχBj,

where {A1, . . . , An} and {B1, . . . , Bm} are partitions of X by ele-ments of S, then(ii)

s =n∑

i=1

ai

m∑

j=1

χAi∩Bj=

m∑

j=1

bj

n∑

i=1

χAi∩Bj.

(ii)n∑

i=1

aiµ(Ai) =m∑

j=1

bjµ(Bj).

(iii)∫

s(x)dµ(x) is independent of the representation of the func-tion s(x) =

∑ni=1 aiχAi

.

(4.2) Let A,B ∈ S. Express the functions |χA−χB | and χA +χB −χA∩B

as indicator functions of sets in S and hence deduce that theybelong to L+

0 .

(4.3) Let s1, s2 ∈ L+0 . Prove the following:

(i) If s1 ≥ s2, then∫

s1dµ ≥∫

s2dµ.(ii) Let ∀ x ∈ X,

(s1 ∨ s2)(x) := max{s1(x), s2(x)} and (s1 ∧ s2)(x) := min{s1(x), s2(x)}.

Then s1 ∧ s2 and s1 ∨ s2 ∈ L+0 with

∫

(s1 ∧ s2)dµ ≤

∫

sidµ ≤

∫

(s1 ∨ s2)dµ, i = 1, 2.

38 4. Integration

(4.4) Express the functions χA ∧χB and χA ∨χB , for A,B ∈ S, in termsof the functions χA and χB .

(4.5) Let X = (0, 1], S = B(0,1], the σ-algebra of Borel subsets of (0, 1]and µ = λ, the Lebesgue measure restricted to S. For x ∈ (0, 1], ifx has non-terminating dyadic expansion x =

∑∞n=1 xn/2

n. Let

fi(x) :=

{

+1 if xi = 1,−1 if xi = 0, i = 1, 2, . . . .

Show that for every i, there exists simple function si ∈ L+0 such

that fi = si − 1. Compute∫

sidλ.

(4.6) Let s : X −→ R∗ be any nonnegative function such that the rangeof s is a finite set. Show that s ∈ L+

0 iff s−1{t} ∈ S for every t ∈ R∗.

(4.7) For s1, s2 ∈ L+0 show that {x | s1(x) ≥ s2(x)} ∈ S. Can you say

that the sets {x ∈ X | s1(x) > s2(x)}, {x ∈ X | s1(x) ≤ s2(x)} and{x ∈ X | s1(x) = s2(x)} are also elements of S?

(4.8) Let s1, s2 ∈ L+0 be real valued and s1 ≥ s2. Let φ = s1 − s2. Show

that φ ∈ L+0 . Can you say that

∫

φdµ =

∫

s1dµ−

∫

s2dµ?

(4.9) Let {sn}n≥1 and {s′

n}n≥1 be sequences in L+0 such that for each

x ∈ X, both {sn(x)}n≥1 and {s′

n(x)}n≥1 are increasing and

limn→∞

sn(x) = limn→∞

s′

n(x).

Show that

limn→∞

∫

sndµ = limn→∞

∫

s′

ndµ.

(Hint: Apply exercise (4.3) and proposition 4.1.5 to {sn ∧ s′

m}n for

all fixed m to deduce that∫

s′

mdµ ≤ limn→∞

∫

sndµ.)

(4.10) Show that in general L+0 need not be closed under limiting oper-

ations. For example, consider the Lebesgue measure space (R,L, λ)and construct a sequence {sn}n≥1 in L+

0 such that limn→∞

sn(x) = f(x)

exists but f 6∈ L+0 .

4.2. Integral of nonnegative measurable

functions


Having defined the integral for functions s ∈ L+0 , i.e., nonnegative simple

measurable functions, we would like to extend it to a larger class.

4.2. Integral of nonnegative measurable functions 39

4.2.1. Definition:

(i) A nonnegative function f : X −→ R∗ is said to be S-measurable ifthere exists an increasing sequence of functions {sn}n≥1 in L+

0 such thatf(x) = lim

n→∞sn(x) ∀ x ∈ X.

If the underlying σ-algebra is clear from the context, a S-measurablefunction is also called measurable. We denote the set of all nonnegativemeasurable functions by L+.

(ii) For a function f ∈ L+, we define the integral of f with respect to µ by∫

f(x)dµ(x) := limn→∞

∫

sn(x)dµ(x).

It follows from exercise (4.9) that for f ∈ L+,∫

fdµ is well-defined. Clearly,

L+0 ⊆ L+ and

∫

sdµ for an element s ∈ L+0 is the same as

∫

sdµ, for s as anelement of L+. The next proposition gives a characterization of functions inL+ and the integrals of its elements. Another (intrinsic) characterization ofL+ will be given in the next section.

4.2.2. Proposition:Let f : X −→ R∗ be a nonnegative function. Then the following hold:

(i) f ∈ L+ iff there exist functions sn ∈ L+0 , n ≥ 1, such that 0 ≤ sn ≤ f ∀ n

and f(x) = limn→∞

sn(x) ∀ x ∈ X.

(ii) If f ∈ L+ and s ∈ L+0 is such that 0 ≤ s ≤ f, then

∫

sdµ ≤∫

fdµ and∫

fdµ = sup

{∫

sdµ 0 ≤ s ≤ f, s ∈ L+0

}

.

4.2.3. Definition:Let (X,S, µ) be a measure space and Y ∈ S. We say a property P holdsalmost everywhere on Y with respect to the measure µ if the set E = {x ∈Y | P does not hold at x} ∈ S and µ(E) = 0. We write this as P for a.e.x(µ) or P for a.e. (µ)x ∈ Y. If the set Y and µ are clear from the context,we shall simply write P a.e. For example if f : X −→ R is a function,then f(x) = 0 for a.e. x(µ) means that E = {x ∈ X | f(x) 6= 0} ∈ S andµ(E) = 0.

We describe next the properties of∫

fdµ, for f ∈ L+.

4.2.4. Proposition:Let f, f1, f2 ∈ L+. Then the following hold:

(i)∫

fdµ ≥ 0 and for f1 ≥ f2∫

f1dµ ≥

∫

f2 dµ.

40 4. Integration

(ii) For α, β ≥ 0 we have (αf1 + βf2) ∈ L+ and∫

(αf1 + βf2)dµ = α

∫

f1dµ+ β

∫

f2dµ.

(iii) For every E ∈ S we have χEf ∈ L+. If

ν(E) :=

∫

χEfdµ, E ∈ S,

then ν is a measure on S and ν(E) = 0 whenever µ(E) = 0.

The integral∫

fχEdµ is also denoted by∫

E fdµ and is called the integralof f over E.

(iv) If f1(x) = f2(x) for a.e. x(µ), then∫

f1dµ =

∫

f2dµ.

Since the class L+0 is not closed under limiting operations (exercise 4.10),

we defined the class L+ by taking limits of sequences in L+0 . Naturally, we

expect L+ to be closed under limits. The next theorem discusses this andthe behavior of

∫

fdµ under increasing limits, extending proposition 4.1.5to functions in L+.

4.2.5. Theorem (Monotone convergence):Let {fn}n≥1 be an increasing sequence of functions in L+, and f(x) :=

limn→∞

fn(x), x ∈ X. Then f ∈ L+ and∫

fdµ = limn→∞

∫

fndµ.

4.2.6. Remark:If {fn}n≥1 is a sequence in L+ decreasing to a function f ∈ L+, then theequality

∫

fdµ = limn→∞

∫

fndµ need not hold. For example, let X = R,

S = L and µ = λ, the Lebesgue measure. Let fn = χ[n,∞) . Then fn ∈ L+0 ⊆

L+, and {fn}n≥1 decreases to f ≡ 0. Clearly,∫

fndλ = +∞ for every n and∫

fdλ = 0. In fact, at this stage it is not clear whether f ∈ L+ whenever{fn}n≥1 decreases to f , with each fn ∈ L+. That this is true will be shownas a consequence of the characterization of L+ proved in the next section.

4.2. Integral of nonnegative measurable functions 41

Exercises:

(4.11) Let f ∈ L+ and let {sn}n≥1 be in L+0 and such that {sn(x)}n≥1

is decreasing and ∀ x ∈ X, limn→∞

sn(x) = f(x). Can you conclude

that∫

fdµ = limn→∞

∫

sndµ?

(4.12) Let f ∈ L+. Show that∫

fdµ = sup

{∫

sdµ 0 ≤ s(x) ≤ f(x) for a.e. x(µ), s ∈ L+0

}

.

(4.13) Let {fn}n≥1 be an increasing sequence of functions in L+ such thatf(x) := lim

n→∞fn(x) exists for a.e. x(µ). Show that f ∈ L+ and

∫

fdµ = limn→∞

∫

fndµ,

where f(x) is defined as an arbitrary constant for all those x forwhich lim

n→∞fn(x) does not converge.

(4.14) Let µ(X) < ∞, and let f ∈ L+ be a bounded function. Let P :={E1, E2, . . . , En} be such that

⋃ni=1Ei = X, Ei ∩ Ej = ∅ for i 6= j

and Ei ∈ S ∀ i. Such a P is called a measurable partition of X.Given a measurable partition P = {E1, . . . , En}, define

Mi := sup{f(x) |x ∈ Ei} and mi := inf{f(x) |x ∈ Ei}.

Let

ΦP :=n∑

i=1

miχEiand ΨP :=

n∑

i=1

MiχEi.

Prove the following:(i) For every partition P, show that ΦP ,ΨP ∈ L+

0 and ΦP ≤ f ≤ΨP .

(ii)∫

fdµ = sup{∫

ΦPdµ |P is a measurable partition of X}

,= inf

{∫

ΨPdµ |P is a measurable partition of X}

.

(This gives an equivalent way of defining∫

fdµ, in a way sim-ilar to that for the Riemann integral.)

(iii) Let

α = sup

{∫

0sdµ | s ∈ L+

0 , s ≤ f

}

and

β = inf

{∫

sdµ | s ∈ L+0 , f ≤ s

}

.

42 4. Integration

Show that

α = sup

{∫

Φpdµ | P is a measurable partition of X

}

and

β = inf

{∫

Ψpdµ | P is a measurable partition of X

}

.

(iv) Deduce that f ∈ L+ implies α =∫

fdµ = β.

Note:Exercise 4.14 tells us that for f ∈ L+, in defining

∫

fdµ it is enoughto consider approximations of f from below, as the approximationsfrom above will also give the same value for

∫

fdµ. This is becausef ∈ L+, i.e., f is nonnegative measurable. The converse is alsotrue, i.e., if α = β, for f : X → [0,∞], then f ∈ L+. To see this,first note that

α = β < M(µ(X)) < ∞,

where M is such that |f(x)| ≤ M ∀ x ∈ X. Thus for every n wecan choose functions φn, ψn ∈ L+

0 such that

φn ≤ f ≤ ψn and

∫

(ψn − φn)dµ <1

n.

Let φ := supφn and ψ := inf ψn. Then φ, ψ ∈ L+ (see corollary4.3.11) and ∀ n,

∫

(ψ − φ)dµ ≤

∫

(ψn − φn)dµ <1

n.

Thus∫

(ψ − φ)dµ = 0

and by proposition 4.3.3, ψ = φ = f a.e. µ. Hence f ∈ L+.

4.3. Intrinsic characterization of nonnegative

measurable functions


Recall, f ∈ L+ means f : X −→ R∗ is a nonnegative function with theproperty that there exists a sequence {sn}n≥1 of functions in L+

0 such that{sn(x)}n≥1 increases to f(x) for every x ∈ X. Note that the measure µplays no part in the definition of the functions in L+. It is only to definethe integral of functions f ∈ L+ that we need the measure µ. We want tocharacterize the functions in L+ intrinsically.

4.3. Intrinsic characterization of nonnegative measurable functions 43

4.3.1. Proposition:Let s : X −→ R∗ be such that s takes only finitely many distinct nonnegativevalues. Then the following are equivalent:

(i) s ∈ L+ (and hence s ∈ L+0 ).

(ii) s−1{t} ∈ S ∀ t ∈ R∗.

(iii) s−1[t,∞] ∈ S ∀ t ∈ R∗.

(iv) s−1(I) ∈ S for every interval I in R∗.

In view of the above proposition it is natural to ask the following ques-tion: does proposition 4.3.1 remain true if s ∈ L+

0 is replaced by any f ∈ L+?The answer is yes, as proved in the next proposition.

4.3.2. Proposition:Let f : X −→ R∗ be a nonnegative function. Then the following are equiva-lent:

(i) f ∈ L+.

(ii) f−1(c,+∞] ∈ S for every c ∈ R.

(iii) f−1[c,+∞] ∈ S for every c ∈ R.

(iv) f−1[−∞, c) ∈ S for every c ∈ R.

(v) f−1[−∞, c] ∈ S for every c ∈ R.

(vi) f−1{+∞}, f−{−∞} and f−1(E) ∈ S for every E ∈ BR.

The statements (ii) to (vi) of proposition 4.3.2 describe the elements ofL+ intrinsically. This proposition is used very often to check the measura-bility of nonnegative functions. Once again we emphasize the fact that fora nonnegative function f : X −→ R∗ to be measurable, i.e., for f to be inL+, the measure µ plays no part. As is clear from the above proposition, itis only the σ-algebra S of subsets of X that is important . The notion ofmeasurability is similar to the concept of continuity for topological spaces.As an immediate application of proposition 4.3.2, we have the following:

4.3.3. Proposition:Let f ∈ L+ and E ∈ S be such that

∫

E fdµ = 0. Then f(x) = 0 for a.e.(µ)x ∈ E.

Next, we extend the concept of measurability of nonnegative functionsto functions f : X −→ R∗ which are not necessarily nonnegative.

4.3.4. Definition:Let (X,S) be a measurable space.

44 4. Integration

(i) A function f : X −→ R∗ is said to be S-measurable if both f+ and f−

are S-measurable.We denote by L the class of all S-measurable functions on X. If the

underlying σ-algebra is clear from the context, we call a S-measurablefunction to be measurable.

(ii) If f ∈ L is such that both f+, f− ∈ L+0 , we call f a simple measurable

function.We denote the class of all simple measurable functions by L0. Note

that s ∈ L0 iff both s+, s− ∈ L+0 .

4.3.5. Proposition:Let (X,S) be a measurable space and f : X → R∗ be any function. Thenfollowing statements are equivalent:

(i) f is S-measurable.

(ii) There exists a sequence {sn}n≥1 of real valued functions on X such that∀ n, s+n and s−n are both nonnegative simple measurable functions on (X,S)and lim

n→∞sn(x) = f(x) ∀ x ∈ X.

(iii) f satisfies any one (and hence all) of the statements (ii) to (vi) of propo-sition 4.3.2.

4.3.6. Examples:

(i) Let f : X −→ R∗ be a constant function, f(x) = α ∀x ∈ X. Then f ismeasurable ∀c ∈ R, since

{x ∈ X | f(x) > c} =

{

∅ if c ≤ α,X if c > α.

(ii) Let f : X → R∗ be measurable, and α ∈ R. Then αf is also measurable,since

{x ∈ X | α(x) > c} =

{x ∈ X | f(x) < c/α} if α > 0,X if α = 0, c ≤ 0,∅ if α = 0, c > 0,

{x ∈ X | f(x) < c/α} if α < 0.

(iii) Let X be a topological space and S be the σ-algebra of Borel subsetsof X, i.e., the σ-algebra generated by the open sets. Let f : X −→ R beany continuous function. Then f is S-measurable.

We next describe some more properties of measurable functions.

4.3.7. Proposition:Let f, g be measurable functions. Then each of the sets {x ∈ X | f(x) >


g(x)}, {x ∈ X | {f(x) ≤ g(x)}, {x ∈ X | f(x) < g(x)}, {x ∈ X | f(x) ≥ g(x)}and {x ∈ X | f(x) = g(x)} ∈ S.

4.3.8. Proposition:Let f, g : X → R∗ be measurable functions and let β ∈ R∗ be arbitrary. Let

A := {x ∈ X | f(x) = +∞, g(X) = −∞}∪{x ∈ X|f(x) = −∞, g(x) = +∞}.

Define ∀ x ∈ X

(f + g)(x) :=

{

f(x) + g(x) if x 6∈ A,β if x ∈ A.

Then f + g : X → R∗ is a well-defined measurable function.

4.3.9. Proposition:Let f : X −→ R∗ be measurable and let Φ : R∗ −→ R∗ be such that R∩{x ∈R∗ |Φ(x) ≥ α} ∈ BR, ∀ α ∈ R. Then Φ ◦ f is also measurable.

4.3.10. Proposition:Let fn : X −→ R∗, n = 1, 2, . . ., be measurable functions. Then each of thefunctions sup

nfn, inf

nfn, lim sup

n→∞fn and lim inf

n→∞fn is a measurable function.

In particular, if {fn}n≥1 converges to f , then f is a measurable function.

4.3.11. Corollary:Let {fn}n≥1 be a sequence in L+. Then each of the functions sup

nfn, inf

nfn,

lim supn→∞

fn and lim infn→∞

fn is in L+. In particular, if limn→∞

fn =: f exists, then

f ∈ L+.

Recall that in theorem 4.2.5 we analyzed the limit of∫

fndµ for an in-creasing sequence of nonnegative measurable functions. We analyze next thebehavior of

∫

fndµ when {fn}n≥1 is not necessarily an increasing sequenceof nonnegative measurable functions.

4.3.12. Theorem (Fatou’s lemma):Let {fn}n≥1 be a sequence of nonnegative measurable functions. Then

∫

(

lim infn→∞

fn

)

dµ ≤ lim infn→∞

∫

fndµ.

4.3.13. Proposition:Let (X,S) be a measurable space and let f : X −→ R∗ be S-measurable. Letµ be a measure on (X,S). Let g : X −→ R∗ be such that {x ∈ X | f(x) 6=g(x)} is a µ-null set. If (X,S, µ) is a complete measure space, then g is alsoS-measurable.

46 4. Integration

Exercises:

(4.15)(i) Let f ∈ L+ and E ∈ S be such that f(x) > 0 for every x ∈ Eand µ(E) > 0. Show that

∫

E fdµ > 0.(ii) Let f, g ∈ L+ be such that∫

fdµ =

∫

gdµ < +∞ and

∫

Efdµ =

∫

Egdµ, ∀ E ∈ S.

Show that f(x) = g(x) a.e. (x)µ.(iii) Let f, g be nonnegative measurable functions on (R,L) such

that∫ b

afdµ =

∫ b

agdµ < +∞ for every a < b

show that then∫

Efdµ =

∫

Egdµ, ∀ E ∈ L,

and deduce that f(x) = g(x) a.e. (x)λ.

(4.16) Let f : X −→ R∗ be a bounded measurable function. Then thereexists a sequence {sn}n≥1 of simple measurable functions such that{sn}n≥1 converges uniformly to f.(Hint: If 0 ≤ f(x) ≤M ∀ x, then |sn(x) − f(x)| < 1/2n ∀ n ≥ n0,where n0 ≥M and the sn’s are as in proposition 4.3.2.)

(4.17) Let f : X −→ R∗ be a nonnegative measurable function. Showthat there exist sequences of nonnegative simple functions {sn}n≥1

and {sn}n≥1 such that

0 ≤ · · · ≤ sn(n) ≤ sn+1(x) ≤ · · · ≤ f(x) ≤ · · · ≤ sn+1(x) ≤ sn(x) · · ·

and limn→∞

sn(x) = f(x) = limn→∞

sn(x) ∀ x ∈ X.

(4.18) Let f and g : X −→ R∗ be measurable functions, p and α ∈ R withp > 1, and let m be any positive integer. Use proposition 4.3.9toprove the following:(i) f + α is a measurable function.(ii) Let β and γ ∈ R∗ be arbitrary. Define for x ∈ R,

fm(x) :=

(f(x))m if f(x) ∈ R,β if f(x) = +∞,γ if f(x) = −∞.

Then fm is a measurable function.(iii) Let |f |p be defined similarly to fm, where p is a nonnegative

real number. Then |f |p is a measurable function.


(iv) Let β, γ, δ ∈ R∗ be arbitrary. Define for x ∈ R,

(1/f)(x) :=

1/f(x) if f(x) 6∈ {0,+∞,−∞},β if f(x) = 0,γ if f(x) = −∞,δ if f(x) = +∞.

Then 1/f is a measurable function.(v) Let β ∈ R∗ be arbitrary and A be as in proposition 4.3.8.

Define for x ∈ R,

(fg)(x) :=

{

f(x)g(x) if x 6∈ A,β if x ∈ A.

Then fg is a measurable function.

(4.19) Let f : X → R∗ be S-measurable. Show that |f | is also S-measurable. Give an example to show that the converse need notbe true.

(4.20) Let (X,S) be a measurable space such that for every function f :X −→ R, f is S-measurable iff |f | is S-measurable. Show thatS = P(X).

(4.21) Let fn ∈ L, n = 1, 2, . . . . Show that the sets

{x ∈ X | {fn(x)}n is convergent}

and

{x ∈ X | {fn(x)}n≥1is Cauchy}

belong to S.

(4.22) Give an example to show that strict inequality can occur in Fatou’slemma.

(4.23) Let {fn}n≥1 be a sequence of functions in L+ and let∑∞

n=1 fn(x) =:f(x), x ∈ X. Show that f ∈ L+ and

∫

fdµ =

∞∑

n=1

∫

fndµ.

(4.24) Show that each of the functions f : R −→ R defined below is L-measurable, and compute

∫

fdλ :

(i) f(x) :=

{

0 if x ≤ 0,1/x if x > 0.

(ii) f(x) := χQ(x), the indicator function of Q, the set of rationals.

48 4. Integration

(iii) f(x) :=

0 if x > 1 or x < 0 or x ∈ [0, 1] and x is rational,n if x is an irrational, 0 < x < 1 and in the

decimal expansion of x, the first nonzeroentry is at the (n+ 1)th place.

(4.25) Let f , g ∈ L+ with f ≥ g. Show that (f−g) ∈ L+ and∫

fdµ ≥∫

gdµ.Can you conclude that

∫

(f − g)dµ =

∫

fdµ−

∫

gdµ?

(4.26) Let f , fn ∈ L+, n = 1, 2, . . ., be such that 0 ≤ fn ≤ f. Iflim

n→∞fn(x) = f(x), can you deduce that

∫

fdµ = limn→∞

∫

fndµ?

(4.27) Let f ∈ L. For x ∈ X and n ≥ 1, define

fn(x) :=

f(x) if |f(x)| ≤ n,n if f(x) > n,−n if f(x) < −n.

Prove the following:(i) fn ∈ L and |fn(x)| ≤ n ∀ n and ∀ x ∈ X.(ii) lim

n→∞fn(x) = f(x) ∀ x ∈ X.

(iii) |fn(x)| := min{|fn(x)|, n} := (|f | ∧ n)(x) is an element of L+

and

limn→∞

∫

|fn|dµ =

∫

|f |dµ.

Note:For f ∈ L, the sequence {fn}n≥1 as defined in exercise 4.27 is calledthe truncation sequence of f. The truncation sequence is usefulin proving results about functions in the class L.

(4.28) Let f ∈ L and

ν(E) := µ{x ∈ X | f(x) ∈ E}, E ∈ BR.

Show that ν is a measure on (R,BR). Further, if g : R −→ R is anynonnegative BR-measurable function, i.e., g−1(A) ∈ BR ∀ A ∈ BR,then g ◦ f ∈ L and

∫

g dν =

∫

(g ◦ f) dµ.

The measure ν is usually denoted by µf−1and is called the distri-bution of the measurable function f.


(4.29) Let f ∈ L+ be a bounded function, say f(x) ≤ N ∀ x ∈ X and forsome N ∈ N. Show that∫

f dµ = limn→∞

N2n∑

k=1

k − 1

2nµ

{

x |k − 1

2n≤ f(x) <

k

2n

}

. (4.1)

a b

m

M

Figure: Area below the curve y = f(x) from inside

Note that here, for every n ∈ N, we are taking the partition{0, 1/2n, 2/2n, . . . , N} of the range of f and using it to induce themeasurable partition Pn of the domain |X| of f , given by

{Xnk | 1 ≤ k ≤ N2n}, where Xn

k := f−1[(k − 1)/2n, k/2n).

Using this partition Pn, we form the sums on the right hand sideof (4.1), approximating the area below the curve y = f(x) from‘inside’ (see Figure). Thus in our integral we use partitions of therange, whereas in Riemann integration we use partitions of thedomain. In the words of H. Lebesgue, the situation is similar tothe problem of finding the total amount in a cash box containingsay n coins, the ith coin being of denomination αi, 1 ≤ i ≤ n.One method of finding the total amount in the box is to add thevalue of each coin, i.e., the sum

∑ni=1 αi. The other method is to

separate out coins of similar denominations. If there are kj coinsof denominations αj , then the total amount is

∑

j kjαj .

(4.30) Let (X,S, µ) be a measure space and let (X, S, µ) be its completion.Let f : X −→ R be an S-measurable function. Show that thereexists an S-measurable function f : X −→ R such that f(x) = f(x)for a.e. x(µ).

(4.31) Let (X,S, µ) be a complete measure space and {fn}n≥1 be a se-quence of S-measurable functions on X. Let f be a function on

50 4. Integration

X such that f(x) = limn→∞

fn(x) for a.e. x(µ). Show that f is S-

measurable.

4.4. Integrable functions


Given a measure space (X,S, µ) in section 4.2, we defined∫

fdµ, the integralfor functions f ∈ L+, i.e., f : X −→ R∗, f nonnegative measurable. If fis measurable, but not necessarily nonnegative, we can write f = f+ − f−.Since both f+ and f− are nonnegative measurable,

∫

f+dµ and∫

f−dµ aredefined. Thus it is reasonable (as the integral is expected to be linear) todefine the integral of f to be

∫

fdµ :=∫

f+dµ−∫

f−dµ. The problem canarise in the case

∫

f−dµ =∫

f+dµ = +∞. To overcome this difficulty, weintroduce the following definition:

4.4.1. Definition:A measurable function f : X −→ R∗ is said to be µ-integrable if both∫

f+dµ and∫

f−dµ are finite, and in that case we define the integral of fto be

∫

fdµ :=

∫

f+dµ−

∫

f−dµ.

We denote by L1(X,S, µ) (or simply by L1(X) or L1(µ)) the space of allµ-integrable functions on X.

The properties of the space L1(X,S, µ) and of the map f 7−→∫

fdµ,f ∈ L1(X,S, µ), are given in the next proposition.

4.4.2. Proposition:For f, g ∈ L and a, b ∈ R, the following hold:

(i) If |f(x)| ≤ g(x) for a.e. x(µ) and g ∈ L1(µ), then f ∈ L1(µ).

(ii) If f(x) = g(x) for a.e. x(µ) and f ∈ L1(µ), then g ∈ L1(µ) and∫

fdµ =

∫

gdµ.

(iii) If f ∈ L1(µ), then af ∈ L1(µ) and∫

(af)dµ = a

∫

fdµ.

(iv) If f and g ∈ L1(µ), then f + g ∈ L1(µ) and∫

(f + g)dµ =

∫

fdµ+

∫

gdµ.

4.4. Integrable functions 51

(v) If f , g ∈ L1(µ), then (af + bg) ∈ L1(µ) and∫

(af + bg)dµ = a

∫

fdµ+ b

∫

gdµ.

4.4.3. Proposition:Let f ∈ L1(µ). For every E ∈ S, let

ν(E) :=

∫

χE |f |dµ and ν(E) :=

∫

χEfdµ.

Then the following hold:

(i) If µ(E) = 0, then ν(E) = 0.

(ii) If µ(E) = 0, then ν(E) = 0.

(iii) limµ(E)→0 ν(E) = 0, i.e., given any ǫ > 0, there exists δ > 0 such thatν(E) < ǫ whenever, for E ∈ S, µ(E) < δ.

(iv) If ν(E) = 0 ∀ E, then f(x) = 0 for a.e. x(µ) on E.

4.4.4. Remark:It is easy to see that (iii) in the above proposition implies (ii). In fact (ii)also implies (iii). To see this, suppose (iii) does not hold. Then there existan ǫ > 0 and sets En ∈ S, n ≥ 1, such that µ(En) < 2−n but ν(En) ≥ ǫ. LetAn =

⋃∞k=nEk. Then {An}n≥1 is a decreasing sequence in S and µ(An) ≤

µ(En) ≤ 2−n. Thus by theorem 2.3.2,

µ

(

∞⋂

n=1

An

)

= limn→∞

µ(An) = 0.

On the other hand, µ(An) ≥ µ(En) ≥ ǫ, ∀ ǫ, contradicting (ii).

We prove next the most frequently used theorem which allows us tointerchange the operations of integration and limits.

4.4.5. Theorem (Lebesgue’s dominated convergence theorem):Let {fn}n≥1 be a sequence of measurable functions and let g ∈ L1(µ) be

such that ∀ n, |fn(x)| ≤ g(x) for a.e. x(µ). Let {fn(x)}n≥1 converge tof(x) for a.e. x(µ). Then the following hold:

(i) f ∈ L1(µ).

(ii)∫

f dµ = limn→∞

∫

fn dµ.

(iii) limn→∞

∫

|fn − f |dµ = 0.

We state another version of this theorem, which is applicable to seriesof functions.

52 4. Integration

4.4.6. Corollary:Let {fn}n≥1 be a sequence of functions in L1(µ) such that

∑∞n=1

∫

|fn|dµ< +∞. Then f(x) :=

∑∞n=1 fn(x) exists for a.e. x(µ), f ∈ L1(µ) and

∫

fdµ =∞∑

n=1

∫

fndµ.

The following is a variation of the dominated convergence theorem forfinite measure spaces. See exercise (4.38) also.

4.4.7. Theorem (Bounded convergence):Let (X,S, µ) be a finite measure space and f, f1, f2, . . . be measurable func-

tions. Suppose there exists M > 0 such that |fn(x)| ≤ M a.e. x(µ) andfn(x) → f(x) a.e. x(µ). Then f, fn ∈ L1(X,S, µ) and

∫

fdµ = limn→∞

∫

fndµ.

4.4.8. Notes:

(i) The monotone convergence theorem and the dominated convergence theo-rem (along with its variations and versions) are the most important theoremsused for the interchange of integrals and limits.

(ii) Simple function technique: This is an important technique (similarto the σ-algebra technique) used very often to prove results about integrableand nonnegative measurable functions. Suppose we want to show that acertain claim (∗) holds for all integrable functions. Then technique is thefollowing:

(1) Show that (∗) holds for nonnegative simple measurable functions.

(2) Show that (∗) holds for nonnegative measurable / integrable func-tions by approximating them by nonnegative simple measurablefunctions and using (1).

(3) Show that (∗) holds for integrable functions f, by using (2) and thefact that for f ∈ L1, f = f+ − f− and both f+, f− ∈ L1.

The proof of the next proposition is an illustration of this technique.

4.4.9. Proposition:Let (X,S, µ) be a σ-finite measure space and f ∈ L1(X,S, µ) be nonnegative.For every E ∈ S, let

ν(E) :=

∫

Efdµ.

4.4. Integrable functions 53

Then ν is a finite measure on S. Further, fg ∈ L1(X,S, µ) for every g ∈L1(X,S, ν), and

∫

fdν =

∫

fgdµ.

We shall see some more applications of the dominated convergence the-orem in the remaining sections. In the next section we look at some specialproperties of L1(X,S, µ) in the particular case when X = R, S = L, theσ-algebra of Lebesgue measurable sets, and µ = λ, the Lebesgue measure.

As an application of the dominated convergence theorem, we exhibit thepossibility of interchanging the order of integration and differentiation inthe next theorem.

4.4.10. Theorem:Let ft ∈ L1(µ) for every t ∈ (a, b) ⊆ R. Let t0 ∈ (a, b) be such that fora.e. x(µ), t 7−→ ft(x) is differentiable in a neighborhood U of t0 and there

exists a function g ∈ L1(µ) such thatdft

dt(x) ≤ g(x) for a.e. x(µ) and for

every t ∈ U. Then φ(t) :=∫

ft(x)dµ(x) is differentiable at t0 and

φ′

(t0) =

∫

(

[

dft

dt(x)

]

t0

)

dµ(x).

Exercises:

(4.32) For f ∈ L, prove the following:

(i) f ∈ L1(µ) iff |f | ∈ L1(µ). Further, in either case∣

∣

∣

∣

∫

fdµ

∣

∣

∣

∣

≤

∫

|f |dµ.

(ii) If f ∈ L1(µ), then |f(x)| < +∞ for a.e. x(µ).

(4.33) Let µ(X) < +∞ and let f ∈ L be such that |f(x)| ≤M for a.e. x(µ)and for some M. Show that f ∈ L1(µ).

(4.34) Let f ∈ L1(µ) and E ∈ S. Show that χEf ∈ L1(µ), where∫

Efdµ :=

∫

χEfdµ.

Further, if E,F ∈ S are disjoint sets, show that∫

E∪Ffdµ =

∫

Efdµ+

∫

Ffdµ.

54 4. Integration

(4.35) Let f ∈ L1(µ) and Ei ∈ S, i ≥ 1, be such that Ei ∩ Ej = ∅ fori 6= j. Show that the series

∑∞i=1

∫

Eifdµ is absolutely convergent,

and if E :=⋃∞

i=1Ei, then∞∑

i=1

∫

Ei

fdµ =

∫

Efdµ.

(4.36) (i) For every ǫ > 0 and f ∈ L1(µ), show that

µ {x ∈ X | |f(x)| ≥ ǫ} ≤1

ǫ

∫

|f |dµ < ∞.

This is called Chebyshev’s inequality.

(ii) Let f ∈ L1(µ), and let there exist M > 0 such that

∣

∣

1

µ(E)

∫

Efdµ

∣

∣ ≤ M

for every E ∈ S with 0 < µ(E) < ∞. Show that |f(x)| ≤ M fora.e. x(µ).

(4.37) Let λ be the Lebesgue measure on R, and let f ∈ L1(R,L, λ) besuch that

∫

(−∞,x)f(t)dλ(t) = 0, ∀ x ∈ R.

Show that f(x) = 0 for a.e. (λ)x ∈ R.

(4.38) Let I ⊆ R be an interval and ∀ t ∈ I, let ft ∈ L. Let g ∈ L1(µ)be such that ∀ t, |ft(x)| ≤ g(x) for a.e. x(µ). Let t0 ∈ R∗ be anyaccumulation point of I and let f(x) := lim

t→t0ft(x) exist for a.e. x(µ).

Then f ∈ L1(µ) and∫

fdµ = limt→t0

∫

ft(x)dµ(x).

Further, if ∀ x the function t 7−→ ft(x) is continuous, then so isthe function h(t) :=

∫

ft(x) dµ, t ∈ I.(Hint: Apply theorem 4.4.5 to every sequence tn → t0.)

(4.39) Let {fn}n≥1 and {gn}n≥1 be sequences of measurable functions suchthat |fn| ≤ gn ∀ n. Let f and g be measurable functions such thatlim

n→∞fn(x) = f(x) for a.e. x(µ) and lim

n→∞gn(x) = g(x) for a.e. x(µ).

If

limn→∞

∫

gn dµ =

∫

g dµ < +∞,

show that

limn→∞

∫

fn dµ =

∫

fdµ.

(Hint: Apply Fatou’s lemma to (gn − fn) and (gn + fn).)

4.5. The Lebesgue integral and its relation with the Riemann integral 55

(4.40) Let {fn}n≥0 be a sequence in L1(X,S, µ). Show that{∫

|fn|dµ}

n≥1

converges to∫

|f0|dµ iff{∫

|fn − f0|dµ}

n≥1converges to zero.

(Hint: Use exercise 4.38.)

(4.41) Let (X,S) be a measurable space and f : X −→ R be S-measurable.Prove the following:(i) S0 := {f−1(E) |E ∈ BR} is the σ-algebra of subsets of X, and

S0 ⊆ S.(ii) If φ : R −→ R is Borel measurable, i.e., φ−1(E) ∈ BR ∀ E ∈

BR, then φ ◦ f is an S0-measurable function on X.(iii) If ψ : X −→ R is any S0-measurable function, then there exists

a Borel measurable function φ : R −→ R such that ψ = φ ◦ f .(Hint: Use the simple function technique and note that if ψ isa simple S0-measurable function, then

ψ =n∑

i=1

anχf−1(Ei)

for some positive integer n, ai ∈ R for each i, and Ei ∈ BR,then

ψ =n∑

i=1

ai(χEi◦f).

(4.42) Let {fn}n≥1 be a decreasing sequence of nonnegative functions inL1(µ) such that fn(x) → f(x). Show that

limx→∞

∫

fndµ = 0 iff f(x) = 0 a.e. x(µ).

(4.43) Let µ, ν be as in proposition 4.4.9. Let Sν denote the σ-algebra ofall ν∗-measurable subsets of X. Prove the following:(i) S ⊆ Sν .(ii) There exist examples such that S is a proper subclass of Sν .

Show that S = Sν if µ∗{x ∈ X | f(x) = 0} = 0.

(4.44) Let (X,S, µ) be a finite measure space and {fn}n≥1 be a sequencein L1(µ) such that fn → f uniformly. Show that f ∈ L1(µ) and

limn→∞

∫

|fn − f | dµ = 0.

Can the condition of µ(X) < +∞ be dropped?

4.5. The Lebesgue integral and its relation with

the Riemann integral


56 4. Integration

In this section we analyze the integral, as constructed in the previous section,for the particular situation when X = R, S = L (the σ-algebra of Lebesguemeasurable sets) and µ = λ, the Lebesgue measure. The space L1(R,L, λ),also denoted by L1(R) or L1(λ), is called the space of Lebesgue integrablefunctions on R, and

∫

fdλ is called the Lebesgue integral of f. For anyset E ∈ L, we write L1(E) for the space of integrable functions on themeasure space (E,L ∩ E, λ), where λ is restricted to L ∩ E. In the specialcase when E = [a, b], we would like to show that the new notion of integralfor f ∈ L1[a, b] indeed extends the notion of Riemann integral. To be precise,we have the following theorem:

4.5.1. Theorem:Let f : [a, b] −→ R be a Riemann integrable function. Then f ∈ L1[a, b] and

∫

fdλ =

∫ b

af(x)dx.

4.5.2. Remark:In fact, the proof of the above theorem includes a proof of the following: Iff ∈ R[a, b], then f is continuous a.e. x(λ). This is because

f(x) = limn→∞

Ψn(x) = limn→∞

Φn(x) a.e. x(λ).

Thus, if we put

E := {x ∈ [a, b] | f(x) = limn→∞

Φn(x) = limn→∞

Ψn(x)},

then E is a Lebesgue measurable set and λ([a, b] \E) = 0. For x ∈ E, givenan arbitrary ǫ > 0, we can choose n0 such that

Ψn0(x) − Φn0

(x) < ǫ. (4.2)

Further, if x is not a point in any partition Pn, then we can choose δ >0 such that whenever y ∈ [a, b] and |x − y| < δ, then y belongs to thesame subinterval of the partition Pn0

to which x belongs. Thus by (4.2),|f(x) − f(y)| < ǫ, showing that x is a point of continuity of f . Thus theset of discontinuity points of f forms a subset of ([a, b]) \ E) ∪ P, whereP is the set of partition points of Pn, n = 1, 2, . . . . Hence f is continuousalmost everywhere. Another proof of the converse is given in the one of theexercises.

Exercises:

(4.45) Let f : [a, b] −→ R be bounded and continuous for a.e. x(λ).

4.5. The Lebesgue integral and its relation with the Riemann integral 57

(i) Let {Pn}n≥1 be any sequence of partitions of [a, b] such thateach Pn+1 is a refinement of Pn and ‖Pn‖ −→ 0 as n → ∞.Let Φn,Ψn be as constructed in theorem 4.5.1. Let x ∈ (a, b)be a point of continuity of f . Show that

limn→∞

Φn(x) = f(x) = limn→∞

Ψn(x).

(ii) Using (i) and the dominated convergence theorem, deduce thatf ∈ L1([a, b]) and∫

fdλ = limn→∞

∫

Φndµ = limn→∞

∫

Ψndµ.

(iii) Show that f ∈ R[a, b] and∫

fdλ =

∫ b

af(x)dx.

(4.46) Let f : [0, 1] −→ [0,∞) be Riemann integrable on [ǫ, 1] for all ǫ > 0.

Show that f ∈ L1[0, 1] iff limǫ→0

∫ 1ǫ f(x)dx exists, and in that case

∫

f(x)dλ(x) = limǫ→0

∫ 1

ǫf(x)dx.

(4.47) Let f(x) = 1/xp if 0 < x ≤ 1, and f(0) = 0. Find necessary and suf-

ficient condition on p such that f ∈ L1[0, 1]. Compute∫ 10 f(x)dλ(x)

in that case.(Hint: Use exercise 4.46.)

(4.48) (Mean value property):Let f : [a, b] −→ R be a continuousfunction and let E ⊆ [a, b], E ∈ L, be such that λ(E) > 0. Showthat there exists a real number α such that

∫

Ef(x)λ(x) = αλ(E).

(4.49) Let f ∈ L1(R), and let g : R −→ R be a measurable function suchthat α ≤ g(x) ≤ β for a.e. x(λ). Show that fg ∈ L1(R) and thereexists γ ∈ [α, β] such that

∫

|f |gdλ = γ

∫

|f |dλ.

(4.50) Let f ∈ L1(R,L, λ) and let a ∈ R be fixed. Define

F (x) :=

∫

[a,x] f(t) dλ(t) for x ≥ a,

∫

[x,a] f(t)dλ(t) for x ≥ a.

Show that F is continuous.

58 4. Integration

(Hint: Without loss of generality take f ≥ 0 and show that F iscontinuous from the left and right. In fact, F is actually uniformlycontinuous.)

(4.51) Let f ∈ L1(R,L, λ) and let c ∈ R be a point of continuity of f.Show that

limn→∞

n

∫

[c, c+ 1/n]f(x) dλ(x) = f(c).

(4.52) (Arzela’s theorem):Let {fn}n≥1 be a sequence of Riemann integrable functions on [a, b]such that for some M > 0, |fn(x)| < M ∀ x ∈ [a, b] and ∀ n =1, 2, . . . . Let fn(x) −→ f(x) ∀ x ∈ [a, b] and let f be Riemannintegrable on [a, b]. Then

limn→∞

∫ b

afn(x)dx =

∫ b

af(x)dx.

(4.53) Let f : [a, b] −→ R be any constant function. Show that f ∈L1[a, b].

(4.54) Let f : [a, b] −→ R be any bounded measurable function. Showthat f ∈ L1[a, b].

(4.55) Let f : [a, b] −→ R be any continuous function. Show that f ∈L1[a, b].

(4.56) Let f ∈ L1(R) be such that∫

K fdλ = 0 for every compact setK ⊆ R. Show that f(x) = 0 for a.e. x(λ).

(4.57) Let {fn}n≥1 be a decreasing sequence of nonnegative functions inC(a, b) and let f1 ∈ L1(a, b). If

∑∞n=1(−1)n−1fn ∈ C(a, b), show

that∫ b

a

(

∞∑

n=1

(−1)n−1fn(x)

)

dx =∞∑

n=1

(

(−1)n−1

(∫ b

afn(x)dx

))

.

(Hint: For every n,∑n

k=1(−1)kfk ≤ f ∈ L1(a, b)).

(4.58) Give examples to show that analogues of the monotone convergencetheorem and the dominated convergence theorem do not hold forthe Riemann integral.

(4.59) Let f ∈ L1(R) and, ∀ t ∈ [0,∞),

g(t) = sup

{∫

|f(x+ y) − f(x)| dµ(x) − t ≤ y ≤ t

}

.

Show that g is continuous at t = 0.(Hint: Use the simple function technique).

4.6. L1[a, b] as the completion of R[a, b] 59

4.6. L1[a, b] as the completion of R[a, b]


Recall that, R[a, b] is not a complete metric space under the L1-metric

d(f, g) :=

∫ b

a|f(x) − g(x)|dx,

for f, g ∈ R[a, b]. Since every metric space has a completion, the questionarises: what is the completion of R[a, b] under this metric? You might haveseen the abstract construction of the completion of a metric space. We shallshow that for R[a, b] this completion is nothing but L1[a, b]. Thus L1[a, b]is the concrete realization of the completion of R[a, b]. We shall first showthat L1[a, b] is a complete metric space, and then show that R[a, b] is densein L1[a, b].

4.6.1. Definition:For f, g ∈ L1[a, b], we say f is equivalent to g, and write f ∼ g, if the set{x ∈ [a, b] | f(x) 6= g(x)} has Lebesgue measure zero.

It is easy to see that the relation ∼ is an equivalence relation on L1[a, b].We denote the set of equivalence classes again by L1[a, b]. In other words,we identify two functions f, g with each other if f ∼ g. Thus, for g, f ∈L1[a, b], f = g iff f(x) = g(x) for a.e. x(λ), as functions.

4.6.2. Definition:For f ∈ L1[a, b], we define

‖f‖1 :=

∫

|f(x)|dλ(x).

Clearly ‖f‖1 is well-defined, and it is easy to check that the function f 7−→‖f‖1, f ∈ L1[a, b], has the following properties:

(i) ‖f‖1 ≥ 0 ∀ f ∈ L1[a, b].

(ii) ‖f‖1 = 0 iff f = 0.

(iii) ‖af‖1 = |a| ‖f‖1 ∀ a ∈ R and f ∈ L1[a, b].

(iv) ‖f + g‖1 ≤ ‖f‖1 + ‖g‖1 ∀ f, g ∈ L1[a, b].

The function ‖ · ‖1 is called a norm on L1[a, b]. (Geometrically it is thedistance of the ‘vector’ f ∈ L1[a, b] from the ‘vector’ 0 ∈ L1[a, b].) Forf, g ∈ L1[a, b], if we define

d(f, g) := ‖f − g‖1,

60 4. Integration

then d is a metric on L1[a, b], called the L1-metric . The most importantproperty of this metric is given by the next theorem.

4.6.3. Theorem (Riesz-Fischer):L1[a, b] is a complete metric space in the L1-metric.

We show next that C[a, b], the space of continuous function on [a, b],(and hence R[a, b]) is dense in L1[a, b].

4.6.4. Theorem:The space C[a, b] is a dense subset of L1[a, b].

Since C[a, b] ⊆ R[a, b], theorems 4.6.3 and 4.6.4 together prove the fol-lowing theorem:

4.6.5. Theorem:L1[a, b] is the completion of R[a, b].

4.6.6. Notes:

(i) In the proof of theorem 4.6.3, one does not use require anywhere thefact that the functions are defined on an interval. One can define ‖ · ‖1

for functions defined on (E,L ∩ E, λ), where E ∈ L is arbitrary, and theproof of theorem 5.6.1 will show that L1(E) := L1(E, L ∩ E, λ) is also acomplete metric space under the metric

‖f − g‖1 :=

∫

E|f(x) − g(x)|dλ(x).

A closer look will show that the metric d makes sense on L1(X,S, µ),where (X,S, µ) is any complete measure space and we identify functionswhich agree for a.e. (µ). Further, theorem 4.6.3 also remains true forL1(X,S, µ).

(ii) Parts of the proof of theorem 4.6.4 also exhibit the following facts whichare of independent interest. Let f ∈ L1[a, b] and ǫ > 0 be given. Thenthere exists a simple function Φ such that ‖f−Φ‖1 < ǫ and a step functionh such that ‖f −h‖1 < ǫ. Thus simple functions in L1[a, b] are dense in it.

Exercises:

(4.60) Let f ∈ L1(R) and let ǫ > 0 be given. Prove the following:(i) There exists a positive integer n such that

‖f − χ[−n,n]f‖1 < ǫ.

4.6. L1[a, b] as the completion of R[a, b] 61

(ii) There exists a continuous function g on R such that g is zerooutside some finite interval and ‖fχ[−n,n] − g‖1 < ǫ.

(iii) For f : R −→ R, let

supp (f) := closure {x ∈ R | f(x) 6= 0}.

The set supp(f) is called the support of f and is the smallestclosed subset of R outside which f vanishes. Let the space ofcontinuous functions on R be denoted by C(R) and let

Cc(R) := {f : R −→ R | f ∈ C(R) and supp(f) is compact}.

Then Cc(R) is a dense subset of L1(R).

(4.61) Let f ∈ L1(R) and f(x) := f(−x) ∀ x ∈ R. Prove the following:

(i)∫

f(x)dλ(x) =∫

f(x)dλ(x).(ii) fg ∈ L1(R) for any bounded measurable function g : R → R.

Dense subspaces of L1(R) are very useful in proving resultsabout L1-functions. As an application of exercise 4.59 we have thefollowing:

(4.62) Let f ∈ L1(R) and for every h, k ∈ R, let

fh(x) := f(x+ h) and φ(x) := f(kx+ h), x ∈ R.

Prove the following:(i) fh, φ ∈ L1(R) with

∫

φ(x)dλ(x) = |k|

∫

f(x)dλ(x) and

∫

fh(x)dλ(x) =

∫

f(x)dλ(x).

(ii) For every h ∈ R, fh ∈ L1(R) and limh→0

‖fh − f‖1 = 0.

(Hint : Use exercise 4.60 to get a function g ∈ Cc(R) with‖g − f‖1 < ǫ, and note that ‖gh − g‖1 < 2(b − a + 1) whensupp(g) ⊆ [a, b] and |h| is sufficiently small.)

(iii) The function h 7−→ ‖fh‖1 is continuous.

(4.63) (Riemann-Lebesgue lemma): Let f ∈ L1(R), and let g : R −→R be any bounded measurable function such that g(x+p)−g(x) = 0for every x ∈ R and p ∈ R fixed. Show that ∀ t 6= 0,

∣

∣

∣

∣

∫

f(x)g(tx)dx

∣

∣

∣

∣

≤ M ‖fp/t − f‖1,

where M = sup |g(x)|/2. Hence deduce (using exercise 4.60) that

lim|t|→∞

∫

f(x)g(tx)dx = 0.

62 4. Integration

(In the special case when f ∈ L1[0, 2π] and g(x) = cosx or sinx,we have

limn→∞

∫ 2π

0f(x) cosnxdx = 0 = lim

n→∞

∫ 2π

0f(x) sinnxdx.

This finds applications in the theory of Fourier series.)

Chapter 5

Measure and

integration on product

spaces

5.1. Introduction


The intuitive notion of length, originally defined for intervals in R, leads oneto the class of Lebesgue measurable sets which included not only the intervalsand all the topologically nice subsets of R, but also BR – the σ-algebra ofBorel subsets of R. In a similar manner, one would like to extend the notionof area in R2 (volume in R3, and so on) to a larger class of subsets whichincludes BR2 (BR3) – the σ-algebra generated by open subsets of R2 (R3).In the abstract setting, given measure spaces (X,A, µ) and (Y,B, ν), onewould like to define a measure η on the σ-algebra generated by sets of theform {A × B |A ∈ A, B ∈ B} in the ‘natural’ way: η(A × B) = µ(A)ν(B).We call this natural for, when X = Y = R,A = B = BR and µ = ν = λ, theLebesgue measure, then for intervals I and J, η(I×J) = λ(I)λ(J) is the areaof the rectangle with sides the intervals I and J . Thus η will automaticallybe an extension of the notion of area in R2. We note that the collectionR := {A× B |A ∈ A, B ∈ B} is only a semi-algebra of subsets of A× B ingeneral. Let A⊗ B denote the σ-algebra of subsets of X × Y generated byR.

63

64 5. Measure and integration on product spaces

5.1.1. Definition:Let (X,A) and (Y,B) be measurable spaces. A subset E ⊆ X×Y is calleda measurable rectangle if E = A×B for some A ∈ A and B ∈ B. Wedenote by R the class of all measurable rectangles. The σ-algebra of subsetsof X×Y generated by the semi-algebra R is called the product σ-algebraand is denoted by A⊗B.

5.1.2. Proposition:Let pX : X × Y −→ X and pY : X × Y −→ Y be defined by

pX (x, y) = x and pY (x, y) = y,

∀ x ∈ X, y ∈ Y . Then the following hold:

(i) The maps pX and pY are measurable, i.e., ∀ A ∈ A, B ∈ B wehave p−1

X(A) ∈ A⊗ B and p−1

Y(B) ∈ A⊗ B.

(ii) The σ-algebra A⊗B is the smallest σ-algebra of subsets of X × Ysuch that (i) holds.

5.1.3. Proposition:Let X and Y be nonempty sets and let C,D be families of subsets of X andY, respectively. Let C × D := {C ×D |C ∈ C, D ∈ D}. Then the followinghold:

(i) S(C × D) ⊆ S(C) ⊗ S(D).

(ii) Let C and D have the property that there exist increasing sequences{Ci}i≥1 and {Di}i≥1 in C and D respectively such that

∞⋃

i=1

Ci = X and∞⋃

i=1

Di = Y .

Then

S(C × D) = S(C) ⊗ S(D).

Exercises:

(5.1) Let (X,A) be a measurable space. Let α, β ∈ R and E ∈ A ⊗ BR.Show that {(x, t) ∈ X × R | (x, αt+ β) ∈ E} ∈ A ⊗ BR.(Hint: Use the σ-algebra technique.)

(5.2) Let E ∈ BR. Show that {(x, y) ∈ R2 |x + y ∈ E} and {(x, y) ∈R2 |x− y ∈ E} are elements of BR ⊗ BR.

(5.3) Let X and Y be nonempty sets and C, D be nonempty families ofsubsets of X and Y , respectively, as in proposition 5.1.3. Is it truethat S(C×D) = S(C)⊗S(D) in general? Check in the case when

5.2. Product of measure spaces 65

C = {∅} and D is a σ-algebra of subsets of Y containing at leastfour elements.

(5.4) Let BR2 denote the σ-algebra of Borel subsets of R2, i.e., the σ-algebra generated by the open subsets of R2. Show that

BR2 = BR ⊗ BR.

(Hint: Use proposition 5.1.3.)

5.2. Product of measure spaces


For the rest of the chapter, let (X,A, µ) and (Y,B, ν) be fixed measurespaces. Subsets of X × Y of the form A × B,A ∈ A, B ∈ B, are calledmeasurable rectangles. As before, let R denote the class of all measurablerectangles. The σ-algebra of subsets of X × Y generated by R, denoted byA⊗ B, is called the product σ-algebra.

The problem we want to analyze in this section is the following: howcan we construct a measure η : A ⊗ B −→ [0,+∞] such that η(A × B) =µ(A)ν(B) for every A ∈ A, B ∈ B? The answer is given by the next theorem.

5.2.1. Theorem:Let η : R −→ [0,∞] be defined by

η(A×B) := µ(A)ν(B), A ∈ A, B ∈ B.

Then η is a well-defined measure on R. Further, if µ, ν are σ-finite, thenthere exists a unique measure η : A⊗ B → [0,+∞] such that

η(A×B) = η(A×B) for every A×B ∈ R.

5.2.2. Definition:The measure η on A ⊗ B given by theorem 5.2.1 is called the productof the measures µ and ν and is denoted by µ × ν. The measure space(X × Y,A ⊗ B, µ × ν) is called the product of the measure space (X,A, µ)and (Y,B, ν), or just the product measure space.

We note that µ× ν is uniquely defined on A×B when µ, ν are σ-finite.So from now on we shall assume that µ and ν are σ-finite. As is clear, µ× νis defined on A⊗B via the extension theory . So, the natural question arises:how to compute (µ× ν)(E) for a general element E ∈ A⊗ B?


5.2.3. Definition:Let E ⊆ X × Y, x ∈ X and y ∈ Y . Let

Ex := {y ∈ Y | (x, y) ∈ E} and Ey := {x ∈ X | (x, y) ∈ E}.

The set Ex is called the section of E at x (or x-section of E) and theset Ey is called the section of E at y (or y-section of E).

5.2.4. Examples:

(i) Let E = A× B, where A ∈ A and B ∈ B. Clearly, Ex = B if x ∈ A andEx = ∅ if x 6∈ A. Similarly, Ey = A if y ∈ B and Ey = ∅ if y 6∈ B.

(ii) Let (X,A) be a measurable space and let A ∈ A. Let

E = {(x, t) ∈ X × R | 0 ≤ t < χA(x)}.

It is easy to see that

E = (A× [0, 1)) ∪ (Ac × {0}).

Thus

Ex =

{

[0, 1) if x ∈ A,{0} if x 6∈ A.

Similarly,

Ey =

X if y = 0,A if y ∈ (0, 1),∅ if y 6∈ [0, 1).

5.2.5. Proposition:For E,F,Ei ∈ A⊗B and i ∈ I, any indexing set, the following hold ∀ x ∈X, y ∈ Y :

(i) (⋃

i∈I Ei)x =⋃

i∈I(Ei)x and (⋃

i∈I Ei)y =

⋃

i∈I(Ei)y.

(ii) (⋂

i∈I Ei)x =⋂

i∈I(Ei)x and (⋂

i∈I Ei)y =

⋂

i∈I(Ei)y.

(iii) (E \ F )x = Ex \ Fx and (E \ F )y = Ey \ F y.

(iv) If E ⊆ F, then Ex ⊆ Fx and Ey ⊆ F y.

5.2.6. Theorem:Let E ∈ A⊗ B. Then the following hold:

(i) Ex ∈ B and Ey ∈ A for every x ∈ X, y ∈ Y.

(ii) The functions x 7−→ ν(Ex) and y 7−→ µ(Ey) are measurable func-tions on X and Y, respectively.

(iii)∫

Xν(Ex)dµ(x) = (µ× ν)(E) =

∫

Yµ(Ey)dν(y).

5.2. Product of measure spaces 67

5.2.7. Remark:Given σ-finite measure spaces (X,A, µ) and (Y,B, ν), we showed that µ×ν isthe unique measure on A⊗B such that (µ×ν)(A×B) = µ(A)ν(B). Note thatthe measure space (X×Y,A⊗B, µ×ν) need not be a complete measure spaceeven if the measure spaces (X,A, µ) and (Y,B, ν) are complete. For example,if A ⊂ X,A 6∈ A and ∅ 6= B ∈ B with µ(B) = 0, then (µ× ν)∗(A×B) = 0,but A × B 6∈ A ⊗ B. In fact, theorem 5.2.1 itself gives a complete measurespace (X × Y,A⊗ B, µ× ν), where A⊗ B is the σ-algebra of η∗-measurablesubsets of X × Y , η being as in theorem 5.2.1, and µ × ν is the restrictionof η∗ to A⊗ B. The measure space (X × Y,A⊗ B, µ × ν) is nothing butthe completion of the measure space (X × Y,A⊗B, µ× ν). It is easy to seethat theorem 5.2.6 holds for E ∈ A⊗ B also, as claimed in the next exercise(5.7).

Exercises:

(5.5) Let (X,A, µ) be a σ-finite measure space. For any nonnegativefunction f : X −→ R, let

E∗(f) := {(x, t) ∈ X × R | 0 ≤ t ≤ f(x)}

and

E∗(f) := {(x, t) ∈ X × R | 0 ≤ t < f(x)}.

Then the following hold:(i) If {fn}n≥1 is an increasing sequence of nonnegative functions

on X increasing to f(x), show that {E∗(fn)}n≥1 is an increas-ing sequence of sets in A⊗BR with

∞⋃

n=1

E∗(fn) = E∗(f).

(ii) Using (i) and the ‘simple function technique’, show that E∗(f) ∈A ⊗ BR, whenever f : X −→ R is a nonnegative measurablefunction. Deduce that E∗(f) ∈ A⊗ BR.(Hint: E∗(f) =

⋂∞n=1E∗(f + 1/n).)

(iii) Let f : X −→ R be a nonnegative function such that E∗(f) ∈A⊗ BR. Using exercise (5.1) and the following equality

A := {(x, t) ∈ X × R | f(x) > c, t > 0}

=∞⋃

n=1

{(x, t) ∈ X × R | (x, t/n+ c) ∈ E∗(f), t > 0},

show that A ∈ A⊗ BR, and deduce that f is measurable.


(iv) Let f : X −→ R be any measurable function. Show thatG(f) ∈ A⊗ BR, where

G(f) := {(x, t) ∈ X × R | f(x) = t}.

The set G(f) is called the graph of the function.(v) Let f ∈ L1(X,A, µ), where µ(X) < +∞. Show that

(µ× ν)(E∗|f |) =

∫

|f |dµ = (µ× ν)(E∗|f |). (5.1)

(Hint: First prove this for the case when f is bounded. For thegeneral case, consider fn = |f | ∧ n and note that {E∗(fn)}n≥1

increases to E∗|f |.)(vi) Extend (iv) to the case when µ is σ-finite.

The identity (5.1) shows that for nonnegative functions∫

fdµrepresents the area below the curve y = f(x).

(5.6) Let X be any nonempty set and P(X) be its power set. Let A :=P(X) ⊗ P(X) and let D := {(x, y) ∈ X × X |x = y}. SupposeD ∈ A. Prove the following statements:(i) There exist sets Ai, Bi ∈ P(X) such that D belongs to the

σ-algebra generated by (Ai ×Bi), i = 1, 2, . . . .(Hint: See exercise (1.19).)

(ii) Let B = S({Ai | i = 1, 2, . . .}). Then card (B) ≤ c. For everyx, y ∈ X,Dx 6= Dy if x 6= y and Dx ∈ B ∀ x ∈ X. Deducethat card (X) ≤ c.

(iii) If card (X) > c, then D 6∈ P(X) ⊗ P(X), even though Dx ∈P(X) and Dy ∈ P(X), ∀ x, y ∈ X. Hence in general,

P(X) ⊗ P(X) 6= P(X ⊗X).

(5.7) Let E ∈ A⊗ B. Then the functions x 7−→ ν(Ex) and y 7−→ µ(Ey)are measurable and

∫

Xν(Ex)dµ(x) = (µ× ν)(E) =

∫

Yµ(Ey)dν(y).

(Hint: E = F ∪N,F ∈ A ⊗ B and (µ × ν)∗(N) = 0 by theorem3.4.7.)

(5.8) Let E ∈ A⊗B be such that µ(Ey) = 0 for a.e. (ν)y ∈ Y . Show thatµ(Ex) = 0 for a.e. (µ)x ∈ X. What can you say about (µ× ν)(E)?

5.3. Integration on product spaces: Fubini’s theorems 69

5.3. Integration on product spaces: Fubini’s

theorems


Let (X,A, µ) and (Y,B, ν) be σ-finite measure spaces and (X×Y,A⊗B, µ×ν)the product measure space. Theorem 5.2.6 can be interpreted as follows: forevery E ∈ A⊗ B,

∫

X×YχE (x, y)d(µ× ν)(x, y) =

∫

X

(∫

YχE (x, y)dν(y)

)

dµ(x)

=

∫

Y

(∫

XχE (x, y)dµ(x)

)

dν(y).

This allows us to compute the integral of the function χE (x, y) by integratingone variable at a time. So, the natural question arises: does the above holdwhen χE is replaced by a nonnegative measurable function on X × Y ? Theanswer is yes, and is made precise in the next theorem.

5.3.1. Theorem (Fubini):Let f : X × Y −→ R be a nonnegative A ⊗ B-measurable function. Then

the following statements hold:

(i) For x0 ∈ X and y0 ∈ Y fixed, the functions x 7−→ f(x, y0) andy 7−→ f(x0, y) are measurable on X and Y, respectively.

(ii) The functions y 7−→∫

X f(x, y)dµ(x) and x 7−→∫

Y f(x, y)dν(y) arewell-defined nonnegative measurable functions on Y and X, respec-tively.

(iii)

∫

X

(∫

Yf(x, y)dν(y)

)

dµ(x) =

∫

Y

(∫

Xf(x, y)dµ(x)

)

dν(y)

=

∫

X×Yf(x, y)d(µ× ν)(x, y).

In view of theorem 5.3.1, it is natural to expect a similar result forf ∈ L1(µ × ν), µ and ν being σ-finite measures. This is given by thefollowing theorem.

5.3.2. Theorem (Fubini):Let f ∈ L1(µ× ν). Then the following statements are true:

(i) The functions x 7−→ f(x, y) and y 7−→ f(x, y) are integrable fora.e. y(ν) and for a.e. x(µ), respectively.


(ii) The functions

y 7−→

∫

Xf(x, y)dµ(x) and x 7−→

∫

Yf(x, y)dν(y)

are defined for a.e. y(ν) and a.e. x(µ) , and are ν, µ-integrable,respectively.

(iii)

∫

Y

(∫

Xf(x, y)dµ(x)

)

dν(y) =

∫

X×Yf(x, y) d(µ× ν)

=

∫

X

(∫

Yf(x, y)dν(y)

)

dµ(x).

Theorem 5.3.1, exercise (5.9) and theorem 5.3.2 together give us thefollowing theorem, which enables us to check the integrability of a functionof two variables and compute its integral.

5.3.4. Theorem:Let (X,A, µ) and (X,B, ν) be σ-finite measure spaces. Let f : X × Y −→ R

be an A⊗B-measurable function such that f satisfies any one of the followingstatements:

(i) f is nonnegative.

(ii) f ∈ L1(µ× ν).

(iii)

∫

X

(∫

Y|f(x, y)|dν(y)

)

dµ(x) < +∞.

(iv)

∫

Y

(∫

X|f(x, y)|dµ(x)

)

dν(y) < +∞.

Then

∫

X×Yf(x, y)d(µ× ν) =

∫

X

(∫

Yf(x, y)dν(y)

)

dµ(x)

=

∫

Y

(∫

Xf(x, y)dµ(x)

)

dν(y),

in the sense that all the integrals exist and are equal.

We give next some examples which illustrate the necessity of the condi-tions on µ, ν and f for the conclusions of theorem 5.3.4 to hold.

5.3.5. Example:Let X = Y = [0, 1] and A = B = B[0,1], the σ-algebra of Borel subsets of[0, 1]. Let µ be the Lebesgue measure on A and ν be the counting measureon B, i.e., ν(E) := number of elements in E if E is finite and ν(E) := +∞


otherwise. Let D := {(x, y) |x = y}. Let, for n ≥ 1,

Dn :=n⋃

j=1

( [(j − 1)/n, j/n] × [(j − 1)/n, j/n] ).

Then D = ∩∞n=1Dn. Thus D ∈ A ⊗ B. In fact, D is a closed subset of

[0, 1] × [0, 1]. Further,∫

XχD(x, y)dµ(x) = 0 ∀ y ∈ Y and

∫

YχD(x, y)dν(y) = 1 ∀x ∈ X.

Hence∫

Y

(∫

XχD(x, y)dµ(x)

)

dν(y) = 0 and

∫

X

(∫

YχD(x, y)dν(y)

)

dµ(x) = 1.

This does not contradict theorem 5.3.1, since ν is not σ-finite.

5.3.6. Example:Let X = Y = [0, 1],A = B = B[0,1], and let µ = ν be the Lebesgue measureon [0,1]. Let

f(x, y) :=

x2 − y2

(x2 + y2)2if (x, y) 6= (0, 0),

0 if x = y otherwise.

Noting that for fixed x, f(x, y) is a Riemann integrable function on [0, 1] and

∂

∂y

(

y

x2 + y2

)

= f(x, y),

it is easy to see that∫ 1

0

(∫ 1

0f(x, y)dµ(x)

)

dν(y) = −

∫ 1

0

(∫ 1

0f(x, y)dν(y)

)

dµ(x) = −π/4.

This does not contradict theorem 5.3.2, because f 6∈ L1(X × Y ) as

∫

[0,1]×[0,1]|f(x, y)|d(µ× ν) =

∫ 1

0

(∫ 1

0|f(x, y)|dν(y)

)

dµ(x)

≥

∫ 1

0

(∫ x

0|f(x, y)|dν(y)

)

dµ(x)

=

∫ 1

0

1

x

(

∫ π/4

0cos 2θdθ

)

dµ(x)

=

∫ 1

0

1

2xdµ(x) = +∞.


*5.3.7. Example:Assume the continuum hypothesis and let < be the well-order on [0, 1] suchthat ∀x ∈ [0, 1], {y ∈ [0, 1] | y < x} is at most a countable set. Let

E := {(x, y) ∈ [0, 1] × [0, 1] |x < y}.

Then ∀ y fixed and ∀ x ∈ [0, 1], χE (x, y) = 1 if x < y and 0 otherwise.Hence χE (x, y) is the indicator function of a countable set for y ∈ [0, 1]fixed. Thus it is a Borel measurable function. Similarly, for x ∈ [0, 1] fixed,y 7−→ χE (x, y) is Borel measurable, for it is the indicator function of [0, 1]\C,a countable set. However, χE is not Borel measurable on [0, 1]×[0, 1]. To seethis, note that B[0,1]×[0,1] = B[0,1]⊗B[0,1]. Thus if χE were Borel measurable,then by theorem 5.3.1 we should have

∫

[0,1]×[0,1]χ

E(x, y)d(λ× λ)(x, y)

=

∫

[0,1]

(

∫

[0,1]χE (x, y)dλ(x)

)

dλ(y)

=

∫

[0,1]

(

∫

[0,1]χE (x, y)dλ(y)

)

dλ(x). (5.2)

However, it is easy to see that for x, y ∈ [0, 1],∫

[0,1]χE (x, y)dλ(y) = 1 and

∫

[0,1]χE (x, y)dλ(x) = 0.

This contradicts (5.2). Hence χE is not Borel measurable.

Exercises:

(5.9) Let f : X×Y −→ R be A⊗B-measurable. Show that the followingstatements are equivalent:(i) f ∈ L1(µ× ν) := L1(X × Y,A⊗ B, µ× ν).

(ii)

∫

Y

(∫

X|f(x, y)| dµ(x)

)

dν(y) < +∞.

(iii)

∫

X

(∫

Y|f(x, y)|dν(y)

)

dµ(x) < +∞.

(5.10) Let (X,A, µ) and (Y,B, ν) be complete σ-finite measure spaces andlet (X×Y,A⊗ B, µ×ν) be the completion of (X×Y,A⊗B, µ×ν).Let f : X × Y −→ R be any nonnegative extended real valuedA⊗ B-measurable function. Show that:(i) The function x 7−→ f(x, y) is A-measurable for a.e. y(ν) and

the function y 7−→ f(x, y) is B-measurable for a.e. x(µ).


(ii) The function y 7−→∫

X f(x, y)dµ(x) is B-measurable and thefunction x 7−→

∫

Y f(x, y)dν(y) is A-measurable.

(iii)

∫

X×Yf(x, y)d(µ× ν) =

∫

X

(∫

Yf(x, y)dν(y)

)

dµ(x)

=

∫

Y

(∫

Xf(x, y)dµ(x)

)

dν(y).

(5.11) Let X = Y = [−1, 1],A = B = B[−1,1], and let µ = ν be theLebesgue measure on [−1, 1]. Let

f(x, y) :=

{ xy

(x2 + y2)2if (x, y) 6= (0, 0),

0 otherwise.

Show that∫ 1

−1

(∫ 1

−1f(x, y)dν(y)

)

dµ(x) = 0 =

∫ 1

−1

(∫ 1

−1f(x, y)dµ(x)

)

dν(y).

Can you conclude that∫ 1

−1

(∫ 1

−1f(x, y)dν(y)

)

dµ(x) =

∫

X×Yf(x, y)d(µ× ν)(x, y)?

(5.12) Let f ∈ L1(X,A, µ) and g ∈ L1(Y,B, ν). Let

φ(x, y) := f(x)g(y), x ∈ X and y ∈ Y.

Show that φ ∈ L1(X × Y,A⊗ B, µ× ν) and∫

X×Yφ(x, y)d(µ× ν) =

(∫

Xfdµ

)(∫

Ygdν

)

.

(5.13) Let f ∈ L1(0, a) and let

g(x) :=

∫ a

x(f(t)/t)dλ(t),0 < x ≤ a.

Show that g ∈ L1(0, a), and compute

∫ a

0g(x)dλ(x).

*(5.14) Let (X,A, µ), and (X,B, ν) be as in example 5.3.7. Define, forx, y ∈ [0, 1],

f(x, y) :=

{

1 if x is rational,2y if y is irrational.

Compute∫ 1

0

(∫ 1

0f(x, y)dν(y)

)

dµ(x) and

∫ 1

0

(∫ 1

0f(x, y)dµ(x)

)

dν(y).

Is f in L1(µ× ν)?


(5.15) Let (X,A, µ) be as in example 5.3.7. Let Y = [1,∞),B = LR ∩[1,∞), and let ν be the Lebesgue measure restricted to [1,∞).Define, for (x, y) ∈ X × Y,

f(x, y) := e−xy − 2e−2xy.

Show that f 6∈ L1(µ× ν).

(5.16) Let X be a topological space and let BX be the σ-algebra of Borelsubsets of X. A function f : X −→ R is said to be Borel measur-able if f−1(E) ∈ BX ∀ E ∈ BR. Prove the following:(i) f is Borel measurable iff f−1(U) ∈ BX for every open set

U ⊆ R.(Hint: Use the ‘σ-algebra technique’.)

(ii) Let f : X −→ R be continuous. Show that f is Borel measur-able.

(iii) Let {fn}n≥1 be a sequence of Borel measurable functions onX such that f(x) := lim

n→∞fn(x) exists ∀ x ∈ X. Show that f

also is Borel measurable.(iv) Consider R2 with the product topology and let f, g be Borel

measurable functions on R. Show that the function φ on R2

defined by

φ(x, y) := f(x)g(y), x ∈ X, y ∈ Y,

is Borel measurable.

(5.17) Let f : R2 −→ R be Borel measurable. Show that for x ∈ X fixed,y 7−→ f(x, y) is a Borel measurable function on R. Is the functionx 7−→ f(x, y), for y ∈ Y fixed, also Borel measurable?

(5.18) Let f : R2 −→ R be such that for x ∈ X fixed, y 7−→ f(x, y) isBorel measurable and for y ∈ Y fixed, x 7−→ f(x, y) is continuous.(i) For every n ≥ 1 and x, y ∈ R, define

fn(x, y) := (i− nx)f((i− 1)/n, y) + (nx− i+ 1)f(i/n, y),

whenever x ∈ [(i− 1)/n, i/n), i ∈ Z. Show that each fn : R2 −→R is continuous and hence is Borel measurable.

(ii) Show that fn(x, y) → f(x, y) as n → ∞ for every (x, y) ∈ R2,and hence f is Borel measurable.

(5.19) Let (X,A) and (Y,B) be measurable spaces and let f : X×Y −→ R

be a nonnegative A ⊗ B-measurable function. Let µ be a σ-finitemeasure on (Y,B). For any E ∈ B and x ∈ X, let

η(x,E) :=

∫

Ef(x, y)dµ(y).

Show that η(x,E) has the following properties:

5.4. Lebesgue measure on R2 and its properties 75

(i) For every fixed E ∈ B, 7−→ η(x,E) is an A-measurable func-tion.

(ii) For every fixed x ∈ X,E 7−→ η(x,E) is a measure on (Y,B).A function η : X×B −→ [0,∞) having properties (i) and (ii) aboveis called a transition measure.

5.4. Lebesgue measure on R2 and its properties


We now specialize the construction of (X×Y,A⊗B, µ×ν) to the case whenX = Y = R, A = B = LR and µ = ν = λ, the Lebesgue measure. Wehave already seen in exercise 5.4 that BR ⊗ BR = BR2 and that (R2,LR ⊗LR, λ× λ) is not complete. The completion of this measure space, denotedby (R2,LR2 , λR2), is called the Lebesgue measure space. Elements ofLR2 are called the Lebesgue measurable subsets of R2, and λR2 is calledthe Lebesgue measure on R2. The following proposition ensures that λR2

is the unique extension of the natural concept of area in R2.

5.4.1. Proposition:Let I denote the collection of left-open, right-closed intervals in R, and letI2 := {I × J | I, J ∈ I}. Then the following hold:

(i) I2 is a semi-algebra of subsets of R2, and S(I2) = BR2 .

(ii) λR2(I × J) = λ(I)λ(J), ∀ I, J ∈ I.

(iii) The measure space (R2,LR2 , λR2) is the completion of the measurespaces (R2,LR ⊗ LR, λ× λ) and (R2,BR2 , λR2).

We describe next some properties of the Lebesgue measure λR2 . ForE ⊆ R2 and x ∈ R2, let E + x := {y + x | y ∈ E}.

5.4.2. Theorem:The Lebesgue measure λR2 has the following properties:

(i) Let E ∈ BR2 and x ∈ R2. Then E + x ∈ BR2 and

λR2(E) = λR2(E + x).

(This property of λR2 is called translation invariance.)

(ii) For every nonnegative Borel measurable function f on R2 and y ∈R2,

∫

f(x + y) dλR2(x) =

∫

f(x) dλR2(x) =

∫

f(−x) dλR2(x).


(iii) Let µ be any σ-finite measure on BR2 such that µ(E + x) = µ(E)∀ E ∈ BR2 ,x ∈ R2. Suppose

0 < µ(E0) = CλR2(E0) < +∞,

for some E0 ∈ B and for some C ≥ 0. Then

µ(E) = CλR2(E), ∀ E ∈ BR.

5.4.3. Theorem:Let T : R2 −→ R2 be a linear transformation and E ∈ LR2 . Then T (E) ∈LR2 and

λR2(T (E)) = |detT |λR2(E).

5.4.4. Theorem (Integration of ‘radial’ functions):Let f : [0,∞) −→ (0,∞) be a nonnegative measurable function. Then

∫

R2

f(|x|)dλR2(x) = 2π

∫ ∞

0f(r)rdλ(r).

Proof: The proof is once again an application of the ‘simple function tech-nique’. We outline the steps

Step 1: The theorem holds for f = χ(a,b) , 0 ≤ a < b < +∞.

Step 2: Let {En}n≥1 be a sequence of sets from [0,∞) ∩ LR such thateither the En’s are pairwise disjoint, or the En’s are increasing. If thetheorem holds for each χEn

, then the theorem holds for χE also, where

E =⋃∞

n=1En.

Step 3: The theorem holds for f = χU , U being any open subset of [0,∞).

Step 4: The theorem holds for f = χN , where N ⊂ [0,∞) and λ(N) = 0.

Step 5: The theorem holds when f = χE , E ∈ LR and E ⊆ [0,∞).

Step 6: The theorem holds for any nonnegative measurable function.

Exercises:

(5.19) (Regularity of λR2): Prove the following:(i) λR2(U) > 0 for every nonempty open subset U of R2.(ii) A set E ∈ LR2 iff ∀ ǫ > 0, there exists an open set U such

that E ⊆ U and λ(U \ E) < ǫ.(Hint: Proceed as in theorem 3.5.2.)

(iii) λR2(K) < +∞ for every compact subset K of R.

5.4. Lebesgue measure on R2 and its properties 77

(iv) λR2(U) = sup{λR2(K) | K compact , K ⊆ U}.(Hint: Given U open, there exists a sequence of compact setsKn such that Kn ↑ U .)

(v) If E ∈ BR2 is such that λR2(E) < +∞, then

λR2(E) = sup{λR2(k) | K ⊆ E,K compact}.

(5.19) Show that for f ∈ L1(R2,LR2 , λR2),x ∈ R2, the function y 7−→

f(x + y) is integrable and∫

f(x + y)dλR2(x) =

∫

f(x)dλR2(x).

(Hint: Use exercise (4.30) and theorem 5.4.2.)

(5.20) Let E ∈ LR2 and x = (x, y) ∈ R2. Let

xE := {(xt, yr) | (t, r) ∈ E}.

Prove the following:(i) xE ∈ LR2 for every x ∈ R, E ∈ LR2 , and λR2(xE) = |xy|λR2(E).(ii) For every nonnegative Borel measurable function f : R2 −→ R,

∫

f(xt)dλR2(t) = |xy|

∫

f(t)dλR2(t),

where for x = (x, y) and t = (s, r),xt := (xs, yr).(iii) Let λR2{x ∈ R2 | |x| ≤ 1} =: π. Then

λR2{x ∈ R2| |x| < 1} = π and λR2{x ∈ R2| |x| < r} = πr2.

(iv) Let E be a vector subspace of R2. Then λR2(E) = 0 if E hasdimension less than 2.

(5.21) Let T : R2 → R be a linear map.(i) If N ⊆ R2 is such that λ∗

R2(N) = 0, show that λ∗R2(T (N)) = 0.

(ii) Use (i) above and proposition 5.4.1(ii) to complete the proofof theorem 5.4.3 for sets E ∈ LR2 .

(5.14) Consider the vectors (a1, b1), (a2, b2) ∈ R2 and let

P := {(α1a1 + α2a2, α1b1 + α2b2) ∈ R2 |α1, α2 ∈ R, 0 ≤ αi ≤ 1},

called the parallelogram determined by these vectors. Show that

λR2(P ) = |a1b2 − a2b1|.

(5.15) Let E ∈ LR2 with 0 < λ(E) < ∞. Show that there exists a uniquepoint (c1, c2) ∈ R2 such that

∫

xχE (x+ c1, y + c2)dλR2(x, y) = 0

and∫

yχE (x+ c1, y + c2)dλR2(x, y) = 0.


In fact,

c1 =1

λ(E)

∫

xχE (x, y)dλR2(x, y)

and

c2 =1

λ(E)

∫

yχE (x, y)dλR2(x, y).

(c = (c1, c2) is called the centroid of E.)

(5.16) Let f : [0,∞) −→ R be a measurable function. Show that, if eitherof the two integrals in theorem 5.4.4 exists, then so does the otherand the equality holds.

(5.17) Let f : (a, b) × (c, d) −→ R be such that(i) f is continuous;

(ii)∂f

∂xexists and is continuous;

(iii) for some t ∈ (a, b),d

dyf(t, y) exists ∀ y;

(iv)∂

∂y

(

∂f

∂x

)

exists and is continuous.

Show that∂f

∂yand

∂

∂x

(

∂f

∂y

)

exist. Moreover,

∂

∂x

(

∂f

∂y

)

=∂

∂y

(

∂f

∂x

)

.

(Hint: Use Fubini’s theorem and the fundamental theorem of cal-culus to show that ∀ ξ ∈ (a, b), η ∈ (c, d) and s < y

f(ξ, η) − f(t, η) − f(ξ, s) + f(t, s) =

∫ η

s

∫ ξ

t

∂

∂y

(

∂t

∂x

)

dx dy,

and note that the inner integral on the right is a continuous functionof ξ.)

Chapter 6

Lp-spaces

Throughout this chapter, we shall work with a fixed σ-finite measure space(X,S, µ), which is assumed to be complete.

The main aim of this chapter is to analyze the convergence of sequencesof measurable functions. Given a sequence {fn}n≥1 of measurable functionson (X,S, µ), we have already come across concepts like pointwise conver-gence of {fn}n≥1 to a measurable function f, i.e., when {fn(x)} convergesto f(x) ∀ x ∈ X. When a sequence does not converge pointwise, one wouldlike to find other methods of analyzing the behavior of the sequence {fn}n≥1

for large n. Analysis of these methods, and the relations between them, isthe main aim of this chapter. But before we do that, we extend the notionof measurability and integration to complex-valued functions.

6.1. Integration of complex-valued functions


Let (X,S, µ) be a measure space and let C denote the field of complexnumbers. For a function f : X −→ C, consider the functions Re (f) andIm (f) defined by: ∀ x ∈ X,

Re (f)(x) := Real part of f(x)

and

Im (f)(x) := Imaginary part of f(x).

The functions Re (f) and Im (f) are called, respectively, the real part andthe imaginary part of the function f. Note that Re (f) and Im (f) arereal-valued functions on X.

79

80 6. Lp-spaces

6.1.1. Definition:A complex-valued function f : X −→ C is said to be measurable if bothRe (f) and Im (f) are measurable functions. We say f is µ-integrable ifboth Re (f) and Im (f) are integrable. In that case we define the integralof f, denoted by

∫

fdµ, to be∫

fdµ :=

∫

Re (f)dµ+ i

∫

Im (f)dµ.

We denote the set of all complex-valued µ-integrable functions on Xby L1(X,S, µ) itself. Whenever we restrict ourselves to only real-valuedµ-integrable functions on X, we shall specify it by Lr

1(X,S, µ). Our nexttheorem tells us that L1(X,S, µ) is as nice a space as Lr

1(X,S, µ) is.

6.1.2. Theorem:

(i) Let f, g ∈ L1(X,S, µ) and α, β ∈ C. Then αf + βg ∈ L1(X,S, µ)and

∫

(αf + βg)dµ = α

∫

fdµ+ β

∫

gdµ.

(ii) Let f : X −→ C be a measurable function. Then f ∈ L1(X,S, µ)iff |f | ∈ Lr

1(X,S, µ). Further, in either case,

∣

∣

∫

fdµ∣

∣ ≤

∫

|f |dµ.

(iii) Let f ∈ L1(X,S, µ) and E ∈ S. Then χEf ∈ L1(X,S, µ). We write∫

Efdµ :=

∫

χEfdµ.

If E1, E2 ∈ S and E1 ∩ E2 = ∅, then∫

E1∪E2

fdµ =

∫

E1

fdµ+

∫

E2

fdµ.

(iv) Let f ∈ L1(X,S, µ), and let {En}n≥1 be a sequence of pairwisedisjoint sets from S. Then the series

∑∞n=1(

∫

Enfdµ) is absolutely

convergent. Also, (χEf) ∈ L1(X,S, µ), where E :=⋃∞

n=1En, and

∫

Efdµ =

∞∑

n=1

∫

En

fdµ.

That the integral of complex-valued functions behaves as nicely withrespect to limiting operations as the integral of real-valued functions is thecontent of the next theorem.

6.1. Integration of complex-valued functions 81

6.1.3. Theorem (Complex form of Lebesgue’s dominated conver-gence):Let {fn}n≥1 be a sequence in L1(X,S, µ). Let f(x) := lim

n→∞fn(x) exist for

a.e. x ∈ X and let there exist a function g ∈ Lr1(X,S, µ) such that ∀ n,

|fn(x)| ≤ g(x) for a.e. x(µ). Then f ∈ L1(X,S, µ) and

limn→∞

∫

fn dµ =

∫

f dµ.

6.1.4. Theorem:Let f ∈ L1(X,S, µ), where µ(X) < ∞. Let S be a closed subset of C suchthat ∀ E ∈ S with µ(E) > 0,

(

1

µ(E)

∫

Ef dµ

)

∈ S.

Then f(x) ∈ S for a.e. x ∈ X.

Exercises:

(6.1 ) Let f ∈ L1(X,S, µ) be such that∣

∣

∣

∣

∫

f dµ

∣

∣

∣

∣

=

∫

|f | dµ.

What can you conclude about f?

(6.2) Let f : X → C. Show that f is measurable iff f−1(E) ∈ S for everyBorel set E ⊆ X.

(6.3) Extend the claim of exercise (4.30) to complex-valued measurablefunctions.

(6.4) Let {fn}n≥1 be a sequence of S-measurable complex-valued func-tions on (X,S, µ). Show that the following statements are equiva-lent:(i) (

∑∞n=1 |fn|) ∈ Lr

1(X,S, µ).(ii)

∑∞n=1

∫

|fn|dµ < +∞.Further, if either of the above is true, then (

∑∞n=1 fn) ∈ L1(X,S, µ)

and∫

(

∞∑

n=1

fn

)

dµ =∞∑

n=1

∫

fn dµ.

(6.5) State and prove the Riesz-Fischer theorem (theorem 4.6.3) for complex-valued integrable functions.

82 6. Lp-spaces

(6.6) Let f ∈ L1(X,S, µ) and∫

Efdµ = 0 for every E ∈ S.

Show that f(x) = 0 for a.e. x(µ) (see proposition 5.4.6).

6.2. Lp-spaces


In section 4.6, we analyzed L1[a, b], the space of Lebesgue integrable func-tions on [a, b]. We saw that L1[a, b] is a vector space over R, and that forf ∈ L1[a, b], we can define the notion of absolute value of f, i.e., ‖f‖1. Fur-ther, this notion of absolute value can be used to define the L1-metric onL1[a, b] so that L1[a, b] becomes a complete metric space. In this sectionwe look at examples of a family of spaces of this type, called Lp-spaces.Throughout this section, we fix a measure space (X,S, µ) which is σ-finiteand complete.

6.2.1. Definition:Let 0 < p < ∞. Let Lp(µ) := Lp(X,S, µ) denote the space of all complex-valued S-measurable functions on X such that

∫

|f |pdµ < +∞.

The space Lp(µ) is called the space of pth-power integrable functions.

6.2.2. Proposition:The space Lp(X,S, µ) is a vector space over C.

6.2.3. Definition:Let f ∈ Lp(X,B, µ). Define ‖f‖p, called the pth-norm of f, as follows:

‖f‖p :=

(∫

|f |pdµ

)1/p

.

Since ‖f‖p = ‖g‖p if f(x) = g(x) for a.e. x(µ), we treat such f and g asthe same element of Lp(X,B, µ). To show that ‖f‖p has the properties of ametric (as was the case for p = 1 in section 4.6), we need some inequalities.

6.2.4. Lemma:For nonnegative real numbers a, b and 0 < t < 1, the following inequalitieshold:

(i) atb1−t ≤ ta+ (1 − t)b.

6.2. Lp-spaces 83

(ii) (a+ b

2)1/t ≤

1

2(a1/t + b1/t).

6.2.5. Theorem (Holder’s inequality):Let p > 1 and q > 1 be such that 1/p + 1/q = 1. Let f ∈ Lp(µ) and

g ∈ Lq(µ). Then fg ∈ L1(µ) and∫

|fg|dµ ≤

(∫

|f |pdµ

)1/p(∫

|g|qdµ

)1/q

.

6.2.6. Corollary:Let p, q > 1 be real numbers with 1/p + 1/q = 1. Let {an}n≥1 and {bn}n≥1

be sequences of complex numbers such that∞∑

n=1

|an|p <∞ and

∞∑

n=1

|bn|q <∞.

Then∑∞

n=1 |anbn| < +∞ and(

∞∑

n=1

|anbn|

)

≤

(

∞∑

n=1

|an|p

)1/p( ∞∑

n=1

|bn|q

)1/q

.

In particular, if an = bn = 0 ∀ n ≥ k, then

k∑

n=1

|anbn| ≤

(

k∑

n=1

|an|p

)1/p( k∑

n=1

|bn|q

)1/q

.

6.2.7. Note:In the special case when p = q = 2, Holder’s inequality is known as theCauchy-Schwarz inequality.

6.2.8. Theorem (Minkowski’s inequality):Let 1 ≤ p <∞ and f, g ∈ Lp(µ). Then f + g ∈ Lp(µ) and

‖f + g‖p ≤ ‖f‖p + ‖g‖p.

6.2.9. Theorem (Riesz-Fischer):The space Lp(X,S, µ), for 1 ≤ p <∞, is a complete metric space with the

metric defined by

dp(f, g) := ‖f − g‖p ∀ f, g ∈ Lp(X,S, µ).

84 6. Lp-spaces

Of course, here we identify f, g ∈ Lp(X,S, µ) if f(x) = g(x) for a.e.x(µ).

6.2.10. Note:In theorem 6.2.9, we showed that Lp(µ), 1 ≤ p < ∞, is a complete metricspace under the metric

dp(f, g) := ‖f − g‖p :=

(∫

|f − g|pdµ

)1/p

.

It is natural to ask the same question for 0 < p < 1. If 0 < p < 1, define forf, g ∈ Lp(µ)

dp(f, g) :=

∫

|f − g|pdµ.

Using the inequality (see exercise (6.7))

1 + tp ≥ (1 + t)p ∀ t ≥ 0,

it is easy to see that dp is a metric on Lp(µ). Also, proceeding as in theorem6.2.9, we can show that Lp(µ) is a complete metric space for 0 < p < 1.In the case 1 ≤ p < ∞, f 7−→ ‖f‖p is a real-valued map on Lp(µ) with theproperties: ∀ f, g ∈ Lp(µ) and α ∈ C

(i) ‖f‖p ≥ 0, and ‖f‖p = 0 iff f = 0.

(ii) ‖αf‖p = |α|‖f‖p.

(iii) ‖f + g‖p ≤ ‖f‖p + ‖g‖p, called the triangle inequality.

Such a function can be defined on any vector space, and is called a norm.A vector space with a norm is called a normed linear space. Thus for1 ≤ p < ∞, the Lp-spaces are examples of normed linear spaces whichare also complete under the metric ‖f − g‖p, the metric induced by thenorm. Such normed linear spaces are called Banach spaces. However,when 0 < p < 1, f 7−→ ‖f‖p is no longer a norm, as it fails to satisfy thetriangle inequality. To see this, consider a measure space (X,S, µ) such that∃ A,B ∈ S with A ∩B = ∅ and 0 < µ(A) < ∞, 0 < µ(B) <∞. Let α, β bepositive real numbers. Then

‖αχA‖p =

(∫

|α|pχA(x)dµ(x)

)1/p

= α(µ(A))1/p.

6.2. Lp-spaces 85

Similarly, ‖βχB‖p = β(µ(B))1/p. Further,

‖αχA + βχB‖p =

[∫

(αχA + βχB )p dµ

]1/p

=

[∫

(αpχA + βpχB ) dµ

]1/p

= (αpµ(A) + βpµ(B))1/p .

Now, using exercise 6.7(ii) with t = (β/α)p(µ(B)/µ(A)), we have

(

1 +

(

β

α

)p µ(B)

µ(A)

)1/p

> 1 +

(

β

α

)(

µ(B)

µ(A)

)1/p

,

i.e.,

(α1/pµ(A) + β1/pµ(B))1/p > α(µ(A))1/p + β(µ(B))1/p,

i.e.,

‖αχA + βχB‖p > ‖αχA‖p + ‖βχB‖p.

This is the reason that Lp-spaces for 0 < p < 1 are not very interesting tostudy.

Exercises:

(6.7) Let 0 < t <∞ and 0 < p < 1. Show that(i) (1 + t)p < 1 + tp.

(ii) (1 + t)1/p > 1 + t1/p.

(6.8) Let 0 < p < 1 and −∞ < q < 0 be such that 1/p + 1/q = 1. Letf, g be positive functions such that f ∈ Lp(µ) and g ∈ Lq(µ). Showthat

‖f‖p‖g‖q <

∫

fgdµ.

(Hint: Apply Holder’s inequality to (fg)p ∈ L1/p and g−p ∈ L−p/q.)

(6.9) If µ(X) <∞ and 1 ≤ p ≤ q <∞, show that Lq(µ) ⊆ Lp(µ).(Hint: Apply Holder’s inequality to |f |p ∈ Lq/p(µ) and 1 ∈ L(q−p)/q(µ).)

(6.10) Let f ∈ Lp(X,S, µ), where 1 ≤ p <∞. Show that ∀ ǫ > 0,

µ({x ∈ X| |f(x)| ≥ 0}) ≤ ‖f‖p/ǫp.

This is called Chebyshev’s inequality for Lp-functions.

86 6. Lp-spaces

6.3. L∞(X,S, µ)


6.3.1. Definition:A measurable function f : X −→ R∗ or C is said to be essentially boundedif there exists some real number M such that µ({x ∈ X | |f(x)| > M}) = 0.

We denote by L∞(X,S, µ) the set of all essentially bounded functions.For f ∈ L∞(X,S, µ) define

‖f‖∞ := inf {M |µ{x ∈ X | |f(x)| > M} = 0}.

We call ‖f‖∞ the essential supremum of f.

6.3.2. Theorem:Let f, g ∈ L∞(X,S, µ) and let α, β be scalars. Then the following hold:

(i) αf ∈ L∞(X,S, µ) and

‖αf‖∞ = |α| ‖f‖∞.

(ii) (f + g) ∈ L∞(X,B, µ) and

‖f + g‖∞ ≤ ‖f‖∞ + ‖g‖∞.

(iii) ‖f‖∞ = 0 iff f(x) = 0 a.e. x(µ).

(iv) There exists a set E ∈ S such that µ(E) = 0 and

‖f‖∞ = supx∈X\E

|f(x)|.

Thus µ({x ∈ X| |f(x)| > ‖f‖∞}) = 0.

(v) ‖f‖∞ = sup{N | µ({x ∈ X| |f(x)| > N}) > 0}.

6.3.3. Corollary:For f, g ∈ L∞(X,S, µ), let

d∞(f, g) := ‖f − g‖∞.

Then for f, g, h ∈ L∞(X,S, µ), the following hold:

(i) d∞(f, g) ≥ 0, and = 0 iff f(x) = g(x) for a.e. x(µ).

(ii) d∞(f, g) ≤ d∞(f, h) + d∞(h, g).

(iii) If {fn}n≥1 is a sequence in L∞(X,S, µ), then d∞(fn, f) → 0 asn→ ∞ iff fn → f uniformly a.e.

(iv) If {fn}n≥1 is a Cauchy sequence in L∞(X,S, µ), i.e., d∞(fn, fm)→ 0 as n,m → ∞, then ∃ h ∈ L∞(X,S, µ) such that d∞(fn, h)→ 0 as n→ ∞.

6.4. L2(X,S, µ) 87

6.3.4. Remark:For f, g ∈ L∞(X,S, µ), let us write

f ∼ g if f(x) = g(x) for a.e. x(µ).

Then ‘∼’ is an equivalence relation. We denote the set of equivalence classesby L∞(X,B, µ) itself. For any element f ∈ L∞(X,S, µ), which is in factan equivalence class, we can define ‖f‖∞ by choosing any element in theequivalence class f. Then ‖f‖∞ is well-defined, and L∞(X,S, µ) is a Banachspace under this norm.

Exercises:

(6.11 ) Let f ∈ L∞(X,S, µ), where (X,S, µ) is a finite measure space.Prove the following statements:(i) For 1 ≤ p <∞, f ∈ Lp(X,S, µ) and ‖f‖p ≤ ‖f‖∞(µ(X))1/p.(ii) If N > 0 is such that µ({x ∈ X| |f(x)| > N}) > 0, show that

N(µ(X))1/p ≤ ‖f‖p.

(iii) Let {pn}n≥1 be any monotonically increasing sequence of realnumbers such that pn ≥ 1 ∀ n and {pn}n≥1 is not boundedabove. Using (i), (ii) above and theorem 6.3.2(v), show that

‖f‖∞ = limn→∞

‖f‖pn .

6.4. L2(X,S, µ)


The reason for analyzing the space L2(X,S, µ) in detail is the following.We have already seen that for 1 ≤ p ≤ ∞, the spaces Lp(X,S, µ) are linearspaces and have a notion of distance (norm) under which they are complete.Let us recall, for f ∈ L2(X,S, µ), that

‖f‖2 :=

(∫

|f(x)|2dµ(x)

)1/2

.

If we treat f ∈ L2(X,S, µ) as a ‘vector’ with uncountable components, f(x)being the xth component, x ∈ X, then ‖f‖2 can be viewed as a generalizationof the Euclidean magnitude of a vector in Rn with summation being replacedby integration. We recall that on Rn, we have the notion of dot product ofvectors, which is related to the magnitude of vectors and helps us to definethe notion of angles on Rn. That such a dot product can also be defined onL2(X,S, µ) and enables one to do geometry on L2(X,S, µ), is the reason wediscuss this space in detail.

88 6. Lp-spaces

6.4.1. Definition:For f, g ∈ L2(X,S, µ), we define

〈f, g〉 :=

∫

f(x) g(x) dµ(x),

whenever it exists, where g denote the complex conjugate function: g(x) :=

g(x) ∀ x ∈ X.

6.4.2. Proposition (Cauchy-Schwarz inequality):For every f, g ∈ L2(X,S, µ), 〈f, g〉 is a well-defined scalar and

|〈f, g〉| ≤ ‖f‖2‖g‖2.

The scalar 〈f, g〉 is called the inner product of f and g.

6.4.3. Proposition:For f, g, h ∈ L2(X,S, µ) and α, β ∈ C, the following hold:

(i) 〈f, f〉 ≥ 0, and equality holds iff f = 0.

(ii) 〈f, g〉 = 〈g, f〉.

(iii) 〈αf + βg, h〉 = α〈f, h〉 + β〈g, h〉.

(iv) 〈f, αg + βh〉 = α〈f, g〉 + β〈f, h〉.

(v) ‖f‖2 = 〈f, f〉1/2.

6.4.4.Note:In the Cauchy-Schwarz inequality, the equality holds iff f and g are linearlydependent.

6.4.5. Note:Given an arbitrary vector space H over the field R (or C), if there is a map〈· , ·〉 : H × H −→ R (or C) having the properties (i) to (iv) as given inproposition 6.4.3, then it is called an inner product space. On everyinner product space H, it is easy to show that ‖u‖ := 〈u, u〉1/2, u ∈ H,is indeed a norm on H, called the norm induced by the inner product.Further, the Cauchy-Schwarz inequality holds:

|〈u, v〉| ≤ ‖u‖ ‖v‖, ∀u, v ∈ H.

This can be proved as follows: For u ∈ H, let ‖u‖ = (〈u, u〉)1/2. If eitheru = 0 or v = 0, then clearly |〈u, v〉| = 0 = ‖u‖ ‖v‖. So, let u, v ∈ H be

such that u 6= 0 and v 6= 0. Then ‖u‖ > 0 and ‖v‖ > 0. Let u′

= u/‖u‖ and

6.4. L2(X,S, µ) 89

v′

= v/‖v‖. Then 〈u′

, u′

〉 = 1 = 〈v′

, v′

〉 and

0 ≤ 〈u′

− 〈u′

, v′

〉v′

, u′

− 〈u′

, v′

〉v′

〉

= 〈u′

, u′

〉 + |〈u′

, v′

〉|2〈v′

, v′

〉 − 2|〈u′

, v′

〉|2

= 1 − |〈u′

, v′

〉|2.

Hence

|〈u′

, v′

〉| ≤ 1 = ‖u′

‖‖v′

‖,

i.e.,

|〈u, v〉| ≤ ‖u‖‖v‖.

If H is also complete under the norm induced by the inner product, thenH is called a Hilbert space. Thus L2(X,S, µ) is an example of a Hilbertspace. We shall not go into the general theory of Hilbert spaces. We discusssome results for L2(X,S, µ) which are in fact true, without any change inthe arguments, for general Hilbert spaces also.

6.4.6. Definition:Let f, g ∈ L2(X,S, µ). We say that f and g are orthogonal, and writef ⊥ g, if 〈f, g〉 = 0. For a subset S of L2(X,B, µ), we write f ⊥ S if〈f, h〉 = 0 ∀ h ∈ S.

6.4.7. Definition:Let S be a nonempty subset of L2(X,S, µ).

(i) We say S is a subspace of L2(X,S, µ) if ∀ α, β ∈ C and f, g ∈L2(X,B, µ), we have αf + βg ∈ S.

(ii) We say S is a closed subspace if it is closed under the ‖·‖2 metric,i.e., for every sequence {fn}n≥1 in S with lim

n→∞‖fn − f‖2 = 0 for

some f ∈ L2(X,S, µ), we have f ∈ S.

6.4.8. Proposition:Let S be any nonempty subset of L2(X,S, µ) and let

S⊥ := {g ∈ L2(X,S, µ) | 〈f, g〉 = 0 ∀ f ∈ S}.

Then S⊥ is a closed subspace of L2(X,S, µ).

The set S⊥ is called the orthogonal complement of S.

We next prove a result which seems geometrically obvious.

90 6. Lp-spaces

6.4.9. Theorem:Let f ∈ L2(X,S, µ), and let S be a closed subspace of L2(X,S, µ). Let

α := inf {‖f − g‖2 | g ∈ S}.

Then there exists a unique function f0 ∈ S such that α = ‖f−f0‖2. Further,if f 6∈ S then 0 6= (f − f0) ⊥ S.

6.4.10. Corollary:Let S be a proper closed subspace of L2(X,B, µ). Then S⊥ 6= {0}.

Next we give two applications of theorem 6.4.9. Our first applicationsays that if S is any closed subspace of L2(X,S, µ), then L2(X,S, µ) =S + S⊥ := {f + g|f ∈ S, g ∈ S⊥} with S ∩ S⊥ = {0}.

6.4.12. Theorem:Let S1, S2 be subsets of L2(X,S, µ). Then the following hold:

(i) S⊥1 is a closed subspace of L2(X,S, µ) and S1 ∩ S

⊥1 ⊆ {0}. If S1 is

also a subspace, then S1 ∩ S⊥1 = {0}.

(ii) S⊥1 ⊆ S⊥

2 if S2 ⊆ S1.

(iii) S1 ⊆ (S⊥1 )⊥, with equality iff S1 is a closed subspace of L2(X,S, µ).

(iv) If S1 and S2 are closed subspaces and f ⊥ g ∀ f ∈ S1 and ∀ g ∈ S2,then S1 + S2 := {f + g | f ∈ S1, g ∈ S2} is also a closed subspace.

(v) If S1 is a closed subspace, then S1 ∩ S⊥1 = {0} and L2(X,S, µ) =

S1 + S⊥1 . Thus every f ∈ L2(X,S, µ) can be uniquely expressed as

f = g + h, where g ∈ S1 and h ∈ S⊥1 .

(This is also expressed as L2(X,S, µ) = S1 ⊕ S⊥1 , and is called the

projection theorem.)

As our second application of theorem 6.4.9, we characterize all boundedlinear functions on L2, as defined next.

6.4.12. Definition:A map T : L2(X,S, µ) −→ C is called a bounded linear functional if ithas the following properties:

(i) For every f, g ∈ L2(X,S, µ) and α, β ∈ C,

T (αf + βg) = αT (f) + βT (g).

(ii) There exists a real number M such that

|T (f)| ≤M‖f‖2, ∀ f ∈ L2(X,S, µ).

6.4. L2(X,S, µ) 91

6.4.13. Proposition:Let T : L2(X,S, µ) −→ C be such that ∀ f, g ∈ L2(X,S, µ) and α, β ∈ C

T (αf + βg) = αT (f) + βT (g).

Then the following are equivalent:

(i) T is bounded, i.e., ∃M ∈ R such that

|T (f)| ≤ M‖f‖2, ∀ f ∈ L2(X,S, µ).

(ii) T is continuous.

(iii) T is continuous at 0 ∈ L2(X,S, µ).

6.4.14. Example:Let g ∈ L2(X,S, µ) be fixed. Define the map

Tg : L2(X,B, µ) −→ C

as follows:

Tg(f) = 〈f, g〉 =

∫

f g dµ ∀ f ∈ L2(X,S, µ).

It is easy to see that Tg is linear, i.e.,

Tg(αf1 + βf2) = αTg(f1) + βTg(f2)

∀ α, β ∈ C and f1, f2 ∈ L2(X,S, µ). Also, by the Cauchy-Schwarz inequal-ity,

|Tg(f)| = |〈f, g〉| ≤ ‖g‖2‖f‖2.

Hence Tg is a bounded linear functional on L2(X,S, µ). We note that g ∈(Ker(Tg))

⊥, where

Ker(Tg) := {f ∈ L2(X,B, µ) |Tg(f) = 0}.

Ker(Tg) is called the kernel of Tg. Our next theorem tell us that everybounded linear functional on L2(X,S, µ) is of this type.

6.4.15. Theorem (Riesz representation):Let T : L2(X,B, µ) −→ C be a bounded linear functional. Then there is a

unique g0 ∈ L2(X,S, µ) such that

T (f) = 〈f, g0〉 ∀ f ∈ L2(X,S, µ).

6.4.16. Note:In theorem 6.4.15, we showed that, given a closed subspaceM of L2(X,S, µ),every element f ∈ L2(X,S, µ) can be written uniquely as f = g + h, whereg ∈M and h ∈M⊥. Let us denote g by PM (f), called the projection of fonto M . Geometrically, PM (f) is the unique best approximator of f in M.

92 6. Lp-spaces

The properties of the map PM : L2(X,S, µ) → L2(X,S, µ) are given in thenext theorem.

6.4.17. Theorem:PM : L2(X,S, µ) → L2(X,S, µ), as defined in note 6.4.16 above, has thefollowing properties: ∀ f, g ∈ L2(X,S, µ)

(i) PM is linear and PM (f) = f whenever f ∈M .

(ii) f − PM (f) ∈M⊥.

(iii) 〈PM (f), g〉 = 〈f, PM (g)〉 = 〈PM (f), PM (g)〉.

(iv) PM (PM (f)) = f.

(v) ‖f‖22 = ‖PM (f)‖2

2 + ‖f − PM (f)‖22.

(vi) PM is continuous.

Exercises:

(6.12) Let f ∈ L2(X,S, µ) be such that f ⊥ g ∀ g ∈ L2(X,S, µ). Whatcan you conclude about f?

(6.13) Pythagoras identity: Let f, g ∈ L2(X,S, µ) and f ⊥ g. Showthat

‖f + g‖22 = ‖f‖2

2 + ‖g‖22.

(6.14) (Bessel’s inequality):Let f1, f2, . . . , fn be elements of L2(X,S, µ)such that ‖fi‖2 = 1 ∀ i and fi ⊥ fj for i 6= j. Show that

n∑

j=1

|〈f, fj〉|2 ≤ ‖f‖2

2.

(Hint: Consider 〈z, z〉, where z = f −∑n

j=1〈f, fj〉fj .)

(6.15) Let {fn}n≥1, {gn}n≥1 be sequences in L2(X,S, µ) such that limn→∞

‖fn →

f‖2 = 0, limn→∞

‖gn → g‖2 = 0 for f, g ∈ L2(X,S, µ). Show that

〈fn, h〉 → 〈f, h〉, 〈h, gn〉 → 〈h, g〉, 〈fn, gn〉 → 〈f, g〉

for every h ∈ L2(X,S, µ). In other words, the inner product map〈· , ·〉 : L2 × L2 −→ C is continuous in each variable and jointly.

(6.16) (Parallelogram identity): Let f, g ∈ L2(X,S, µ). Then

‖f + g‖22 + ‖f − g‖2

2 = 2‖f‖22 + 2‖g‖2

2.

6.5. L2-convergence of Fourier series 93

6.5. L2-convergence of Fourier series


The problem of investigating the convergence of Fourier series motivatedmathematicians to look for an extension of the integral concept. It is natu-ral to ask the question: did the extended integral, i.e., the Lebesgue integral,achieve success in this direction? One can say without any doubt that oneof the pioneering applications of the Lebesgue integral lies in the study ofFourier series. We shall not go into the general theory of Fourier series,but only prove an important result (the Riesz-Fischer theorem) which hasapplications in many branches of mathematics, physics and electrical engi-neering. For the general theory of Fourier series, one can consult Bhatia [4]and other references given there.

6.5.1. Definition:For f ∈ Lr

1[−π, π], let

an :=1

π

∫ π

−πf(x) cosnx dλ(x), n = 0, 1, 2, . . . ,

and

bn :=1

π

∫ π

−πf(x) sinnx dλ(x), n = 1, 2, . . . .

The scalars an, bn are called the Fourier coefficients of f , and the series

a0

2+

∞∑

n=1

(an cosnx+ bn sinnx)

is called the Fourier series of f. Let

sn(x) :=a0

2+

n∑

k=1

(ak cos kx+ bn sin kx), x ∈ [−π, π].

The function sn(x) is called the nth-partial sum of the Fourier series of f atx. The main problem analyzed in the theory of Fourier series is the following:when does {sn(x)}n≥1 converge to f(x), x ∈ [−π, π]? This is known as thepointwise convergence problem of Fourier series. For f ∈ L1[−π, π] andx ∈ [−π, π], if lim

n→∞sn(x) = f(x), we say that f has pointwise representation

by its Fourier series at x. The answer to the pointwise convergence problemis not easy, and, given the nature and scope of this text, we shall not go intoit. For details one can consult any one of the texts listed above. We statebelow an important relation between functions and their Fourier coefficients.

94 6. Lp-spaces

6.5.2. Theorem (Uniqueness of the Fourier series):If f ∈ Lr

1[−π, π] is such that all its Fourier coefficients are zero, then f(x) =0 for a.e. x, i.e., a function is uniquely determined by its Fourier coefficients.

Though the pointwise representation of a function f ∈ Lr1[−π, π] by its

Fourier series is undeniably of great intrinsic interest, it has its limitations.First of all, not every function f ∈ Lr

1[−π, π] can have a pointwise represen-tation one has to put extra conditions on f to get a pointwise representation.Secondly, f ∈ Lr

1[−π, π] need be defined only a.e. Thus pointwise represen-tation makes sense only a.e. Finally, from the point of view of many applica-tions, pointwise representations have very little utility. Such considerationmotivated mathematicians to look for some other methods of analyzing theconvergence problem. We consider below one such method, namely the con-vergence of Fourier series in the L2-metric. For other methods we refer tothe texts cited earlier.

Let f ∈ Lr2[−π, π]. Using the Cauchy-Schwarz inequality, it follows that

f ∈ Lr1[−π, π] also. Let an, bn be its Fourier coefficients and let sn(x) be the

nth-partial sum of the Fourier-series of f. In the terminology of the previoussection , if we write

φk(x) := cos kx and ψk(x) := sin kx,

then

ak = 〈f, φk〉/π and bk = 〈f, ψk〉/π.

6.5.3. Lemma:For nonnegative integers n and m, the following relations hold:

〈ψn, φm〉 = 0;

〈ψn, ψm〉 = 0 if n 6= m and = π if n = m;

〈φn, φm〉 = 0 if n 6= m and = π if n = m.

6.5.4. Theorem (Bessel’s inequality):For f ∈ Lr

2[−π, π],

|a0|2

2+

∞∑

n=1

(|an|2 + |bn|

2) ≤1

π‖f‖2

2.

The above theorem has a converse, as given in the next theorem.

6.5. L2-convergence of Fourier series 95

6.5.5. Theorem (Riesz-Fischer):Let {an}n≥0 and {bn}n≥1 be sequences of real numbers such that

|a0|2

2+

∞∑

n=1

(|an|2 + |bn|

2) < +∞.

Then there exists a unique function f ∈ Lr2[−π, π] such that an, bn are its

Fourier coefficients.

6.5.6. Corollary ( Parseval’s identity):Let f ∈ Lr

2[−π, π]. Then the Fourier series of f converges to f in the L2-norm, i.e., ‖sn − f‖2 −→ 0 as n −→ ∞. Further,

1

π‖f‖2

2 =|a0|

2

2+

∞∑

n=1

(|an|2 + |bn|

2).

6.5.7 Note:A careful observation of the above arguments will tell the reader that Fouriercoefficients can be defined for Riemann integrable functions also (as was donehistorically first), and Bessel’s inequality remains valid. However, Riesz-Fischer critically uses the fact that Lr

2[−π, π] is complete under the L2-metric, which is not true for Riemann integrable functions.

96 6. Lp-spaces

Exercises:

(6.17) (Euler’s identity):Show that

∞∑

ℓ=0

1

(2ℓ+ 1)2=

π2

8.

Appendix A

Extended real numbers

A.1. Extended real numbers

Let R denote the set of all real numbers and let

R∗ := R ∪ {+∞} ∪ {−∞},

where +∞ and −∞ are two symbols read as plus infinity and minusinfinity. We extend the algebraic operations and the order relation of R toR∗ as follows:

(1) For every x ∈ R, −∞ < x < +∞.

(2) For every x ∈ R,

(−∞) + x = −∞ and (+∞) + x = +∞;(+∞) + (+∞) = +∞ and (−∞) + (−∞) = −∞.

(3) For every x ∈ R,

x(+∞) = (+∞)x = +∞x(−∞) = (−∞)x = −∞

}

if x > 0,

x(+∞) = (+∞)x = −∞x(−∞) = (−∞)x = +∞

}

if x < 0.

Further,

(+∞)0 = (−∞)0 = 0, (±∞)(+∞) = (±∞) and (±∞)(−∞) = (∓∞).

Note that the relations −∞ + (+∞) and (+∞) + (−∞) are not defined.

97

98 Appendix A

The set R∗, also denoted as [−∞,+∞], with the above properties iscalled the set of extended real numbers. The symbol +∞ is also denotedby ∞, when no confusion arises.

A.2. sup(A) and inf(A)

For a nonempty set A ⊆ R∗, we write sup(A) := +∞ if A∩R is not boundedabove, and inf(A) := −∞ if A ∩ R is not bounded below. Thus

• sup(A) and inf(A) always exist for every nonempty subset A of R∗.

A.3. Limits of sequences in R∗

For {xn}n≥1 any monotonically increasing sequence in R∗ which is notbounded above, we say {xn}n≥1 is convergent to +∞ and write

limn→∞

xn = +∞.

Similarly, if {xn}n≥1 is a monotonically decreasing sequence which is notbounded below, we say {xn}n≥1 is convergent to −∞ and write

limn→∞

xn = −∞.

Hence

• every monotone sequence in R∗ is convergent.

Thus for any sequence {xn}n≥1 in R∗, the sequences {supk≥j xk}j≥1 and{infk≥j xk}j≥1 always converge. We write

lim supn→∞

xn := limj→∞

(supk≥j

xk)

and

lim infn→∞

xn := limj→∞

(infk≥j

xk).

limn→∞ supxn is called the limit superior of the sequence {xn}n≥1 andlim infn→∞ xn is called the limit inferior of the sequence {xn}n≥1. Notethat

lim infn→∞

xn ≤ lim supn→∞

xn.

We say a sequence {xn}n≥1 is convergent to x ∈ R∗ if lim infn→∞ xn =lim supn→∞ xn =: x, say. In that case we write limn→∞ xn := x.

A.4. Series in R∗

Let {xk}k≥1 be a sequence in R∗ such that for every n ∈ N, sn :=∑n

k=1 xk iswell-defined. We say that the series

∑∞k=1 xk is convergent to x if {sn}n≥1

is convergent. We write this as x =∑∞

k=1 xk, x being called the sum of

Extended real numbers 99

the series∑∞

k=1 xk. For example, if each xk ≥ 0, then the series∑∞

k=1 xk

is always convergent.

Let {xk}k≥1 be a sequence in R∗, xn ≥ 0 for every n. Let σ : N → N beany bijective map. Then the series

∑∞k=1 xσ(k) is called an arrangement

of the series∑∞

k=1 xk.

• For every rearrangement σ, the series∑∞

k=1 xσ(k) and∑∞

k=1 xk con-verge to the same sum.

Clearly, both the series are convergent in R∗. Let α :=∑∞

k=1 xk and β :=∑∞

k=1 xσ(k). To prove α = β, in view of the symmetry, it is enough to provethat β ≤ α. Now

β =n∑

k=1

xσ(k) ≤m∑

k=1

xk ≤∞∑

k=1

xk = α,

where in the the middle term m := max{σ(1), · · · , σ(k)}. Hence β ≤ α,proving the required claim.

Similar arguments apply to {xn,m}n,m≥1, any double-indexed sequence ofnonnegative elements of R∗. For every fixed n, the series

∑∞m=1 xn,m is con-

vergent in R∗. Let yn :=∑∞

m=1 xn,m. Further,∑∞

n=1 yn is also convergent inR∗. Let y :=

∑∞n=1 yn. We can also define for all fixed m, zm :=

∑∞n=1 xn,m,

and z :=∑∞

m=1 zm in R∗. We show that both these processes lead to thesame sum, i.e., y = z, which is written as

∞∑

m=1

(

∞∑

n=1

xn,m

)

=∞∑

n=1

(

∞∑

m=1

xn,m

)

.

To see this, we note that ∀ r, s ∈ N,r∑

n=1

s∑

m=1

xn,m ≤s∑

m=1

∞∑

n=1

xn,m ≤∞∑

m=1

∞∑

n=1

xn,m.

Hence

y :=∞∑

n=1

∞∑

m=1

xn,m ≤∞∑

m=1

∞∑

n=1

xn,m =: z.

The reverse inequality follows from symmetry.

A similar result holds for rearrangement of double-indexed series of non-negative terms in R∗ :

• Let σ : N → N×N be any bijective map, and let {xn,m}n≥1,m≥1 bea double-indexed sequence of nonnegative elements of R∗. Then

∞∑

r=1

xσ(r) =∞∑

n=1

∞∑

m=0

xn,m =∞∑

m=1

∞∑

n=1

xn,m.

Appendix B

Axiom of choice

Let A,B be two sets. One defines A×B, the Cartesian product of A andB, to be the empty set if either A or B or both are empty, and to be the setof all ordered pairs (a, b), a ∈ A, b ∈ B, when both A and B are nonempty.Similarly, for a finite family of nonempty sets A1, . . . , An, we define theirCartesian product to be the set

A1 × · · · ×An := {(x1, x2, . . . xn) | xi ∈ Ai, i = 1, 2, . . . , n}.

Obviously, A1×· · ·×An is a nonempty set. We can think of A1×· · ·×An asthe set of all functions f : {1, 2, . . . , n} −→

⋃ni=1Ai with f(i) = xi ∈ Ai for

each i. One can copy this to define the Cartesian product of any arbitraryfamily of sets, say {Aα}α∈I , to be the set

∏

α∈I

Aα :=

{

f : I →⋃

α∈I

Aα f(α) ∈ Aα, ∀ α ∈ I

}

.

However, there is no surety that the set∏

α∈I Aα is nonempty, i.e.,we do notknow that there always exists at least one function f : I →

⋃

α∈I Aα suchthat f(α) ∈ Aα ∀ α ∈ I, although intuitively it seems obvious that such afunction should always exist. However, this cannot be proved with the usualaxioms of set theory. (For a short introduction to axiomatic set theory, seeRana [30]. For detailed account of axiomatic set theory, the axiom of choice,and its history, see Halmos [15] and Fraenkel [12]). The way out of the abovedilemma is to treat this an axiom itself, called the axiom of choice:

• If {Aα}α∈I is a non-empty family of sets such that each Aα is non-empty, then there exists a function f : I −→

⋃

α∈I Aα such thatf(α) ∈ Aα ∀ α ∈ I.

101

102 Appendix B

Such a function is called a choice function.

The axiom of choice has many equivalent formulations; a useful one isthe following:

• If {Aα |α ∈ I} is a nonempty family of pairwise disjoint sets suchthat Aα 6= ∅ for every α ∈ I, then there exists a set E ⊆

⋃

α∈I Aα

such that E ∩Aα consists of precisely one element for each α ∈ I.

The axiom of choice finds applications in many diverse branches of math-ematics. (We have used it in section 4.6 to construct nonmeasurable subsetsof R.)

Appendix C

Continuum hypothesis

Let X and Y be two sets. We say X and Y are equipotent iff there existsa bijection between them. We write this as X ≈ Y. In a sense, equipotentsets have same ‘number’ of elements.

We say a set A is finite if A ≈ {1, 2, . . . , n} for some n ∈ N, and we sayA is countable if A ≈ N, the set of natural numbers. A set which is notcountable is called uncountable. For example, N,Z,Q are all countablesets while R is uncountable. (For a detailed discussion, see Rana [30].)

The statement X ≈ Y means X and Y have the same ‘number of ele-ments’, and can be made precise as follows (see Halmos [15] for details). LetC be a collection of sets such that any two members of C are equipotent toeach other. Then one can assign a symbol, called its cardinal number,to each A ∈ C, denoted by card(A). Thus

card(A) = card(B) iff A ≈ B.

The cardinal number of a setA is also called the cardinality ofA. For exam-ple, for any set A which is equipotent to {1, 2, . . . , n}, we write card(A) = n.For any set A ≈ N, we write card(A) = ℵ0 , called aleph-nought. For anyset A ≈ R, we write card(A) = c, called cardinality of the continuum.For finite sets, it is easy to see that if card(A) = n, then card (P(A)) = 2n,where P(A) is the set of all subsets of A. We define, for any nonempty setX,

card(P(X)) := 2card(X).

For example,

card(P(N)) := 2ℵ0 and card(P(R)) := 2c.

103

104 Appendix C

For a finite set A, we know that

card(P(A)) = 2card(A) > card(A).

Can the same be said about arbitrary sets which are not necessarily finite?For two sets A and B, we say card(A) � card(B) if there exists a one-onemap from A into B. We write card(A) <card(B) if card (A) �card(B) butcard(A) 6=card(B). Now we ask the question:

• Is card(A) < card(P(A)) for any nonempty set A?

The answer is in the affirmative and is due to George Cantor (see Rana[30] for a proof). It can be shown that 2ℵ0 = c, i.e., P(N) ≈ R. Thus wehave the following:

0 < 1 < 2 < · · · < n < · · · < ℵ0 < 2ℵ0 = c < 2c < 22c

< · · · .

This raises the following natural question:

• Does there exist a cardinal number α such that ℵ0 < α < 2ℵ0 = c?

That is, does there exist a set A ⊂ R such that A 6≈ R and N ⊂ A butA 6≈ N? The answer to this question is not known. The statement that theanswer to the above question is in the negative is called the continuumhypothesis:

• There does not exist any cardinal number between ℵ0 and 2ℵ0 = c.

An equivalent formulation of this is the following:

• The set R can be well-ordered in such a way that each element ofR is preceded by only countably many elements.

We used this in section 3.4, to prove Ulam’s theorem. It is known thatthe continuum hypothesis is independent of the Zermelo-Fraenkel axioms ofset theory.

References

[1] Aliprantis, C.D. and Burkinshaw, O. Principles of Real Analysis (3rdEdition). Academic Press, Inc. New York, 1998.

[2] Apostol, T.M. Mathematical Analysis. Narosa Publishing House, NewDelhi (India), 1995.

[3] Bhatia, Rajendra Fourier Series. Hindustan Book Agency, New Delhi(India), 1993.

[4] Billingsley, Patrick Probability and Measure. 3rd Edition, John Wileyand Sons, New York, 1995.

[5] Friedman, A. Foundations of Modern Analysis. Holt, Rinehart andWinston, Inc., New York, 1970.

[6] Halmos, P.R. Measure Theory. Van Nostrand, Princeton, 1950.

[7] Halmos, P.R. Naive Set Theory. Van Nostrand, Princeton, 1960.

[8] Hewitt, E. and Stromberg, K. Real and Abstract Analysis. Springer-Verlag, Heidelberg, 1969.

[9] Kolmogorov, A.N. Foundations of Probability Theory. Chelsea Pub-lishing Company, New York, 1950.

105

106 References

[10] Parthasarathy, K.R. Introduction to Probability and Measure.Macmillan Company of India Ltd., Delhi, 1977.

[11] Parthasarathy, K. R. Probablity Measures on Metric Spaces, AcademicPress, New York, 1967.

[12] Rana, Inder K. An Introduction to Measure and Integration,(First Edition) Narosa Publishers, Delhi, 1997.(Second Edition) Graduate Seies in Mathematics Volume 45, Ameri-can Mathematical Society, USA, 2001.

[13] Rana, Inder K. From Numbers to Analysis,World Scientific Publishers, Singapore, 1998.

[14] Royden, H.L. Real Analysis (3rd Edition). Macmillan, New York,1963.

Index

(X,S, µ), 26

C(R), 61

C[a, b], 60

Cc(R), 61

Ey , 66

Ex, 66

Lr1(X,S, µ), 80

L1(X,S, µ), 50

L1[a, b], 56

L1-metric, 60

L1(E), 56

Lp(X,S, µ), 82

Lp(µ), 82

S⊥, 89

C, 79

L, 44

L+, 39

L+0

, 35

R, 10

I, 2

Im (f), 79

〈f, g〉, 88∫

fdµ, 39, 80∫

sdµ, 36∫

Efdµ, 40

∫

Efdµ, 53

µ∗, 21

µF , 14

Re (f), 79

σ- algebra of Borel subsets of R, 27

σ- finite set function, 17

σ-algebra generated, 5

σ-algebra monotone class theorem, 7

σ-algebra of Borel subsets of X, 7

σ-algebra technique, 7

σ-algebra, product, 64

σ-set, 3

f ∼ g, 59

f−1(C), 3

s1 ∨ s2, 37

s1 ∧ s2, 37

A⊗B, 64

BX , 7

BR, 27

BR2 , 65

C ∩ E, 3

F(C), 3

I, 10

I0, 32

Id, 7

Ir, 7

LF , 24

LR, 27

M(C), 6

S(C), 5

S∗, 23

pth-norm of f , 82

a.e., 39

a.e. (µ)x, 39

a.e. (µ)x ∈ Y , 39

a.e. x(µ), 39

aleph-nought, 103

algebra, 1

algebra generated, 3

almost everywhere, 39

analytic set, 28

arrangement, 99

Arzela’s theorem, 58

107

108 Index

Banach spaces, 84

Bessel’s inequality, 92, 94

Binomial distribution, 10

Borel measurable function, 74

Borel subsets, 27, 33

Borel subsets of R2, 65

bounded convergence theorem, 52

bounded linear functional, 90

cardinal number, 103

cardinality of a set, 103

cardinality of the continuum, 103

Cartesian product, 101

Cauchy-Schwartz inequality, 83, 88

Chebyshev’s inequality, 54, 85

choice function, 102

closed subspace, 89

complete measure space, 26

completion of a measure space, 25, 26

complex numbers, 79

continuity from above, 16

continuity from below, 16

continuum hypothesis, 104

convergent to +∞, 98

convergent to −∞, 98

countable, 103

countably additive, 9

countably subadditive, 10

countably subadditivity, 12

counting measure, 70

cylinder set, 4

discrete measure, 10

discrete probability measure, 10

distribution function, 15, 32

distribution function, probability, 32

distribution of the measurable function, 48

distribution, Binomial, 10

distribution, discrete probability, 10

distribution, Poisson, 10

distribution, uniform, 10

equipotent, 103

equivalent, 59

essential supremum, 86

essentially bounded, 86

Euler’s identity, 96

example, Vitali’s, 34

extended real numbers, 10, 98

extension, 19

extension of a measure, 19, 24

Fatou’s lemma, 45

finite additive property, 11

finite set, 103

finitely additive, 9

Fourier coefficients, 93

Fourier series, 93

Fubini’s theorem, 69

function, choice, 102

function, imaginary part of a, 79

function, integrable, 50

function, length, 11

function, measurable, 44

function, nonnegative measurable, 35, 39

function, nonnegative simple measurable, 35

function, real part of a, 79

function, simple measurable, 44

function, support of a, 61

function,integrable, 80

generated, σ-algebra, 5

generated, algebra, 3

generated, monotone class, 6

graph of the function, 68

Holder’s inequality, 83

Haar measure, 31

Hilbert space, 89

inner product, 88

integrable, 50, 80

integral, 39, 50, 80

integral of nonnegative simple measurable

function, 36

integral over E , 40

integration of ‘radial’ functions, 76

intervals with dyadic endpoints, 7

intervals with rational endpoints, 7

kernel, 91

Lebesgue integrable functions, 56

Lebesgue integral, 56

Lebesgue measurable sets, 27

Lebesgue measurable subsets of R2, 75

Lebesgue measure, 27

Lebesgue measure space, 27, 75

Lebesgue null, 32

Lebesgue outer measure, 27

Lebesgue’s dominated convergence, 81

Lebesgue-Stieltjes measure, 32

left open, right closed intervals, 2

length function, 11

length function, countable additivity of, 12

length function, countable subadditivity of,12

length function, finite additivity of, 11

length function, monotonicity property of,11

length function, translation invariance of, 12

limit inferior, 98

limit superior, 98

Index 109

mean value property, 57

measurable space, 25

measurable cover, 25

measurable function, 44

measurable function, nonnegative, 39

measurable kernel, 25

measurable partition, 41

measurable rectangle, 64, 65

measurable set, 23

measure, 10

measure space, 25

measure space, complete, 26

measure space, completion of a, 26

measure space,completeness of a, 25

measure, counting, 70

measure, discrete, 10

measure, Haar, 31

measure, Lebesgue, 27

measure, Lebesgue-Stieltjes, 24, 32

measure,outer, 21

measure,outer regular, 33

Minkowski’s inequality, 83

monotone class, 5

monotone class generated, 6

monotone convergence theorem, 40

monotonicity property, 11

nonnegative simple measurable function, 35

norm, 59, 84

norm, induced by the inner product, 88

normed linear space, 84

null subset, 26

open intervals, 7, 32

orthogonal, 89

orthogonal complement, 89

outer measure, 21

outer measure induced, 21

outer measure, λ, 27

outer measure, Lebesgue, 27

outer regular, 33

outer-regularity of λ, 30

parallelogram identity, 92

Parseval’s identity, 95

partial sum of the Fourier series, 93

partition, measurable, 41

Poisson distribution, 10

power set, 1

probability, 26

probability distribution function, 32

probability measure, discrete, 10

probability space, 26

product σ-algebra, 64, 65

product measure space, 65

product of the measures µ and ν, 65

projection theorem, 90

Pythagoras identity, 92

real part, 79

regularity of λR2 , 76

representation, 35

Riemann-Lebesgue lemma, 61

Riesz representation, 91

Riesz-Fischer theorem, 83, 95

section of E at x, 66

section of E at y, 66

semi-algebra, 1

set function, 9

set function, σ-finite, 17

set function, countably additive, 9

set function, countably subadditive, 10

set function, finite, 17

set function, finitely additive, 9

set function, induced, 14

set function, monotone, 9

sigma algebra, 5

simple function technique, 52

simple measurable function, 44

simple,nonnegative measurable function, 35

space, Banach, 84

space, Hilbert, 88

space, inner product, 88

space, Lebesgue measure, 27

space, measurable, 25

space, measure, 25

space, normed, 84

space, probability, 26

space, product measure, 65

standard representation, 35

subset,null, 26

subspace, 89

subspace, closed, 89

supp (f), 61

support, 61

Theorem, σ-algebra monotone class, 7

Theorem, Arzela’s, 58

theorem, bounded convergence, 52

Theorem, Fubini, 69

Theorem, Lebesgue’s dominated convergence,

51

Theorem, Monotone Convergence, 40

Theorem, Riesz-Fischer, 60, 95

Theorem, Ulam’s, 20

topological group, 31

totally finite, 17

transition measure, 75

translation invariance, 12, 75

triangle-inequality, 84

truncation sequence, 48

Ulam’s theorem, 20

110 Index

ultrafilter, 13

uncountable, 103

Uniform distribution, 10

Vitali’s example, 34

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Measure and Integration: Concepts, Examples and Exercises · Measure and Integration: Concepts, Examples and Exercises INDER K. RANA Indian Institute of Technology Bombay India Department