Measure and Integration: Concepts, Examples and Exercises INDER K. RANA Indian Institute of Technology Bombay India Department of Mathematics, Indian Institute of Technology, Bombay, Powai, Mumbai 400076, India Current address: Department of Mathematics, Indian Institute of Technology, Powai, Mumbai 400076, India E-mail address:
[email protected]
These notes were specially prepared for the participants of the first Annual Foundation School, May 2004, held at IIT Bombay, India. Abstract. These notes present a quick overview of the theory of Measure 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
§2.1. Set functions
9 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 Riemann integral
55 vii
viii
Contents
§4.6. L1 [a, b] as the completion of R[a, b] Chapter 5.
Measure and integration on product spaces
59 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 Chapter 6.
R2
and its properties
Lp -spaces
75 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 say C 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 Snsets C1 , C2 , . . . , Cn ∈ C such c that Ci ∩ Cj = ∅ for i 6= j and A = i=1 Ci .
1.1.2. Definition: Let X be a nonempty set and F a collection of subsets of X. The collection F 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) Let X 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-algebra of subsets of R. (iii) The collection ( F(I) :=
E⊆R E=
n [
)
Ik , Ik ∈ I, Ik ∩ Iℓ = ∅ for k 6= ℓ, n ∈ N .
k=1
is an algebra of subsets of R. So is the class ) ( n [ ˜ Ik ∩ Iℓ = ∅ for k 6= ℓ, n ∈ N . ˜ := E ⊆ R | E = Ik , Ik ∈ I, F(I) k=1
(iv) Let X be any nonempty set. Let C := {E ⊆ X | either E or E c is finite}. Then C is an algebra of subsets of X. (v) Let X and Y be two nonempty sets, and F and G semi-algebras of subsets of 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 an algebra 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 , . . . ,SFn ∈ F such S that Fi ⊆ Ei for each i, Fi ∩Fj = ∅ for i 6= j and ni=1 Ei = nj=1 Fj .
The next set of exercise describes some methods of constructing algebras and 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 semialgebra (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. Note that C ∩ E is the collection of those subsets of E which are elements of C when E ∈ C. (1.4) Let X, Y be two nonempty sets and f : X −→ Y be any map. For E ⊆ Y, we write f −1 (E) := {x ∈ X | f (x) ∈ E}. Let C be any semi-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 of subsets 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 {F Tα }α∈I be a family of algebras of subsets of a set X. Let F := 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 S a set X. Under what 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 called a σ-set if there S exist sets Ci ∈ C, i = 1, 2, . . . , such that Ci ∩ Cj = ∅ for i 6= j and ∞ i=1 Ci = A. Prove the following: S (i) For any finite number of sets C, C1 , C2 , . . . , Cn in C, C\( ni=1 Ci ) is a finite union of pairwise disjoint sets from C and hence is a σ-set. S (ii) For any sequence {Cn }n≥1 of sets in C, ∞ n=1 Cn 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 a unique algebra F of subsets of X such that C ⊆ F and if A is any other algebra such that C ⊆ A, then F ⊆ A. This unique algebra given is called the algebra generated by C and is denoted by F(C).
(1.10) Show that he algebra generated by I, the class of all intervals, is S {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 set X. ShowSthat F(C), the algebra generated by C, is given by {E ⊆ X | E = ni=1 Ci , Ci ∈ C and Ci ∩ Cj = ∅ for i 6= j, n ∈ N}.
4
1. Classes of sets
Remark: Exercise 1.11 gives a description of F(C), the algebra generated by a semi-algebra C. In general, no description is possible for F(C) when C is not a semi-algebra. S (1.12) Let X be any nonempty set and C = {{x} | x ∈ X} {∅, X}. Is C a semi-algebra of subsets of X? What is the algebra generated by C? 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 sequences with 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 product of 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 X with 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 Concepts and examples: 1.2.1. Definition: Let X be any nonempty set and let S be a class of subsets of X with the following properties: (i) ∅ and X ∈ S. (ii) Ac ∈ S whenever A ∈ S. S (iii) ∞ i=1 Ai ∈ S whenever Ai ∈ S, i = 1, 2, . . . .
Such a class S is called a sigma algebra (written as σ-algebra) of subsets of X. 1.2.2. Examples: (i) Let X be any set. Then {∅, X} and P(X) are obvious examples of σalgebras 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. T (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 such that S ⊇ C (note that P(X) is one such σ-algebra). It is easy to see that S(C) is also a σ-algebra of subsets of X and S(C) ⊇ C. In fact, if S is any σ-algebra of subsets of X such that S ⊇ C, then clearly S ⊇ S(C). Thus S(C) is the smallest σ-algebra of subsets of X containing C, and is called the σ-algebra generated by C. In general it is not possible to represent an 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 a monotone class if S (i) ∞ n=1 An ∈ M, whenever An ∈ M and An ⊆ An+1 for n = 1, 2, . . . , T∞ (ii) n=1 An ∈ 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.
6
1. Classes of sets
(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) isTa monotone class of subsets of X such that C ⊆ P(X). Let M(C) := M, where the intersection is over all those monotone classes M of subsets of X such that C ⊆ M. Clearly, M(C) is itself a monotone class, and if M is any monotone class such that C ⊆ M, then M(C) ⊆ M. Thus M(C) is the smallest monotone class of subsets of X such that C ⊆ M(C). The class M(C) is called the monotone class generated by C.
Exercises:
(1.15) Let S be a σ-algebra of subsets of X and let Y ⊆ X. Show that S ∩ 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 subsets of 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 the σ-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 ⊆ S and 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, and is often used to prove various properties of σ-algebras under consideration, is as follows: The sets satisfying the required property are collected together. One shows that this collection itself is a σ-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 generated σ-algebra. We call this the σ-algebra technique. (1.20) Let X be any topological space. Let U denote the class of all open subsets 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 is denoted by BX . (ii) Let X = R. Let I be the class of all intervals and I˜ the class of all left-open right-closed intervals. Show that I ⊂ S(U), I ⊂ ˜ I˜ ⊂ S(I) and hence deduce that S(I), ˜ = BR . S(I) = S(I) (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 and n). 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 of subsets 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 | E c ∈ 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 ), for F ∈ A.
8
1. Classes of sets
(iv) Using (iii), deduce that M(A) ⊆ L(E) for every E ∈ M(A),i.e., M(A) is closed under unions also. Now use exercise (1.22) to deduce that S(A) ⊆ M(A).
Chapter 2
Measure
2.1. Set functions Concepts and examples: 2.1.1. Definition: Let C be a class of subsets of a set X. A function µ : C −→ [0, +∞] is called a set function. Further, (i) µ is said to be monotone if µ(A) ≤ µ(B) whenever A, B ∈ C and A ⊆ B. (ii) µ is said to be finitely additive if ! n n X [ µ(Ai ). µ Ai = i=1
i=1
whenever A1 , A2 , . . . , An ∈ C are such that Ai ∩ Aj = ∅ for i 6= j S and ni=1 Ai ∈ C.
(iii) µ is said to be countably additive if ! ∞ ∞ [ X µ An = µ(An ) n=1
n=1
whenever A1 , A2 , . . . in C with Ai ∩Aj = ∅ for i 6= j and
S∞
n=1 An
∈ C. 9
10
2. Measure
(iv) µ is said to be countably subadditive if µ(A) ≤
∞ X
µ(An ).
n=1
whenever A ∈ C, A =
S∞
n=1 An
with An ∈ C for every n.
(v) µ is called a measure on C if ∅ ∈ C with µ(∅) = 0 and µ is countably additive on C. Here are some more examples of finitely/countably additive set functions: 2.1.2.Example: Let X be any infinite set and let xn ∈ X, n = 1, 2, . . . . Let {pn }n≥1 be a sequence of nonnegative real numbers. For any A ⊆ X, define X µ(A) := pi . {i|xi ∈A}
It is easy to show that µ is a countably additive set function on the algebra P(X). We say µ is a discrete measure with ‘mass’ P∞ P∞pi at xi . The measure µ 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 that µ({xi }) = pi ∀ i and µ({x}) = 0 if x 6= xi . So, one can regard µ as a set function defined on the subsets of the set Y := {xn : n ≥ 1}. Some of the special cases when X = {0, 1, 2, . . .} are: (a) Binomial distribution: Y := {0, 1, 2, . . . , n} and, for 0 < p < 1, n k p (1 − p)n−k , 0 ≤ k ≤ n. pk = k (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 in R. 2.1.3 Example: We denote the set of real numbers by R. Let R∗ denote the set of extended real numbers. Let I denote the collection of all intervals of R. If an interval I ∈ I has end points a and b we write it as I(a, b). By convention, the open interval
11
2.1. Set functions
(a, a) = ∅ ∀ a ∈ R. Let [0, +∞] := {x ∈ R∗ |x ≥ 0} = [0, +∞) ∪ {+∞}. Define the function λ : I −→ [0, ∞] by |b − a| if a, b ∈ R, λ(I(a, b)) := +∞ if either a = −∞ or b = +∞ or both. The function λ, as defined above, is called the length function and has the 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 is monotone). S Property (3): Let I ∈ I be such that I = ni=1 Ji , where Ji ∩ Jj = ∅ for i 6= j. Then n X λ(Ji ). λ(I) = i=1
This property of λ is called the finite additivity of λ, or one says that λ is finitely additive. S Property (4): Let I ∈ I be a finite interval such that I ⊆ ∞ i=1 Ii , where Ii ∈ I. Then ∞ X λ(Ii ). λ(I) ≤ i=1
Property (5): Let I ∈ I be a finite interval such that I = In ∈ I and In ∩ Im = ∅ for n 6= m. Then λ(I) =
∞ X
S∞
n=1 In ,
where
λ(In ).
n=1
Property (6): Let I ∈ I be any interval. Then ∞ X λ(I) = λ(I ∩ [n, n + 1)). n=−∞
Property (7): Let I ∈ I be any interval such that I = and In ∩ Im = ∅ for n = 6 m. Then λ(I) =
∞ X
n=1
λ(In ).
S∞
n=1 In ,
In ∈ I
12
2. Measure
This property of λ is called the countable additivity of λ, or one says that λ is countably additive. Property (8): Let I ∈ I and I ⊆ λ(I) ≤
S∞
n=1 In ,
∞ X
In ∈ I. Then
λ(In ).
n=1
This property of λ is called the countable subadditivity of λ, or one says that λ is countably subadditive. Property (9): λ(I) = λ(I + x), for every I ∈ I and x ∈ R, where I + x := {y + x | y ∈ I}. This property of the length function is called translation invariance, or one 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 0 if A is finite, µ(A) := 1 if Ac is finite. Show that µ is finitely additive but not countably additive. If X is an 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, #{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 µn (A) :=
C = {A ⊆ X | lim µn (A) exists}. n→∞
Show that C is closed under taking complements, finite disjoint unions and proper differences. Is it an algebra?
13
2.1. Set functions
(2.3) Let µ : I˜ ∩ (0, 1] −→ [0, ∞] be defined by b − a if a 6= 0, 0 < a < b ≤ 1, µ(a, b] := +∞ otherwise.
(Recall that I˜ ∩ (0, 1] is the class of all left-open right-closed intervals 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 ultrafilter in X.) (b) Let U be any ultrafilter in X. Define µ : P(X) −→ [0, ∞) by 1 if A ∈ U, µ(A) := 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) := lim µn (E). n→∞
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 of all those subsets of X which are both open and closed. Show that A is an algebra of subsets of X. Further, every finitely additive set function on A is also countably additive.
14
2. Measure
2.2. Countably additive set functions on intervals Concepts and examples: We saw in the previous section that the length function is a countably additive set function on the class of all intervals. One can ask the question: do there exist countably additive set functions on intervals, other than the length 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 ] := µF (a, ∞) := µF (−∞, ∞) :=
lim [F (b) − F (−x)],
x→∞
lim [F (x) − F (a)],
x→∞
lim [F (x) − F (−x)].
x→∞
˜ Further, µF Then, µF is a well-defined finitely additive set function on I. 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 function F : R −→ R such that µ(a, b] = F (b) − F (a) ∀ a, b ∈ R. If µ is also countably additive, then F is right-continuous. (Hint: Define F as follows: µ(0, x] if x > 0, 0 if x = 0, F (x) := −µ(x, 0] if x < 0.) 2.2.3. Remarks: (i) In case µ(R) < +∞, a more canonical choice for the required function F in proposition 2.2.2 is given by F (x) := µ(−∞, x], x ∈ R.
15
2.3. Set functions on algebras
(ii) Propositions 2.2.1 and 2.2.2 completely characterize the non-trivial countably additive set functions on intervals in terms of functions F : R −→ R which are monotonically increasing and right continuous. Such functions are called distribution functions on R. The set function µF induced by the distribution function F is non-trivial in the sense that it assigns finite non-zero values to bounded intervals. Exercises: (2.7) Let F (x) = [x], the integral part of x, x ∈ R. Describe the set function µF . (2.8) α ∈ R. Show that F1 := F + α is also a distribution function and µF = 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 finitely additive. Is µ monotone? Countably subadditive? (ii) If C be a S semi-algebra, then µ is countably subadditive iff ∀ A ∈ C with A ⊆ ∞ i=1 Ai , Ai ∈ C implies µ(A) ≤
∞ X
µ(Ai ).
i=1
2.3. Set functions on algebras Concepts and examples: In this section, we give some general properties of a set function µ defined on 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 set function. 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 finitely additive 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 additive and countably subadditive.
16
2. Measure
Another characterization of countable additivity of set functions defined on 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 such that µ(∅) = 0. (a) If µ is countably additive thenSthe following hold: (i) For any A ∈ A, if A = ∞ n=1 An , where An ∈ A and An ⊆ An+1 ∀ n, then µ(A) = lim µ(An ). n→∞
This is called the continuity from below of µ at A. T (ii) For any A ∈ A, if A = ∞ A n=1 n , where An ∈ A with An ⊇ An+1 ∀ n and µ(An ) < +∞ for some n, then lim µ(An ) = µ(A).
n→∞
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 countably additive.
Exercises: (2.10 ) (i) In the proofs of part (ii) and part (c) of theorem 2.3.2, where do you think we used the hypothesis that µ(X) < +∞? Do you think this condition is necessary? (ii) Let A be an algebra of subsets of a set X and µ : A → [0, ∞] be a finitely additive set function such that µ(X) < +∞. Show that the following statements are equivalent: (a) lim µ(Ak ) = 0, whenever {Ak }k≥1 is a sequence in A with k→∞ T Ak ⊇ Ak+1 ∀ k, and ∞ k=1 Ak = ∅. (b) µ is countably additive.
(2.11 ) Extend the claim of theorem 2.3.2 when A is only a semi-algebra of subsets of X. (Hint: Use exercise 1.8) (2.12 ) Let A be a σ-algebra and µ : A → [0, ∞] be a measure. For any sequence {En }n≥1 in A, show that
17
2.4. Uniqueness problem for measures
(i) µ (lim inf n→∞ En ) ≤ lim inf n→∞ 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 inf En := n→∞
∞ \ ∞ [
n=1 k=n
Ek ⊆ lim sup En := n→∞
∞ [ ∞ \
Ek . )
n=1 k=n
2.4. Uniqueness problem for measures Concepts and examples: The problem that we want to analyze is the following: Let µ be a σ-finite measure on an algebra A of subsets of X. Let µ1 and µ2 be two measures on 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 of X and let µ : C −→ [0, ∞] be a set function. We say µ is totally finite (or just finite) if µ(A) < +∞ ∀ A ∈ C. The set function µ is said to be sigma finite (written as σ-finite) if there exist pairwise disjoint S∞ sets Xn ∈ C, n = 1, 2, . . . , such that µ(Xn ) < +∞ for every n and X = n=1 Xn . 2.4.2. Examples:
(i) The length function λ on the class of intervals is σ-finite. ˜ the algebra generated by left-open right-closed intervals (ii) Let A = A(I), in R. For A ∈ A, let µ(A) = +∞ if A 6= ∅ and µ(∅) = 0. Then µ is a measure on A and it is not σ-finite. Let ξ ∈ R be chosen arbitrarily and fixed. 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-algebra X ∩S(A) = P(X) and that every nonempty set in the 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 that µc 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 class M = {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-finite measures 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 set X is finite if and only if µ(X) < +∞.
Chapter 3
Construction of measures
Concepts and examples:
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 function µ1 : C1 −→ [0, +∞] and µ2 : C2 −→ [0, +∞] be set functions. The set function 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, in example 2.4.2(iii) each µc is an extension of µ. Above examples show that in general a measure µ on an algebra A can have more than one extension to S(A), the σ-algebra generated by A. Our next theorem describes a method of uniquely extending a measure from 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 ˜ on F(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 = C, define
Sn
i=1 Ei
µ ˜(E) :=
for pairwise disjoint sets E1 , . . . , En ∈
∞ X
µ(Ei ).]
i=1
Exercises: (3.1) 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 finitely additive. Is µ monotone? Countably subadditive? (3.2) If C be a semi-algebra, then µ is countably subadditive iff ∀ A ∈ C S∞ with A ⊆ i=1 Ai , Ai ∈ C implies µ(A) ≤
∞ X
µ(Ai ).
i=1
(3.3) Using theorem 3.1.3, show that length function, which is initially defined on the semi-algebra I of all intervals, can be uniquely extended to a set function on F(I), the algebra generated. It is worth mentioning a result due to S.M. Ulam (1930) which, under the assumption of the “continuum hypothesis”, implies that it is not possible to extend the length function to all subsets of R. Theorem (Ulam): Let µ be a measure defined on all subsets of R such that µ((n, n + 1]) < ∞ ∀ n ∈ Z and µ({x}) = 0 for every x ∈ R. Then µ(E) = 0 for every E ⊆ R. 3.1.3. Remark: Ulam’s theorem shows the impossibility of extending the length function from intervals to all subsets of R, assuming the continuum hypothesis. We shall see later that similar results can be proved if one assumes the ‘axiom of 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 results we 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, our aim 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 measured accurately. The approximate size of any set E ⊆ X is given by the outer measure 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 measure on A. For E ⊆ X, define ) (∞ ∞ [ X ∗ Ai ⊇ E . µ (E) := inf µ(Ai ) Ai ∈ A, i=1
i=1
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) ≤
∞ X
µ∗ (Ai ) whenever A =
i=1
(iv) µ∗ is an extension of µ , i.e.,
∞ [
Ai .
i=1
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 outer measure if ν has properties (i), (ii) and (iii) in proposition 3.2.2. The outer measure µ∗ induced by µ is characterized by the property that if ν is any outer 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 with µ on A. (ii) In the definition of µ∗ (E) the infimum is taken over the all possible countable coverings of E. To see that finite coverings will not suffice, consider E := Q ∩ (0, 1), the set of all rationals in (0, 1), and Snlet I1 , I2 , . . . , In be any finite collection of open intervals such that E ⊆ i=1 Ii . Then it is Pn ∗ easy to see that i=1 λ(Ii ) ≥ 1. This will imply λ (E) ≥ 1 if only finite coverings are considered in the definition of λ∗ , which contradicts the fact that λ∗ (E) = 0, E being a countable set.
22
3. Construction of measures
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 is countable and µ(A) = 1 if Ac is countable. Then, µ is a measure on A. Let µ∗ be the outer measure induced by µ on P(R). It follows from proposition 3.2.2 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 ) (∞ ∞ [ X ∗ Ai ⊇ E . µ (E) = inf µ(Ai ) Ai ∈ A, Ai ∩ Aj = ∅ for i 6= j and i=1
i=1
(3.7) Let X be any nonempty set and let A be any algebra of subsets of X. Let x0 ∈ X be fixed. For A ∈ A, define 0 if x0 6∈ A, µ(A) := 1 if x0 ∈ A.
Show that µ is countably additive. Let µ∗ be the outer measure induced 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 implies x0 ∈ A? Show that this is possible if {x0 } ∈ A.
3.3. Choosing nice sets: Measurable sets
Concepts and examples: In the previous section we defined the notion of µ∗ , the outer measure induced by µ on all subsets of X. We saw that µ∗ (A) = µ(A), ∀ A ∈ A, but in general µ∗ need not be even finitely additive on P(X). Let us try to identify some subclass S of P(X) such that µ∗ restricted to S will be countable additive. 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 next definition. 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 ∩ E c ).
(3.1)
We denote by S ∗ the class of all µ∗ -measurable subsets of X. Note that E ∈ S ∗ iff E c ∈ 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 subset Y of X into two parts, Y ∩ E and Y ∩ E c , so that their sizes µ∗ (Y ∩ E) and µ∗ (Y ∩ E c ) add up to give the size µ∗ (Y ) of Y. Thus a ‘nice’ set is in a sense a ‘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 ∩ E c ). (iii) For every Y ⊆ X, with µ∗ (Y ) < +∞, µ∗ (Y ) ≥ µ∗ (Y ∩ E) + µ∗ (Y ∩ E c ). (iv) For every A ∈ A, µ(A) ≥ µ∗ (A ∩ E) + µ∗ (A ∩ E c ). 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(E c )
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 to S ∗ is finitely additive.
24
3. Construction of measures
(iii) If An ∈ S ∗ , n = 1, 2, . . . , then S ∗ is countably additive.
S∞
n=1 An
∈ S ∗ and µ∗ restricted to
(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, then there exists a unique extension of µ to a measure µ on S(A), the σ-algebra generated by A. 3.3.6. Remark: Theorems 3.2.2.(iv) and 3.3.4 together give us a method of constructing an extension 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 example 3.2.4. (3.9) Let X = [a, b] and let S be the σ-algebra of subsets of X generated by all subintervals of [a, b]. Let µ, ν be finite measures on S such that µ([a, c]) = ν([a, c]), ∀ c ∈ [a, b]. Show that µ(E) = ν(E) ∀ E ∈ S. ˜ as given in proposition (3.10) Let µF be the measure on the algebra A(I) 2.2.1. Let µF itself denote the unique extension of µF to LF , the σ-algebra of µ∗F -measurable sets, as given by theorem 3.3.4. Show that (i) BR ⊆ LF . (ii) µF ({x}) = F (x) − lim F (y). Deduce that the function F is y↑x
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 induced by the distribution function F.
3.4. Completion of a measure space Concepts and examples: Theorem 3.3.4 showed that, given a σ-finite measure µ on an algebra A of subsets of a set X, µ can be extended to a unique measure µ∗ on the σalgebra S ∗ of µ∗ -measurable subsets of X, and S ∗ ⊇ S(A). In this section we 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 S∞measure induced by µ. Let Aσ denote the collection of sets of the form i=1 Ai , 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) = 0 for 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 on S. The pair (X, S) is called a measurable space and the triple (X, S, µ) is called a measure space. Elements of S are normally called measurable sets. Till now what we have done is that, given a measure on an algebra A of subsets 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 ∗ . This property 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 can be 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
26
3. Construction of measures
N ∈ S with µ(N ) = 0}. One says (X, S, µ) is complete if N ⊆ S. Elements of 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 in definition 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 of X. Let µ(E △ N ) = µ(E), ∀ E ∈ S, N ∈ N . Then µ is a measure on S ∪ N and (X, S ∪N , µ) is a complete measure space, called the completion of the measure space (X, S, µ). (The measure space (X, S ∪ N , µ) is also denoted by (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 the induced outer measure. Let E ∈ S ∗ be such that µ∗ (E) < +∞ and let ǫ > 0 be 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 a probability space and the measure µ is called a probability. The reason for this terminology is that the triple (X, S, µ) plays a fundamental role in the axiomatic theory of probability. It gives a mathematical model for analyzing statistical experiments. The set X represents the set of all possible outcomes of the experiment, the σ -algebra S represents the collection of events of interest in that experiment, and for every E ∈ S, the nonnegative number µ(E) is the probability that the event E occurs. For more details see Kolmogorov [9] and Parthasarathy [10]. Exercises: (3.11 ) Let E ⊆ X, and let G1 , G2 be two measurable covers of E. Show that µ∗ (G1 ∆G2 ) = 0. (3.12) Let E1 ⊆ E2 ⊆ E3 ⊆ . . . be subsets of X. Then ! ∞ [ µ∗ En = lim µ∗ (En ). n=1
n→∞
(3.13 ) Let E ⊆ X, and let K1 , K2 be two measurable kernels of E. Show that µ∗ (K1 ∆K2 ) = 0.
27
3.5. The Lebesgue measure
(3.14 ) Let N := {E ⊆ X | µ∗ (E) = 0}. Show that N is closed under countable 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 Concepts and examples: We now apply the extension theory of measures, developed in previous sections, to the particular case when X = R, A = A(I), the algebra generated by 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 subsets of R is called the Lebesgue outer measure and can be described as follows: for E ⊆ R, (∞ ) ∞ X [ ∗ λ (E) := inf λ(Ii ) Ii ∈ I ∀ i, Ii ∩ Ij = ∅ for i 6= j and E ⊆ Ii . i=1 ∗
i=1
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 simply by 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 or BR by λ itself. The measure space (R, L, λ) is called the Lebesgue measure space and λ is called S the Lebesgue measure. We note that since λ 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
28
3. Construction of measures
is at least 2c (here c denotes the cardinality of the real line, also called the cardinality of the continuum). Since L ⊆ P(R), we get the cardinality of L to be 2c. On the other hand, BR is the σ-algebra generated by all open intervals of R with rational endpoints. One can show that the σalgebra 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 is called analytic sets. An analytic set is a set which can be represented as a continuous 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 elements, 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 not possible to define a countably additive set function µ on P(R) such that µ({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 said if 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, assuming that such a set exists, try to see whether one can reach a contradiction. G. Vitali (1905), F. Bernstein (1908), H. Rademacher (1916) and others constructed such sets assuming the ‘axiom of choice’ (see appendix B). The example of Vitali used the translation invariance property of the Lebesgue measure, and that of Bernstein used the regularity properties of the Lebesgue measure. Rademacher proved that every set of positive outer Lebesgue measure includes a Lebesgue nonmeasurable set. Even today, more and more 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 did not accept such constructions. In 1970, R. Solovay [37] proved that if one includes the statement “all subsets of R are Lebesgue measurable” as an axiom in set theory, then it is consistent with the other axioms of set theory if the axiom of choice is not assumed. Construction of a nonmeasurable set(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’ subsets of R, such as open sets, closed sets and compact sets. Also, for E ∈ BR , if we 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 the map y 7−→ x + y is a homeomorphism of R onto R, and hence E + x ∈ B for every open set E. We leave it for the reader to verify (using σ-algebra
29
3.5. The Lebesgue measure
techniques) that this is true for all sets E ∈ BR . The relation of λ on L with λ on topologically nice subsets of R and the question as to whether E + x ∈ L for E ∈ L, x ∈ R, i.e., do the group operations on R preserve the class 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)
(i)
=⇒
(iii)
=⇒
(v)
=⇒
(i).]
and
3.5.3. Note: Theorem 3.5.2 tells us the relation between L, the class of Lebesgue measurable 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 ⊇ E with λ(G \ E) < ǫ can be stated equivalently as: λ(E) = inf{λ(U ) | U open, U ⊇ E}.
30
3. Construction of measures
This is called the outer regularity of λ. Other examples of outer regular measures 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 subsets of R. It is natural to ask the question: does there exist a relation between L and the class of compact subsets of R? Let K be any compact subset of R. 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 , if λ(E) < +∞, from the knowledge of λ(K), K compact in R? The answer is given by the next proposition. 3.5.4. Proposition: Let E ∈ L with 0 < λ(E) < +∞ and let ǫ > 0 be given. Then there exists a compact set K ⊆ E such that λ(E \ K) < ǫ. On the set R, we have the group structure given by the binary operation of the addition of two real numbers. We analyze the behavior of λ on L under the map y 7−→ y + x, y ∈ R and x ∈ R fixed. We saw that A + x is a Lebesgue measurable set whenever A is Lebesgue measurable and x ∈ R. It is 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 function from the class I of intervals to BR , the σ-algebra of Borel subsets of R. This gave 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 regular measure on BR . The question arises: are there other σ-finite measures on BR 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. In fact the following hold:
3.5. The Lebesgue measure
31
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 ‘topological groups’. Let us recall that the set of real numbers R is a group under the binary operation +, the addition of real numbers. Also, there is a topology on R which ‘respects’ the group structure, i.e., the maps (t, s) 7−→ t + s and t −→ −t from R × R −→ R and R −→ R, respectively, are continuous when R × R is given the product topology. In an abstract setting, if G is a set with a binary operation ‘·’ and a topology T such that (G, ·) is a group and the maps G × G −→ G, (g, h) 7−→ g.h and G −→ G, g 7−→ g −1 are continuous with respect to the product topology on G × G, one calls G a topological group. Given a topological group, let BG denote the σ-algebra generated 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 it has 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 topological group 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 multiplication of real numbers. Let R \ {0} be given the subspace topology from R. It is easy to show that R \ {0} is a topological group and, ∀ E ∈ BR\{0} , Z 1 dλ(x) (3.2) µ(E) := |x| E 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 the semi-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-closed intervals in R and consider F : R −→ R as a monotonically increasing right continuous function, then we can construct a countably additive set ˜ as in example 1.1.2. Using theorem function µF on the semi-algebra I,
32
3. Construction of measures
3.4.6, we can construct a complete measure µF on a σ-algebra of subsets of R which includes BR . This measure µF is called the Lebesgue-Stieltjes measure induced by the function F. Note that µF has the property that µF (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 continuous function F : R −→ R such that the unique Lebesgue-Stieltjes measure µF induced by F is nothing but µ (by the uniqueness of the extension). Thus measures µ on BR which have the property that µ(a, b] < +∞ ∀ a < b can be looked upon as a Lebesgue-Stieltjes measure µF for some F. We point out that it is possible to find different F1 , F2 : R −→ R such that both 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 function such that µ = µF . This F is called the distribution function of µ. When µ(R) = 1, µ is called a probability and its distribution function F, which is monotonically increasing and is right continuous with lim [F (x) − F (−x)] 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 (∞ ) ∞ X [ ∗ λ (E) = inf λ(Ii ) Ii ∈ I0 ∀ i, for i 6= j and E ⊆ Ii . i=1
i=1
(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 (∪∞ \ E) < ǫ. Such sets are called Lebesgue null sets.Prove n=1 the 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.
33
3.5. The Lebesgue measure
(iv) Every subset of a null set is also a null set. S (iv) Let A1 , A2 , . . . , An , . . . be null sets. Then ∞ n=1 An 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 null set? (vi) Let E be a null subset of R and x ∈ R. What can you say about the sets E + x := {y + x | y ∈ E} and xE := {xy | y ∈ E}? (vii) Let I be an interval having at least two distinct points. Show that I is not a null set. (viii) If E contains an interval of positive length, show that it is not a 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 necessarily measurable) of R. Then, ! ∞ [ ∗ ∗ En = lim λ (En ). λ n=1
n→∞
(3.22) Let E ⊆ R. Show that the following statements are equivalent: (i) E ∈ L. ∗ ∗ ∗ (ii) λ (I) = λ (E ∩ I) + λ (E c ∩ I) for every interval I. (iii) E ∩ [n, n + 1) ∈ L for every n ∈ Z. ∗ ∗ (iv) λ (E ∩ [n, n + 1)) + λ (E c ∩ [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 , the σ-algebra generated by open subsets of X, called the σ-algebra of Borel 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
34
3. Construction of measures
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 continuous 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 T 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) and λ(−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 the following: (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 element xα ∈ 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 ∞ [ En ⊆ [−1, 2]. [0, 1] ⊆ n=1
(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 Concepts and examples: 4.1.1. Definition: Let s : X −→ [0, ∞] be defined by s(x) =
n X
ai χAi (x),
x ∈ X,
i=1
where n is some positive integer; a1 , a2 , . . . , an are nonnegative extended real S numbers; Ai ∈ S for every i; Ai ∩ Aj = ∅ for i 6= j; and ni=1 Ai = X. Such a function Pn on (X, S) Pn s is called a nonnegative simple measurable function and i=1 ai χAi (x) is called a representation of s. We say i=1 ai χAi is the standard representation of s if a1 , a2 , . . . , an are all distinct. We denote by L+ 0 the class of all nonnegative simple measurable functions 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 P set Ai ∈ S, i = 1, 2, . . . , n. And in that case its standard representation is ni=1 ai χAi . 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 = χA χB ∈ 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: R Pn For s ∈ L+ s(x)dµ(x), 0 with a representation s = i=1 ai χAi , we define the integral of s with respect to µ, by Z n X s(x)dµ(x) := ai µ(Ai ). i=1
R
The integral s(x)dµ(x) is also denoted by exercise (4.1)).
R
sdµ. It is well-defined (see
4.1.4. Proposition: For s, s1 , s2 ∈ L+ 0 and α ∈ R with α ≥ 0, the following hold: R (i) 0 ≤ sdµ ≤ +∞.
(ii) αs ∈ L+ 0 and
(iii) s1 + s2 ∈ L+ 0 and Z
Z
(αs)dµ = α
(s1 + s2 )dµ =
Z
Z
sdµ.
s1 dµ +
Z
s2 dµ.
(iv) For E ∈ S we have sχE ∈ L+ 0 , and the set function Z E 7−→ ν(E) := sχE dµ is a measure on S. Further, ν(E) = 0 whenever µ(E) = 0, E ∈ S. R Before we proceed further, we observe that s(x)dµ(x) is well-defined,(see R 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 Z Z sdµ = lim sn dµ. n→∞
(ii)
R
sdµ = sup
nR
o ′ ′ ′ s dµ | 0 ≤ s ≤ s, s ∈ L+ 0 .
Exercises: (4.1) Show that for s ∈ L+ 0, following: let n X
s=
R
s(x)dµ(x) is well-defined by proving the
ai χAi =
m X
bj χBj ,
j=1
i=1
where {A1 , . . . , An } and {B1 , . . . , Bm } are partitions of X by elements of S, then (ii) s =
n X i=1
ai
m X
χAi ∩Bj =
j=1
m X
bj
j=1
n X
χAi ∩Bj .
i=1
(ii) n X i=1
(iii)
R
ai µ(Ai ) =
m X
bj µ(Bj ).
j=1
s(x)dµ(x) P is independent of the representation of the function s(x) = ni=1 ai χAi .
(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 they belong to L+ 0. (4.3) Let s1 , s2 ∈ L+ 0 . Prove R the following: R (i) If s1 ≥ s2 , then s1 dµ ≥ s2 dµ. (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 Z Z Z (s1 ∧ s2 )dµ ≤ si dµ ≤ (s1 ∨ s2 )dµ, i = 1, 2.
38
4. Integration
(4.4) Express the functions χA ∧ χB and χA ∨ χB , for A, B ∈ S, in terms of 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 PS. For x n∈ (0, 1], if x has non-terminating dyadic expansion x = ∞ n=1 xn /2 . Let +1 if xi = 1, fi (x) := −1 if xi = 0, i = 1, 2, . . . . + Show that for every i, there R exists simple function si ∈ L0 such that fi = si − 1. Compute si dλ.
(4.6) Let s : X −→ R∗ be any nonnegative function such that the range −1 ∗ of s is a finite set. Show that s ∈ L+ 0 iff s {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+ . Can you say that 0 Z Z Z φdµ = s1 dµ − s2 dµ? ′
(4.9) Let {sn }n≥1 and {sn }n≥1 be sequences in L+ 0 such that for each ′ x ∈ X, both {sn (x)}n≥1 and {sn (x)}n≥1 are increasing and ′
lim sn (x) = lim sn (x).
n→∞
n→∞
Show that lim
n→∞
Z
sn dµ = lim
n→∞
Z
′
sn dµ. ′
(Hint: Apply exercise (4.3)Rand proposition 4.1.5 R to {sn ∧ sm }n for ′ all fixed m to deduce that sm dµ ≤ limn→∞ sn dµ.)
(4.10) Show that in general L+ 0 need not be closed under limiting operations. For example, consider the Lebesgue measure space (R, L, λ) and construct a sequence {sn }n≥1 in L+ 0 such that lim sn (x) = f (x) exists but f 6∈ L+ 0.
n→∞
4.2. Integral of nonnegative measurable functions Concepts and examples: 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 if there exists an increasing sequence of functions {sn }n≥1 in L+ 0 such that f (x) = lim sn (x) ∀ x ∈ X. n→∞ If the underlying σ-algebra is clear from the context, a S-measurable function is also called measurable. We denote the set of all nonnegative measurable functions by L+ . (ii) For a function f ∈ L+ , we define the integral of f with respect to µ by Z Z sn (x)dµ(x). f (x)dµ(x) := lim n→∞
R It follows fromRexercise (4.9) that for f ∈ L+ , f dµ is well-defined. Clearly, R + + and L+ ⊆ L sdµ for an element s ∈ L is the same as sdµ, for s as an 0 0 + element of L . The next proposition gives a characterization of functions in L+ and the integrals of its elements. Another (intrinsic) characterization of L+ 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) = lim sn (x) ∀ x ∈ X. n→∞ R R (ii) If f ∈ L+ and s ∈ L+ sdµ ≤ f dµ and 0 is such that 0 ≤ s ≤ f, then Z Z + f dµ = sup sdµ 0 ≤ s ≤ f, s ∈ L0 . 4.2.3. Definition: Let (X, S, µ) be a measure space and Y ∈ S. We say a property P holds almost 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 and µ(E) = 0. R We describe next the properties of f dµ, for f ∈ L+ .
4.2.4. Proposition: Let f, f1 , f2 ∈ L+ . Then the following hold: R (i) f dµ ≥ 0 and for f1 ≥ f2 Z Z f1 dµ ≥ f2 dµ.
40
4. Integration
(ii) For α, β ≥ 0 we have (αf1 + βf2 ) ∈ L+ and Z Z Z (αf1 + βf2 )dµ = α f1 dµ + β f2 dµ.
(iii) For every E ∈ S we have χE f ∈ L+ . If Z ν(E) := χE f dµ, E ∈ S,
then ν is a measure on S and ν(E) = 0 whenever µ(E) = 0. R R The integral f χE dµ is also denoted by E f dµ and is called the integral of f over E.
(iv) If f1 (x) = f2 (x) for a.e. x(µ), then Z Z f1 dµ = f2 dµ.
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 beRclosed under limits. The next theorem discusses this and the behavior of f dµ under increasing limits, extending proposition 4.1.5 to functions in L+ . 4.2.5. Theorem (Monotone convergence): Let {fn }n≥1 be an increasing sequence of functions in L+ , and f (x) := lim fn (x), x ∈ X. Then f ∈ L+ and n→∞ Z Z f dµ = lim fn dµ. n→∞
4.2.6. Remark: + + If {fn }n≥1 R is a sequence inR L decreasing to a function f ∈ L , then the equality f dµ = limn→∞ fn dµ need not hold. For example, let X = R, S = L and µ = λ, the Lebesgue measure. Let fn = χ[n,∞) . Then fn ∈ L+ 0 ⊆ R + n and RL , and {fn }n≥1 decreases to f ≡ 0. Clearly, fn dλ = +∞ for every f dλ = 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 shown as a consequence of the characterization of L+ proved in the next section.
41
4.2. Integral of nonnegative measurable functions
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, lim sn (x) = f (x). Can you conclude n→∞ that Z Z f dµ = lim
n→∞
sn dµ?
(4.12) Let f ∈ L+ . Show that Z Z . f dµ = 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 that f (x) := lim fn (x) exists for a.e. x(µ). Show that f ∈ L+ and n→∞ Z Z fn dµ, f dµ = lim n→∞
where f (x) is defined as an arbitrary constant for all those x for which lim fn (x) does not converge. n→∞
(4.14) Let µ(X) < ∞, and let f ∈ L+Sbe a bounded function. Let P := {E1 , E2 , . . . , En } be such that ni=1 Ei = 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 X
mi χEi and ΨP :=
n X
Mi χEi .
i=1
i=1
Prove the following: (i) For every partition P, show that ΦP , ΨP ∈ L+ 0 and ΦP ≤ f ≤ ΨRP . R (ii) f dµ = supR ΦP dµ | P is a measurable partition of X , = inf ΨP dµ | P is a measurable R partition of X . (This gives an equivalent way of defining f dµ, in a way similar to that for the Riemann integral.) (iii) Let Z + α = sup sdµ | s ∈ L0 , s ≤ f 0
and
β = inf
Z
sdµ | s ∈
L+ 0,
f ≤s .
42
4. Integration
Show that Z Φp dµ | P is a measurable partition of X α = sup and
β = inf
Z
Ψp dµ | P is a measurable partition of X .
(iv) Deduce that f ∈ L+ implies α =
R
f dµ = β.
Note: R Exercise 4.14 tells us that for f ∈ L+ , in defining f dµ it is enough to consider approximations of f from below, Ras the approximations from above will also give the same value for f dµ. This is because f ∈ L+ , i.e., f is nonnegative measurable. The converse is also true, 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 we can choose functions φn , ψn ∈ L+ 0 such that Z 1 φn ≤ f ≤ ψn and (ψn − φn )dµ < . n
Let φ := sup φn and ψ := inf ψn . Then φ, ψ ∈ L+ (see corollary 4.3.11) and ∀ n, Z Z 1 (ψ − φ)dµ ≤ (ψn − φn )dµ < . n Thus Z (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 Concepts and examples: Recall, f ∈ L+ means f : X −→ R∗ is a nonnegative function with the property 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 µ plays no part in the definition of the functions in L+ . It is only to define the integral of functions f ∈ L+ that we need the measure µ. We want to characterize 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 nonnegative values. 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 equivalent: (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 of L+ intrinsically. This proposition is used very often to check the measurability of nonnegative functions. Once again we emphasize the fact that for a nonnegative function f : X −→ R∗ to be measurable, i.e., for f to be in L+ , the measure µ plays no part. As is clear from the above proposition, it is only the σ-algebra S of subsets of X that is important . The notion of measurability 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: R Let f ∈ L+ and E ∈ S be such that E f dµ = 0. Then f (x) = 0 for a.e. (µ)x ∈ E. Next, we extend the concept of measurability of nonnegative functions to 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-measurable function 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. Then following 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 sn are both nonnegative simple measurable functions on (X, S) and lim sn (x) = f (x) ∀ x ∈ X. n→∞
(iii) f satisfies any one (and hence all) of the statements (ii) to (vi) of proposition 4.3.2. 4.3.6. Examples: (i) Let f : X −→ R∗ be a constant function, f (x) = α ∀x ∈ X. Then f is measurable ∀c ∈ R, since ∅ if c ≤ α, {x ∈ X | f (x) > c} = X if c > α.
(ii) Let f : X → R∗ be measurable, and α ∈ R. Then since {x ∈ X | f (x) < c/α} X {x ∈ X | α(x) > c} = ∅ {x ∈ X | f (x) < c/α}
αf is also measurable, if if if if
α > 0, α = 0, c ≤ 0, α = 0, c > 0, α < 0.
(iii) Let X be a topological space and S be the σ-algebra of Borel subsets of X, i.e., the σ-algebra generated by the open sets. Let f : X −→ R be any 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) >
45
4.3. Intrinsic characterization of nonnegative measurable functions
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 the functions sup fn , inf fn , lim sup fn and lim inf fn is a measurable function. n
n
n→∞
n→∞
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 fn , inf fn , n
n
lim sup fn and lim inf fn is in L+ . In particular, if lim fn =: f exists, then
n→∞ f ∈ L+ .
n→∞
n→∞
R Recall that in theorem 4.2.5 we analyzed the limit of fn dµ for an increasing sequence of nonnegative measurable functions. We analyze next the R behavior of fn dµ when {fn }n≥1 is not necessarily an increasing sequence of nonnegative measurable functions. 4.3.12. Theorem (Fatou’s lemma): Let {fn }n≥1 be a sequence of nonnegative measurable functions. Then Z Z lim inf fn dµ ≤ lim inf fn dµ. n→∞
n→∞
4.3.13. Proposition: Let (X, S) be a measurable space and let f : X −→ R∗ be S-measurable. Let µ 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 also S-measurable.
46
4. Integration
Exercises: (4.15)(i) Let f ∈ L+ and E ∈ S beR such that f (x) > 0 for every x ∈ E and µ(E) > 0. Show that E f dµ > 0. (ii) Let f, g ∈ L+ be such that Z Z Z Z f dµ = gdµ < +∞ and f dµ = gdµ, ∀ E ∈ S. E
E
Show that f (x) = g(x) a.e. (x)µ. (iii) Let f, g be nonnegative measurable functions on (R, L) such that Z b Z b f dµ = gdµ < +∞ for every a < b a
a
show that then Z
f dµ =
E
Z
gdµ, ∀ E ∈ L, E
and deduce that f (x) = g(x) a.e. (x)λ. (4.16) Let f : X −→ R∗ be a bounded measurable function. Then there exists 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. Show that there exist sequences of nonnegative simple functions {sn }n≥1 and {˜ sn }n≥1 such that 0 ≤ · · · ≤ sn (n) ≤ sn+1 (x) ≤ · · · ≤ f (x) ≤ · · · ≤ s˜n+1 (x) ≤ s˜n (x) · · · and lim sn (x) = f (x) = lim s˜n (x) ∀ x ∈ X. n→∞
n→∞
R∗
(4.18) Let f and g : X −→ be measurable functions, p and α ∈ R with p > 1, and let m be any positive integer. Use proposition 4.3.9to prove the following: (i) f + α is a measurable function. (ii) Let β and γ ∈ R∗ be arbitrary. Define for x ∈ R, (f (x))m if f (x) ∈ R, m β if f (x) = +∞, f (x) := γ if f (x) = −∞. Then f m is a measurable function. (iii) Let |f |p be defined similarly to f m , where p is a nonnegative real number. Then |f |p is a measurable function.
4.3. Intrinsic characterization of nonnegative measurable functions
47
(iv) Let β, γ, δ ∈ R∗ be arbitrary. Define for x ∈ R, 1/f (x) if f (x) 6∈ {0, +∞, −∞}, β if f (x) = 0, (1/f )(x) := 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, f (x)g(x) if x 6∈ A, (f g)(x) := β if x ∈ A. Then f g is a measurable function.
(4.19) Let f : X → R∗ be S-measurable. Show that |f | is also Smeasurable. Give an example to show that the converse need not be 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 that S = 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≥1 is Cauchy} belong to S. (4.22) Give an example to show that strict inequality can occur in Fatou’s lemma. P (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 Z ∞ Z X f dµ = fn dµ. n=1
(4.24) Show that each of the functions f : R −→ R defined below is LR measurable, and compute f dλ : 0 if x ≤ 0, (i) f (x) := 1/x if x > 0. (ii) f (x) := χQ (x), the indicator function of Q, the set of rationals.
48
4. Integration
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 (iii) f (x) := decimal expansion of x, the first nonzero entry is at the (n + 1)th place. R R (4.25) Let f , g ∈ L+ with f ≥ g. Show that (f −g) ∈ L+ and f dµ ≥ gdµ. Can you conclude that Z Z Z (f − g)dµ = f dµ − gdµ? (4.26) Let f , fn ∈ L+ , n = 1, 2, . . ., be such that 0 ≤ fn ≤ f. If lim fn (x) = f (x), can you deduce that n→∞
Z
f dµ = lim
n→∞
Z
fn dµ?
(4.27) Let f ∈ L. For x ∈ X and n ≥ 1, define f (x) if |f (x)| ≤ n, n if f (x) > n, fn (x) := −n if f (x) < −n.
Prove the following: (i) fn ∈ L and |fn (x)| ≤ n ∀ n and ∀ x ∈ X. (ii) lim fn (x) = f (x) ∀ x ∈ X. n→∞
(iii) |fn (x)| := min{|fn (x)|, n} := (|f | ∧ n)(x) is an element of L+ and Z Z |fn |dµ = |f |dµ. lim n→∞
Note: For f ∈ L, the sequence {fn }n≥1 as defined in exercise 4.27 is called the truncation sequence of f. The truncation sequence is useful in 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 any nonnegative BR -measurable function, i.e., g −1 (A) ∈ BR ∀ A ∈ BR , then g ◦ f ∈ L and Z Z g dν = (g ◦ f ) dµ. The measure ν is usually denoted by µf −1 and is called the distribution of the measurable function f.
4.3. Intrinsic characterization of nonnegative measurable functions
49
(4.29) Let f ∈ L+ be a bounded function, say f (x) ≤ N ∀ x ∈ X and for some N ∈ N. Show that Z N 2n X k−1 k k−1 µ x| ≤ f (x) < n . (4.1) f dµ = lim n→∞ 2n 2n 2 k=1
M
m a
b
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 the measurable partition Pn of the domain |X| of f , given by {Xkn | 1 ≤ k ≤ N 2n }, where Xkn := f −1 [(k − 1)/2n , k/2n ). Using this partition Pn , we form the sums on the right hand side of (4.1), approximating the area below the curve y = f (x) from ‘inside’ (see Figure). Thus in our integral we use partitions of the range, whereas in Riemann integration we use partitions of the domain. In the words of H. Lebesgue, the situation is similar to the problem of finding the total amount in a cash box containing say n coins, the ith coin being of denomination αi , 1 ≤ i ≤ n. One method of finding the total P amount in the box is to add the value of each coin, i.e., the sum ni=1 αi . The other method is to separate out coins of similar denominations. IfPthere are kj coins of denominations αj , then the total amount is j kj αj . ¯ µ) be its completion. (4.30) Let (X, S, µ) be a measure space and let (X, S, ¯ function. Show that there Let f : X −→ R be an S-measurable exists 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 sequence of S-measurable functions on X. Let f be a function on
50
4. Integration
X such that f (x) = lim fn (x) for a.e. x(µ). Show that f is Sn→∞ measurable.
4.4. Integrable functions Concepts and examples: R Given a measure space (X, S, µ) in section 4.2, we defined f dµ, the integral for functions f ∈ L+ , i.e., f : X −→ R∗ , f nonnegative measurable. If f + − is measurable, but not necessarily nonnegative, we can R +write f = R f− − f . + − Since both f and f are nonnegative measurable, f dµ and f dµ are defined. Thus it is reasonableR (as the integral is Rexpected to be linear) to R define the integralR of f to beR f dµ := f + dµ − f − dµ. The problem can arise in the case f − dµ = f + dµ = +∞. To overcome this difficulty, we introduce the following definition: 4.4.1. Definition: f : X −→ R∗ is said to be µ-integrable if both RA measurable R function + − f dµ and f dµ are finite, and in that case we define the integral of f to be Z Z Z + f dµ := f dµ − f − dµ.
We denote by L1 (X, S, µ) (or simply by L1 (X) or L1 (µ)) the space of all µ-integrable functions on X. The properties of the space L1 (X, S, µ) and of the map f 7−→ f ∈ L1 (X, S, µ), are given in the next proposition.
R
f dµ,
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 Z Z f dµ = gdµ.
(iii) If f ∈ L1 (µ), then af ∈ L1 (µ) and Z Z (af )dµ = a f dµ.
(iv) If f and g ∈ L1 (µ), then f + g ∈ L1 (µ) and Z Z Z (f + g)dµ = f dµ + gdµ.
51
4.4. Integrable functions
(v) If f , g ∈ L1 (µ), then (af + bg) ∈ L1 (µ) and Z Z Z (af + bg)dµ = a f dµ + b gdµ. 4.4.3. Proposition: Let f ∈ L1 (µ). For every E ∈ S, let Z ν(E) := χE |f |dµ and
ν˜(E) :=
Z
χE f dµ.
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 that ν(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 exist an ǫ >S0 and sets En ∈ S, n ≥ 1, such that µ(En ) < 2−n but ν(En ) ≥ ǫ. Let An = ∞ k=n Ek . Then {An }n≥1 is a decreasing sequence in S and µ(An ) ≤ µ(En ) ≤ 2−n . Thus by theorem 2.3.2, ! ∞ \ µ An = lim µ(An ) = 0. n=1
n→∞
On the other hand, µ(An ) ≥ µ(En ) ≥ ǫ, ∀ ǫ, contradicting (ii).
We prove next the most frequently used theorem which allows us to interchange 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 to f (x) for a.e. x(µ). Then the following hold: (i) f ∈ L1 (µ). R R (ii) f dµ = limn→∞ fn dµ. R (iii) limn→∞ |fn − f |dµ = 0.
We state another version of this theorem, which is applicable to series of functions.
52
4. Integration
4.4.6. Corollary: P∞ R Let {fn }n≥1 be a sequence of functions in L (µ) such that 1 n=1 |fn |dµ P∞ < +∞. Then f (x) := n=1 fn (x) exists for a.e. x(µ), f ∈ L1 (µ) and Z ∞ Z X fn dµ. f dµ = n=1
The following is a variation of the dominated convergence theorem for finite 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 functions. Suppose there exists M > 0 such that |fn (x)| ≤ M a.e. x(µ) and fn (x) → f (x) a.e. x(µ). Then f, fn ∈ L1 (X, S, µ) and Z Z f dµ = lim fn dµ. n→∞
4.4.8. Notes: (i) The monotone convergence theorem and the dominated convergence theorem (along with its variations and versions) are the most important theorems used for the interchange of integrals and limits. (ii) Simple function technique: This is an important technique (similar to the σ-algebra technique) used very often to prove results about integrable and nonnegative measurable functions. Suppose we want to show that a certain claim (∗) holds for all integrable functions. Then technique is the following: (1) Show that (∗) holds for nonnegative simple measurable functions. (2) Show that (∗) holds for nonnegative measurable / integrable functions by approximating them by nonnegative simple measurable functions and using (1). (3) Show that (∗) holds for integrable functions f, by using (2) and the fact 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 Z ν(E) := f dµ. E
53
4.4. Integrable functions
Then ν is a finite measure on S. Further, f g ∈ L1 (X, S, µ) for every g ∈ L1 (X, S, ν), and Z Z f dν = f gdµ. We shall see some more applications of the dominated convergence theorem in the remaining sections. In the next section we look at some special properties of L1 (X, S, µ) in the particular case when X = R, S = L, the σ-algebra of Lebesgue measurable sets, and µ = λ, the Lebesgue measure. As an application of the dominated convergence theorem, we exhibit the possibility of interchanging the order of integration and differentiation in the next theorem. 4.4.10. Theorem: Let ft ∈ L1 (µ) for every t ∈ (a, b) ⊆ R. Let t0 ∈ (a, b) be such that for a.e. x(µ), t 7−→ ft (x) is differentiable in a neighborhood U of t0 and there dft (x) ≤ g(x) for a.e. x(µ) and for exists a function g ∈ L1 (µ) such that dt R every t ∈ U. Then φ(t) := ft (x)dµ(x) is differentiable at t0 and ! Z dft ′ φ (t0 ) = dµ(x). (x) dt t0
Exercises: (4.32) For f ∈ L, prove the following: (i) f ∈ L1 (µ) iff |f | ∈ L1 (µ). Further, in either case Z Z f dµ ≤ |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 χE f ∈ L1 (µ), where Z Z f dµ := χE f dµ. E
Further, if E, F ∈ S are disjoint sets, show that Z Z Z f dµ = f dµ + f dµ. E∪F
E
F
54
4. Integration
(4.35) Let f ∈ L1 (µ) and Ei ∈ S,P i ≥ 1, R be such that Ei ∩ Ej = ∅ for i 6= j. Show that the series ∞ i=1 Ei f dµ is absolutely convergent, S∞ and if E := i=1 Ei , then Z ∞ Z X f dµ = f dµ. i=1
Ei
E
(4.36) (i) For every ǫ > 0 and f ∈ L1 (µ), show that Z 1 |f |dµ < ∞. µ {x ∈ X | |f (x)| ≥ ǫ} ≤ ǫ This is called Chebyshev’s inequality. (ii) Let f ∈ L1 (µ), and let there exist M > 0 such that Z 1 f dµ ≤ M µ(E) E
for every E ∈ S with 0 < µ(E) < ∞. Show that |f (x)| ≤ M for a.e. x(µ). (4.37) Let λ be the Lebesgue measure on R, and let f ∈ L1 (R, L, λ) be such that Z f (t)dλ(t) = 0, ∀ x ∈ R. (−∞,x)
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 any accumulation point of I and let f (x) := lim ft (x) exist for a.e. x(µ). t→t0
Then f ∈ L1 (µ) and Z Z f dµ = lim ft (x)dµ(x). t→t0
Further, if ∀ x the Rfunction t 7−→ ft (x) is continuous, then so is the 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 such that |fn | ≤ gn ∀ n. Let f and g be measurable functions such that lim fn (x) = f (x) for a.e. x(µ) and lim gn (x) = g(x) for a.e. x(µ). n→∞ n→∞ If Z Z lim gn dµ = g dµ < +∞, n→∞
show that
lim
n→∞
Z
fn dµ =
Z
f dµ.
(Hint: Apply Fatou’s lemma to (gn − fn ) and (gn + fn ).)
4.5. The Lebesgue integral and its relation with the Riemann integral
55
R (4.40) Let {fn }n≥0 be a sequence in L1 (X, S, µ). Show that |fn |dµ n≥1 R R converges to |f0 |dµ iff |fn − f0 |dµ n≥1 converges 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 is a simple S0 -measurable function, then n X ψ= an χf −1 (Ei ) i=1
for some positive integer n, ai ∈ R for each i, and Ei ∈ BR , then n X ai (χEi ◦f ). ψ= i=1
(4.42) Let {fn }n≥1 be a decreasing sequence of nonnegative functions in L1 (µ) such that fn (x) → f (x). Show that Z fn dµ = 0 iff f (x) = 0 a.e. x(µ). lim x→∞
(4.43) Let µ, ν be as in proposition 4.4.9. Let Sν denote the σ-algebra of all 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 sequence in L1 (µ) such that fn → f uniformly. Show that f ∈ L1 (µ) and Z lim |fn − f | dµ = 0. n→∞
Can the condition of µ(X) < +∞ be dropped?
4.5. The Lebesgue integral and its relation with the Riemann integral Concepts and examples:
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 Lebesgue measurable sets) and µ = λ, the Lebesgue measure. The space L1 (R, L, λ), also denoted by L1 (R)R or L1 (λ), is called the space of Lebesgue integrable functions on R, and f dλ is called the Lebesgue integral of f. For any set E ∈ L, we write L1 (E) for the space of integrable functions on the measure space (E, L ∩ E, λ), where λ is restricted to L ∩ E. In the special case when E = [a, b], we would like to show that the new notion of integral for 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 Z Z b f dλ = f (x)dx. a
4.5.2. Remark: In fact, the proof of the above theorem includes a proof of the following: If f ∈ R[a, b], then f is continuous a.e. x(λ). This is because f (x) = lim Ψn (x) = lim Φn (x) a.e. x(λ). n→∞
n→∞
Thus, if we put E := {x ∈ [a, b] | f (x) = lim Φn (x) = lim Ψn (x)}, n→∞
n→∞
then E is a Lebesgue measurable set and λ([a, b] \ E) = 0. For x ∈ E, given an 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 the same 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 the set of discontinuity points of f forms a subset of ([a, b]) \ E) ∪ P, where P is the set of partition points of Pn , n = 1, 2, . . . . Hence f is continuous almost everywhere. Another proof of the converse is given in the one of the exercises.
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 that each Pn+1 is a refinement of Pn and kPn k −→ 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 lim Φn (x) = f (x) = lim Ψn (x).
n→∞
n→∞
(ii) Using (i) and the dominated convergence theorem, deduce that f ∈ L1 ([a, b]) and Z Z Z Ψn dµ. Φn dµ = lim f dλ = lim n→∞
n→∞
(iii) Show that f ∈ R[a, b] and Z Z b f (x)dx. f dλ = a
(4.46) Let f : [0, 1] −→ [0, ∞) be Riemann integrable on [ǫ, 1] for all ǫ > 0. R1 Show that f ∈ L1 [0, 1] iff limǫ→0 ǫ f (x)dx exists, and in that case Z 1 Z f (x)dx. f (x)dλ(x) = lim ǫ→0 ǫ
(4.47) Let f (x) = 1/xp if 0 < x ≤ 1, and f (0) = 0. Find necessary and sufR1 ficient condition on p such that f ∈ L1 [0, 1]. Compute 0 f (x)dλ(x) in that case. (Hint: Use exercise 4.46.)
(4.48) (Mean value property):Let f : [a, b] −→ R be a continuous function and let E ⊆ [a, b], E ∈ L, be such that λ(E) > 0. Show that there exists a real number α such that Z f (x)λ(x) = αλ(E). E
(4.49) Let f ∈ L1 (R), and let g : R −→ R be a measurable function such that α ≤ g(x) ≤ β for a.e. x(λ). Show that f g ∈ L1 (R) and there exists γ ∈ [α, β] such that Z Z |f |gdλ = γ |f |dλ. (4.50) Let f ∈ L1 (R, L, λ) and let a ∈ R be fixed. Define R [a,x] f (t) dλ(t) for x ≥ a, F (x) := R for x ≥ a. [x,a] f (t)dλ(t) Show that F is continuous.
58
4. Integration
(Hint: Without loss of generality take f ≥ 0 and show that F is continuous from the left and right. In fact, F is actually uniformly continuous.) (4.51) Let f ∈ L1 (R, L, λ) and let c ∈ R be a point of continuity of f. Show that Z f (x) dλ(x) = f (c). lim n n→∞ [c, c + 1/n] (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 Riemann integrable on [a, b]. Then Z b Z b lim fn (x)dx = f (x)dx. n→∞ a
a
(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. Show that f ∈ L1 [a, b]. (4.55) Let f : [a, b] −→ R be any continuous function. Show that f ∈ L1 [a, b]. R (4.56) Let f ∈ L1 (R) be such that K f dλ = 0 for every compact set K ⊆ R. Show that f (x) = 0 for a.e. x(λ).
(4.57) Let {fn }n≥1 be a decreasing sequence functions in P∞ of nonnegative n−1 C(a, b) and let f1 ∈ L1 (a, b). If n=1 (−1) fn ∈ C(a, b), show that ! Z b Z b X ∞ ∞ X n−1 n−1 (−1) (−1) fn (x) dx = fn (x)dx . a
(Hint: For every n,
a
n=1
n=1
Pn
k k=1 (−1) fk
≤ f ∈ L1 (a, b)).
(4.58) Give examples to show that analogues of the monotone convergence theorem and the dominated convergence theorem do not hold for the Riemann integral. (4.59) Let f ∈ L1 (R) and, ∀ t ∈ [0, ∞), Z 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] Concepts and examples: Recall that, R[a, b] is not a complete metric space under the L1 -metric Z b ˜ d(f, g) := |f (x) − g(x)|dx, a
for f, g ∈ R[a, b]. Since every metric space has a completion, the question arises: what is the completion of R[a, b] under this metric? You might have seen the abstract construction of the completion of a metric space. We shall show 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 show that L1 [a, b] is a complete metric space, and then show that R[a, b] is dense in 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 kf k1 :=
Z
|f (x)|dλ(x).
Clearly kf k1 is well-defined, and it is easy to check that the function f 7−→ kf k1 , f ∈ L1 [a, b], has the following properties: (i) kf k1 ≥ 0 ∀ f ∈ L1 [a, b]. (ii) kf k1 = 0 iff f = 0. (iii) kaf k1 = |a| kf k1 ∀ a ∈ R and f ∈ L1 [a, b]. (iv) kf + gk1 ≤ kf k1 + kgk1 ∀ f, g ∈ L1 [a, b]. The function k · k1 is called a norm on L1 [a, b]. (Geometrically it is the distance of the ‘vector’ f ∈ L1 [a, b] from the ‘vector’ 0 ∈ L1 [a, b].) For f, g ∈ L1 [a, b], if we define d(f, g) := kf − gk1 ,
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4. Integration
then d is a metric on L1 [a, b], called the L1 -metric . The most important property 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 following 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 the fact that the functions are defined on an interval. One can define k · k1 for functions defined on (E, L ∩ E, λ), where E ∈ L is arbitrary, and the proof of theorem 5.6.1 will show that L1 (E) := L1 (E, L ∩ E, λ) is also a complete metric space under the metric Z kf − gk1 := |f (x) − g(x)|dλ(x). E
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 functions which agree for a.e. (µ). Further, theorem 4.6.3 also remains true for L1 (X, S, µ). (ii) Parts of the proof of theorem 4.6.4 also exhibit the following facts which are of independent interest. Let f ∈ L1 [a, b] and ǫ > 0 be given. Then there exists a simple function Φ such that kf −Φk1 < ǫ and a step function h such that kf − hk1 < 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 kf − χ[−n,n] f k1 < 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 zero outside some finite interval and kf χ[−n,n] − gk1 < 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 smallest closed subset of R outside which f vanishes. Let the space of continuous 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 fR ∈ L1 (R) and fR˜(x) := f (−x) ∀ x ∈ R. Prove the following: (i) f˜(x)dλ(x) = f (x)dλ(x). (ii) f g ∈ L1 (R) for any bounded measurable function g : R → R. Dense subspaces of L1 (R) are very useful in proving results about L1 -functions. As an application of exercise 4.59 we have the following: (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 Z Z Z Z 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 lim kfh − f k1 = 0. h→0
(Hint : Use exercise 4.60 to get a function g ∈ Cc (R) with kg − f k1 < ǫ, and note that kgh − gk1 < 2(b − a + 1) when supp(g) ⊆ [a, b] and |h| is sufficiently small.) (iii) The function h 7−→ kfh k1 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) = 0 for every x ∈ R and p ∈ R fixed. Show that ∀ t 6= 0, Z f (x)g(tx)dx ≤ M kfp/t − f k1 , where M = sup |g(x)|/2. Hence deduce (using exercise 4.60) that Z lim f (x)g(tx)dx = 0. |t|→∞
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4. Integration
(In the special case when f ∈ L1 [0, 2π] and g(x) = cos x or sin x, we have Z 2π Z 2π lim f (x) cos nxdx = 0 = lim f (x) sin nxdx. n→∞ 0
n→∞ 0
This finds applications in the theory of Fourier series.)
Chapter 5
Measure and integration on product spaces
5.1. Introduction Concepts and examples: The intuitive notion of length, originally defined for intervals in R, leads one to the class of Lebesgue measurable sets which included not only the intervals and all the topologically nice subsets of R, but also BR – the σ-algebra of Borel subsets of R. In a similar manner, one would like to extend the notion of area in R2 (volume in R3 , and so on) to a larger class of subsets which includes 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), one would like to define a measure η on the σ-algebra generated by sets of the form {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, the Lebesgue measure, then for intervals I and J, η(I ×J) = λ(I)λ(J) is the area of the rectangle with sides the intervals I and J. Thus η will automatically be an extension of the notion of area in R2 . We note that the collection R := {A × B | A ∈ A, B ∈ B} is only a semi-algebra of subsets of A × B in general. Let A ⊗ B denote the σ-algebra of subsets of X × Y generated by R.
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 called a measurable rectangle if E = A×B for some A ∈ A and B ∈ B. We denote by R the class of all measurable rectangles. The σ-algebra of subsets of X×Y generated by the semi-algebra R is called the product σ-algebra and 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 we have p−1 (A) ∈ A ⊗ B and p−1 (B) ∈ A ⊗ B. X Y (ii) The σ-algebra A ⊗ B is the smallest σ-algebra of subsets of X × Y such that (i) holds. 5.1.3. Proposition: Let X and Y be nonempty sets and let C, D be families of subsets of X and Y, respectively. Let C × D := {C × D | C ∈ C, D ∈ D}. Then the following hold: (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
Then
Ci = X and
∞ [
Di = Y .
i=1
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 of subsets of X and Y , respectively, as in proposition 5.1.3. Is it true that S(C×D) = S(C)⊗S(D) in general? Check in the case when
65
5.2. Product of measure spaces
C = {∅} and D is a σ-algebra of subsets of Y containing at least four elements. (5.4) Let BR2 denote the σ-algebra of Borel subsets of R2 , i.e., the σalgebra generated by the open subsets of R2 . Show that BR2 = BR ⊗ BR . (Hint: Use proposition 5.1.3.)
5.2. Product of measure spaces Concepts and examples: For the rest of the chapter, let (X, A, µ) and (Y, B, ν) be fixed measure spaces. Subsets of X × Y of the form A × B, A ∈ A, B ∈ B, are called measurable rectangles. As before, let R denote the class of all measurable rectangles. The σ-algebra of subsets of X × Y generated by R, denoted by A ⊗ B, is called the product σ-algebra. The problem we want to analyze in this section is the following: how can 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, then there 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 product of 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 × ν is 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?
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5. Measure and integration on product spaces
5.2.3. Definition: Let E ⊆ X × Y, x ∈ X and y ∈ Y . Let Ex := {y ∈ Y | (x, y) ∈ E}
and
E y := {x ∈ X | (x, y) ∈ E}.
The set Ex is called the section of E at x (or x-section of E) and the set E y 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 and Ex = ∅ if x 6∈ A. Similarly, E y = A if y ∈ B and E y = ∅ 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 = Similarly,
[0, 1) {0}
if x ∈ A, if x ∈ 6 A.
X if y = 0, y A if y ∈ (0, 1), E = ∅ 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 : S S S S (i) ( i∈I Ei )x = i∈I (Ei )x and ( i∈I Ei )y = i∈I (Ei )y . T T T T (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 = E y \ F y . (iv) If E ⊆ F, then Ex ⊆ Fx and E y ⊆ F y .
5.2.6. Theorem: Let E ∈ A ⊗ B. Then the following hold: (i) Ex ∈ B and E y ∈ A for every x ∈ X, y ∈ Y. (ii) The functions x 7−→ ν(Ex ) and y 7−→ µ(E y ) are measurable functions on X and Y, respectively. (iii) Z
X
ν(Ex )dµ(x) = (µ × ν)(E) =
Z
Y
µ(E y )dν(y).
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5.2. Product of measure spaces
5.2.7. Remark: Given σ-finite measure spaces (X, A, µ) and (Y, B, ν), we showed that µ×ν is the unique measure on A⊗B such that (µ×ν)(A×B) = µ(A)ν(B). Note that the measure space (X ×Y, A⊗B, µ×ν) need not be a complete measure space even 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 measure space (X × Y, A ⊗ B, µ × ν), where A ⊗ B is the σ-algebra of η ∗ -measurable subsets of X × Y , η being as in theorem 5.2.1, and µ × ν is the restriction of η ∗ to A ⊗ B. The measure space (X × Y, A ⊗ B, µ × ν) is nothing but the completion of the measure space (X × Y, A ⊗ B, µ × ν). It is easy to see that 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 nonnegative function 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 increasing sequence of sets in A⊗BR with ∞ [
E∗ (fn ) = E∗ (f ).
n=1
(ii) Using (i) and the ‘simple function technique’, show that E∗ (f ) ∈ A ⊗ BR , whenever f : X −→ R is a nonnegative measurable function. Deduce that E ∗ (f ) ∈ A ⊗ BR . T ∞ (Hint: E ∗ (f ) = n=1 E∗ (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} ∞ [ {(x, t) ∈ X × R | (x, t/n + c) ∈ E ∗ (f ), t > 0}, = n=1
show that A ∈ A ⊗ BR , and deduce that f is measurable.
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5. Measure and integration on product spaces
(iv) Let f : X −→ R be any measurable function. Show that G(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 Z ∗ (µ × ν)(E |f |) = |f |dµ = (µ × ν)(E∗ |f |).
(5.1)
(Hint: First prove this for the case when f is bounded. For the general 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. R The identity (5.1) shows that for nonnegative functions f dµ represents 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}. Suppose D ∈ 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 every x, y ∈ X, Dx 6= Dy if x 6= y and Dx ∈ B ∀ x ∈ X. Deduce that 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(E y ) are measurable and Z Z µ(E y )dν(y). ν(Ex )dµ(x) = (µ × ν)(E) = X
Y
(Hint: E = F ∪ N, F ∈ A ⊗ B and (µ × ν)∗ (N ) = 0 by theorem 3.4.7.)
(5.8) Let E ∈ A⊗B be such that µ(E y ) = 0 for a.e. (ν)y ∈ Y . Show that µ(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 Concepts and examples: 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: for every E ∈ A ⊗ B, Z
X×Y
Z Z
χE (x, y)d(µ × ν)(x, y) = χE (x, y)dν(y) dµ(x) Y X Z Z = χE (x, y)dµ(x) dν(y). Y
X
This allows us to compute the integral of the function χE (x, y) by integrating one variable at a time. So, the natural question arises: does the above hold when χE is replaced by a nonnegative measurable function on X × Y ? The answer 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 ) and y 7−→ f (x0 , y) are measurable on X and Y, respectively. R R (ii) The functions y 7−→ X f (x, y)dµ(x) and x 7−→ Y f (x, y)dν(y) are well-defined nonnegative measurable functions on Y and X, respectively. Z Z Z Z (iii) f (x, y)dν(y) dµ(x) = f (x, y)dµ(x) dν(y) Y X Z X Y = f (x, y)d(µ × ν)(x, y). X×Y
In view of theorem 5.3.1, it is natural to expect a similar result for f ∈ L1 (µ × ν), µ and ν being σ-finite measures. This is given by the following 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 for a.e. y(ν) and for a.e. x(µ), respectively.
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5. Measure and integration on product spaces
(ii) The functions Z Z y 7−→ f (x, y)dµ(x) and x 7−→ f (x, y)dν(y) X
Y
are defined for a.e. y(ν) and a.e. x(µ) , and are ν, µ-integrable, respectively. Z Z Z (iii) f (x, y)dµ(x) dν(y) = f (x, y) d(µ × ν) X Y X×YZ Z = f (x, y)dν(y) dµ(x). X
Y
Theorem 5.3.1, exercise (5.9) and theorem 5.3.2 together give us the following theorem, which enables us to check the integrability of a function of 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 following statements: (i) f is nonnegative. (ii) f ∈ L1 (µ × ν). Z Z (iii) |f (x, y)|dν(y) dµ(x) < +∞. Y
X
(iv)
Z Z Y
X
|f (x, y)|dµ(x) dν(y) < +∞.
Then Z
X×Y
f (x, y)d(µ × ν) =
Z Z X
f (x, y)dν(y) dµ(x) Y Z Z f (x, y)dµ(x) dν(y), = Y
X
in the sense that all the integrals exist and are equal. We give next some examples which illustrate the necessity of the conditions 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 measure on B, i.e., ν(E) := number of elements in E if E is finite and ν(E) := +∞
5.3. Integration on product spaces: Fubini’s theorems
71
otherwise. Let D := {(x, y) | x = y}. Let, for n ≥ 1, Dn :=
n [
( [(j − 1)/n, j/n] × [(j − 1)/n, j/n] ).
j=1
Then D = ∩∞ n=1 Dn . Thus D ∈ A ⊗ B. In fact, D is a closed subset of [0, 1] × [0, 1]. Further, Z Z χD (x, y)dµ(x) = 0 ∀ y ∈ Y and χD (x, y)dν(y) = 1 ∀ x ∈ X. X
Hence Z Z Y
Y
X
Z Z χD (x, y)dν(y) dµ(x) = 1. χD (x, y)dµ(x) dν(y) = 0 and X
Y
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 measure on [0,1]. Let x2 − y 2 if (x, y) 6= (0, 0), f (x, y) := (x2 + y 2 )2 0 if x = y otherwise.
Noting that for fixed x, f (x, y) is a Riemann integrable function on [0, 1] and ∂ y = f (x, y), ∂y x2 + y 2 it is easy to see that Z 1Z 1 Z f (x, y)dµ(x) dν(y) = − 0
0
0
1Z 1 0
f (x, y)dν(y) dµ(x) = −π/4.
This does not contradict theorem 5.3.2, because f 6∈ L1 (X × Y ) as Z Z 1 Z 1 |f (x, y)|d(µ × ν) = |f (x, y)|dν(y) dµ(x) [0,1]×[0,1]
0
≥
Z
=
Z
0
0 1 Z x 1
0
=
Z
0
1
1 x
|f (x, y)|dν(y) dµ(x) 0 ! Z π/4
cos 2θdθ dµ(x)
0
1 dµ(x) = +∞. 2x
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5. Measure and integration on product spaces
*5.3.7. Example: Assume the continuum hypothesis and let < be the well-order on [0, 1] such that ∀ 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 see this, 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 Z χ (x, y) d(λ × λ)(x, y) [0,1]×[0,1]
E
=
Z
Z
Z
Z
[0,1]
=
[0,1]
[0,1]
[0,1]
!
χE (x, y)dλ(x) dλ(y) !
χE (x, y)dλ(y) dλ(x).
(5.2)
However, it is easy to see that for x, y ∈ [0, 1], Z Z χE (x, y)dλ(y) = 1 and χE (x, y)dλ(x) = 0. [0,1]
[0,1]
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 following statements are equivalent: (i) fZ ∈LZ1 (µ × ν) := L1 (X× Y, A ⊗ B, µ × ν). (ii)
(iii)
ZY ZX X
Y
|f (x, y)| dµ(x) dν(y) < +∞. |f (x, y)|dν(y) dµ(x) < +∞.
(5.10) Let (X, A, µ) and (Y, B, ν) be complete σ-finite measure spaces and let (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 valued A ⊗ 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).
5.3. Integration on product spaces: Fubini’s theorems
73
R (ii) The function yR7−→ X f (x, y)dµ(x) is B-measurable and the function x 7−→ Y f (x, y)dν(y) is A-measurable. Z Z Z f (x, y)dν(y) dµ(x) (iii) f (x, y)d(µ × ν) = X ZY Z X×Y f (x, y)dµ(x) dν(y). = X
Y
(5.11) Let X = Y = [−1, 1], A = B = B[−1,1] , and let µ = ν be the Lebesgue measure on [−1, 1]. Let ( xy if (x, y) 6= (0, 0), 2 (x + y 2 )2 f (x, y) := 0 otherwise. Z
1
−1
Show that Z 1 Z f (x, y)dν(y) dµ(x) = 0 = −1
Z
1
−1
Z
Can you conclude that Z 1 Z 1 f (x, y)dν(y) dµ(x) =
−1
−1
1 −1
f (x, y)dµ(x) dν(y).
f (x, y)d(µ × ν)(x, y)?
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 Z Z Z φ(x, y)d(µ × ν) = f dµ gdν . X×Y
X
Y
(5.13) Let f ∈ L1 (0, a) and let Z a g(x) := (f (t)/t)dλ(t),0 < x ≤ a. x
Show that g ∈ L1 (0, a), and compute
Z
a
g(x)dλ(x). 0
*(5.14) Let (X, A, µ), and (X, B, ν) be as in example 5.3.7. Define, for x, y ∈ [0, 1], 1 if x is rational, f (x, y) := 2y if y is irrational. Z
0
Compute Z 1 Z 1 f (x, y)dν(y) dµ(x) and 0
Is f in L1 (µ × ν)?
0
1 Z 1 0
f (x, y)dµ(x) dν(y).
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5. Measure and integration on product spaces
(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 Borel subsets of X. A function f : X −→ R is said to be Borel measurable 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 measurable. (iii) Let {fn }n≥1 be a sequence of Borel measurable functions on X such that f (x) := lim fn (x) exists ∀ x ∈ X. Show that f n→∞ 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 function x 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) is Borel 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-finite measure on (Y, B). For any E ∈ B and x ∈ X, let Z η(x, E) := f (x, y)dµ(y). E
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 function. (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) above is called a transition measure.
5.4. Lebesgue measure on R2 and its properties Concepts and examples: We now specialize the construction of (X × Y, A ⊗ B, µ × ν) to the case when X = Y = R, A = B = LR and µ = ν = λ, the Lebesgue measure. We have 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, denoted by (R2 , LR2 , λR2 ), is called the Lebesgue measure space. Elements of LR2 are called the Lebesgue measurable subsets of R2 , and λR2 is called the 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 let ˜ Then the following hold: I˜ 2 := {I × J | I, J ∈ I}. (i) I˜ 2 is a semi-algebra of subsets of R2 , and S(I˜ 2 ) = BR2 . ˜ (ii) λR2 (I × J) = λ(I)λ(J), ∀ I, J ∈ I. (iii) The measure space (R2 , LR2 , λR2 ) is the completion of the measure spaces (R2 , LR ⊗ LR , λ × λ) and (R2 , BR2 , λR2 ). We describe next some properties of the Lebesgue measure λR2 . For E ⊆ 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 , Z Z Z f (x + y) dλR2 (x) = f (x) dλR2 (x) = f (−x) dλR2 (x).
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5. Measure and integration on product spaces
(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)) = | det T | λR2 (E). 5.4.4. Theorem (Integration of ‘radial’ functions): Let f : [0, ∞) −→ (0, ∞) be a nonnegative measurable function. Then Z Z ∞ f (|x|)dλR2 (x) = 2π f (r)rdλ(r). R2
0
Proof: The proof is once again an application of the ‘simple function technique’. 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 that either the En ’s are pairwise disjoint, or the En ’s are increasing. If the theorem holds for each χEn , then the theorem holds for χE also, where S E= ∞ n=1 En .
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 sets Kn 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 Z Z 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, Z Z 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 has dimension 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 proof of theorem 5.4.3 for sets E ∈ LR2 . (5.14) Consider the vectors (a1 , b1 ), (a2 , b2 ) ∈ R2 and let P := {(α1 a1 + α2 a2 , α1 b1 + α2 b2 ) ∈ R2 | α1 , α2 ∈ R, 0 ≤ αi ≤ 1}, called the parallelogram determined by these vectors. Show that λR2 (P ) = |a1 b2 − a2 b1 |. (5.15) Let E ∈ LR2 with 0 < λ(E) < ∞. Show that there exists a unique point (c1 , c2 ) ∈ R2 such that Z xχE (x + c1 , y + c2 )dλR2 (x, y) = 0 and
Z
yχE (x + c1 , y + c2 )dλR2 (x, y) = 0.
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5. Measure and integration on product spaces
In fact, c1 = and
1 λ(E)
Z
xχE (x, y)dλR2 (x, y)
Z 1 yχE (x, y)dλR2 (x, y). c2 = λ(E) (c = (c1 , c2 ) is called the centroid of E.) (5.16) Let f : [0, ∞) −→ R be a measurable function. Show that, if either of the two integrals in theorem 5.4.4 exists, then so does the other and the equality holds. (5.17) Let f : (a, b) × (c, d) −→ R be such that (i) f is continuous; ∂f (ii) exists and is continuous; ∂x d (iii) for some t ∈ (a, b), f (t, y) exists ∀ y; dy ∂ ∂f (iv) exists and is continuous. ∂y ∂x ∂ ∂f ∂f and exist. Moreover, Show that ∂y ∂x ∂y ∂ ∂f ∂ ∂f = . ∂x ∂y ∂y ∂x
(Hint: Use Fubini’s theorem and the fundamental theorem of calculus to show that ∀ ξ ∈ (a, b), η ∈ (c, d) and s < y Z η Z ξ ∂t ∂ f (ξ, η) − f (t, η) − f (ξ, s) + f (t, s) = dx dy, ∂x s t ∂y and note that the inner integral on the right is a continuous function of 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 sequences of measurable functions. Given a sequence {fn }n≥1 of measurable functions on (X, S, µ), we have already come across concepts like pointwise convergence of {fn }n≥1 to a measurable function f, i.e., when {fn (x)} converges to f (x) ∀ x ∈ X. When a sequence does not converge pointwise, one would like 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, is the main aim of this chapter. But before we do that, we extend the notion of measurability and integration to complex-valued functions.
6.1. Integration of complex-valued functions Concepts and examples: Let (X, S, µ) be a measure space and let C denote the field of complex numbers. For a function f : X −→ C, consider the functions Re (f ) and Im (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 and the imaginary part of the function f. Note that Re (f ) and Im (f ) are real-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 both Re (f ) and Im (f ) are measurable functions. We say f is µ-integrable if both Re (f ) and Im R (f ) are integrable. In that case we define the integral of f, denoted by f dµ, to be Z Z Z f dµ := Re (f )dµ + i Im (f )dµ. We denote the set of all complex-valued µ-integrable functions on X by L1 (X, S, µ) itself. Whenever we restrict ourselves to only real-valued µ-integrable functions on X, we shall specify it by Lr1 (X, S, µ). Our next theorem tells us that L1 (X, S, µ) is as nice a space as Lr1 (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 Z Z Z (αf + βg)dµ = α f dµ + β gdµ.
(ii) Let f : X −→ C be a measurable function. Then f ∈ L1 (X, S, µ) iff |f | ∈ Lr1 (X, S, µ). Further, in either case, Z Z f dµ ≤ |f |dµ.
(iii) Let f ∈ L1 (X, S, µ) and E ∈ S. Then χE f ∈ L1 (X, S, µ). We write Z Z f dµ := χE f dµ. E
If E1 , E2 ∈ S and E1 ∩ E2 = ∅, then Z Z Z f dµ = f dµ + E1 ∪E2
E1
f dµ.
E2
(iv) Let f ∈ L1 (X, S, µ), and let {En }n≥1 P be aR sequence of pairwise is absolutely disjoint sets from S. Then the series ∞ n=1 ( En f dµ) S∞ convergent. Also, (χE f ) ∈ L1 (X, S, µ), where E := n=1 En , and Z ∞ Z X f dµ. f dµ = E
n=1 En
That the integral of complex-valued functions behaves as nicely with respect to limiting operations as the integral of real-valued functions is the content of the next theorem.
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6.1. Integration of complex-valued functions
6.1.3. Theorem (Complex form of Lebesgue’s dominated convergence): Let {fn }n≥1 be a sequence in L1 (X, S, µ). Let f (x) := lim fn (x) exist for n→∞
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 Z Z lim fn dµ = f dµ. n→∞
6.1.4. Theorem: Let f ∈ L1 (X, S, µ), where µ(X) < ∞. Let S be a closed subset of C such that ∀ E ∈ S with µ(E) > 0, Z 1 f dµ ∈ S. µ(E) E Then f (x) ∈ S for a.e. x ∈ X. Exercises: (6.1 ) Let f ∈ L1 (X, S, µ) be such that Z Z 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 every Borel set E ⊆ X. (6.3) Extend the claim of exercise (4.30) to complex-valued measurable functions. (6.4) Let {fn }n≥1 be a sequence of S-measurable complex-valued functions on (X, S, µ). Show that the following statements are equivalent: P r (i) P ( ∞ n=1 |fn |) ∈ L1 (X, S, µ). ∞ R (ii) n=1 |fn |dµ < +∞. P Further, if either of the above is true, then ( ∞ n=1 fn ) ∈ L1 (X, S, µ) and ! Z ∞ Z ∞ X X fn dµ. fn dµ = n=1
n=1
(6.5) State and prove the Riesz-Fischer theorem (theorem 4.6.3) for complexvalued integrable functions.
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6. Lp -spaces
(6.6) Let f ∈ L1 (X, S, µ) and Z f dµ = 0 for every E ∈ S. E
Show that f (x) = 0 for a.e. x(µ) (see proposition 5.4.6).
6.2. Lp -spaces Concepts and examples: In section 4.6, we analyzed L1 [a, b], the space of Lebesgue integrable functions on [a, b]. We saw that L1 [a, b] is a vector space over R, and that for f ∈ L1 [a, b], we can define the notion of absolute value of f, i.e., kf k1 . Further, this notion of absolute value can be used to define the L1 -metric on L1 [a, b] so that L1 [a, b] becomes a complete metric space. In this section we 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-finite and complete. 6.2.1. Definition: Let 0 < p < ∞. Let Lp (µ) := Lp (X, S, µ) denote the space of all complexvalued S-measurable functions on X such that Z |f |p dµ < +∞. 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 kf kp , called the pth -norm of f, as follows: 1/p Z p . |f | dµ kf kp := Since kf kp = kgkp if f (x) = g(x) for a.e. x(µ), we treat such f and g as the same element of Lp (X, B, µ). To show that kf kp has the properties of a metric (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 inequalities hold: (i) at b1−t ≤ ta + (1 − t)b.
6.2. Lp -spaces
(ii) (
83
1 a + b 1/t ) ≤ (a1/t + b1/t ). 2 2
6.2.5. Theorem (H¨ older’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 f g ∈ L1 (µ) and 1/q 1/p Z Z Z q p . |g| dµ |f | dµ |f g|dµ ≤ 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 ∞ ∞ X X p |bn |q < ∞. |an | < ∞ and n=1
n=1
Then
P∞
n=1 |an bn |
∞ X
< +∞ and !
n=1
|an bn |
≤
∞ X
n=1
|an |
p
!1/p
In particular, if an = bn = 0 ∀ n ≥ k, then !1/p k k X X |an bn | ≤ |an |p n=1
n=1
∞ X
|bn |
n=1
k X
|bn |q
n=1
q
!1/q
!1/q
.
.
6.2.7. Note: In the special case when p = q = 2, H¨ older’s inequality is known as the Cauchy-Schwarz inequality. 6.2.8. Theorem (Minkowski’s inequality): Let 1 ≤ p < ∞ and f, g ∈ Lp (µ). Then f + g ∈ Lp (µ) and kf + gkp ≤ kf kp + kgkp .
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) := kf − gkp ∀ 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 metric space under the metric
dp (f, g) := kf − gkp :=
Z
1/p |f − g| dµ . p
It is natural to ask the same question for 0 < p < 1. If 0 < p < 1, define for f, g ∈ Lp (µ) Z dp (f, g) := |f − g|p dµ. 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 theorem 6.2.9, we can show that Lp (µ) is a complete metric space for 0 < p < 1. In the case 1 ≤ p < ∞, f 7−→ kf kp is a real-valued map on Lp (µ) with the properties: ∀ f, g ∈ Lp (µ) and α ∈ C (i) kf kp ≥ 0, and kf kp = 0 iff f = 0. (ii) kαf kp = |α|kf kp . (iii) kf + gkp ≤ kf kp + kgkp , 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 for 1 ≤ p < ∞, the Lp -spaces are examples of normed linear spaces which are also complete under the metric kf − gkp , the metric induced by the norm. Such normed linear spaces are called Banach spaces. However, when 0 < p < 1, f 7−→ kf kp is no longer a norm, as it fails to satisfy the triangle 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 be positive real numbers. Then
kαχA kp =
Z
1/p |α| χA (x)dµ(x) = α(µ(A))1/p . p
6.2. Lp -spaces
85
Similarly, kβχB kp = β(µ(B))1/p . Further, kαχA + βχB kp =
Z
=
Z
p
(αχA + βχB ) dµ p
1/p
p
(α χA + β χB ) dµ
1/p
= (αp µ(A) + β p µ(B))1/p .
Now, using exercise 6.7(ii) with t = (β/α)p (µ(B)/µ(A)), we have 1/p p µ(B) µ(B) 1/p β β > 1+ , 1+ α µ(A) α µ(A) i.e., (α1/p µ(A) + β 1/p µ(B))1/p > α(µ(A))1/p + β(µ(B))1/p , i.e., kαχA + βχB kp > kαχA kp + kβχB kp . This is the reason that Lp -spaces for 0 < p < 1 are not very interesting to study. 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. Let f, g be positive functions such that f ∈ Lp (µ) and g ∈ Lq (µ). Show that Z kf kp kgkq < f gdµ. (Hint: Apply H¨ older’s inequality to (f g)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 H¨ older’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}) ≤ kf kp /ǫp . This is called Chebyshev’s inequality for Lp -functions.
86
6. Lp -spaces
6.3. L∞ (X, S, µ) Concepts and examples: 6.3.1. Definition: A measurable function f : X −→ R∗ or C is said to be essentially bounded if 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 kf k∞ := inf {M | µ{x ∈ X | |f (x)| > M } = 0}. We call kf k∞ 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 kαf k∞ = |α| kf k∞ . (ii) (f + g) ∈ L∞ (X, B, µ) and kf + gk∞ ≤ kf k∞ + kgk∞ . (iii) kf k∞ = 0 iff f (x) = 0 a.e. x(µ). (iv) There exists a set E ∈ S such that µ(E) = 0 and kf k∞ = sup |f (x)|. x∈X\E
Thus µ({x ∈ X| |f (x)| > kf k∞ }) = 0. (v) kf k∞ = sup{N | µ({x ∈ X| |f (x)| > N }) > 0}. 6.3.3. Corollary: For f, g ∈ L∞ (X, S, µ), let d∞ (f, g) := kf − gk∞ . 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 as n → ∞ 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 classes by L∞ (X, B, µ) itself. For any element f ∈ L∞ (X, S, µ), which is in fact an equivalence class, we can define kf k∞ by choosing any element in the equivalence class f. Then kf k∞ is well-defined, and L∞ (X, S, µ) is a Banach space 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 kf kp ≤ kf k∞ (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 ≤ kf kp . (iii) Let {pn }n≥1 be any monotonically increasing sequence of real numbers such that pn ≥ 1 ∀ n and {pn }n≥1 is not bounded above. Using (i), (ii) above and theorem 6.3.2(v), show that kf k∞ = lim kf kpn . n→∞
6.4. L2 (X, S, µ) Concepts and examples: 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 linear spaces and have a notion of distance (norm) under which they are complete. Let us recall, for f ∈ L2 (X, S, µ), that 1/2 Z 2 . |f (x)| dµ(x) kf k2 :=
If we treat f ∈ L2 (X, S, µ) as a ‘vector’ with uncountable components, f (x) being the xth component, x ∈ X, then kf k2 can be viewed as a generalization of the Euclidean magnitude of a vector in Rn with summation being replaced by integration. We recall that on Rn , we have the notion of dot product of vectors, which is related to the magnitude of vectors and helps us to define the notion of angles on Rn . That such a dot product can also be defined on L2 (X, S, µ) and enables one to do geometry on L2 (X, S, µ), is the reason we discuss this space in detail.
88
6. Lp -spaces
6.4.1. Definition: For f, g ∈ L2 (X, S, µ), we define hf, gi :=
Z
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), hf, gi is a well-defined scalar and |hf, gi| ≤ kf k2 kgk2 .
The scalar hf, gi 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) hf, f i ≥ 0, and equality holds iff f = 0. (ii) hf, gi = hg, f i. (iii) hαf + βg, hi = αhf, hi + βhg, hi. (iv) hf, αg + βhi = αhf, gi + βhf, hi. (v) kf k2 = hf, f i1/2 . 6.4.4.Note: In the Cauchy-Schwarz inequality, the equality holds iff f and g are linearly dependent. 6.4.5. Note: Given an arbitrary vector space H over the field R (or C), if there is a map h· , ·i : H × H −→ R (or C) having the properties (i) to (iv) as given in proposition 6.4.3, then it is called an inner product space. On every inner product space H, it is easy to show that kuk := hu, ui1/2 , u ∈ H, is indeed a norm on H, called the norm induced by the inner product. Further, the Cauchy-Schwarz inequality holds: |hu, vi| ≤ kuk kvk, ∀u, v ∈ H. This can be proved as follows: For u ∈ H, let kuk = (hu, ui)1/2 . If either u = 0 or v = 0, then clearly |hu, vi| = 0 = kuk kvk. So, let u, v ∈ H be ′ such that u 6= 0 and v 6= 0. Then kuk > 0 and kvk > 0. Let u = u/kuk and
6.4. L2 (X, S, µ) ′
89
′
′
′
′
v = v/kvk. Then hu , u i = 1 = hv , v i and ′
′
′
′
′
′
′
′
0 ≤ hu − hu , v iv , u − hu , v iv i ′
′
′
′
′
′
′
′
= hu , u i + |hu , v i|2 hv , v i − 2|hu , v i|2 ′
′
= 1 − |hu , v i|2 . Hence ′
′
′
′
|hu , v i| ≤ 1 = ku kkv k, i.e., |hu, vi| ≤ kukkvk. If H is also complete under the norm induced by the inner product, then H is called a Hilbert space. Thus L2 (X, S, µ) is an example of a Hilbert space. We shall not go into the general theory of Hilbert spaces. We discuss some results for L2 (X, S, µ) which are in fact true, without any change in the 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 write f ⊥ g, if hf, gi = 0. For a subset S of L2 (X, B, µ), we write f ⊥ S if hf, hi = 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 k·k2 metric, i.e., for every sequence {fn }n≥1 in S with lim kfn − f k2 = 0 for some f ∈ L2 (X, S, µ), we have f ∈ S.
n→∞
6.4.8. Proposition: Let S be any nonempty subset of L2 (X, S, µ) and let S ⊥ := {g ∈ L2 (X, S, µ) | hf, gi = 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 {kf − gk2 | g ∈ S}. Then there exists a unique function f0 ∈ S such that α = kf −f0 k2 . 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 application says 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) S1⊥ is a closed subspace of L2 (X, S, µ) and S1 ∩ S1⊥ ⊆ {0}. If S1 is also a subspace, then S1 ∩ S1⊥ = {0}. (ii) S1⊥ ⊆ S2⊥ if S2 ⊆ S1 . (iii) S1 ⊆ (S1⊥ )⊥ , 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 ∩ S1⊥ = {0} and L2 (X, S, µ) = S1 + S1⊥ . Thus every f ∈ L2 (X, S, µ) can be uniquely expressed as f = g + h, where g ∈ S1 and h ∈ S1⊥ . (This is also expressed as L2 (X, S, µ) = S1 ⊕ S1⊥ , and is called the projection theorem.) As our second application of theorem 6.4.9, we characterize all bounded linear 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 it has 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 kf k2 , ∀ 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 kf k2 , ∀ 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 ) = hf, gi =
Z
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 inequality, |Tg (f )| = |hf, gi| ≤ kgk2 kf k2 . 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 every bounded 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 ) = hf, g0 i ∀ f ∈ L2 (X, S, µ).
6.4.16. Note: In theorem 6.4.15, we showed that, given a closed subspace M of L2 (X, S, µ), every element f ∈ L2 (X, S, µ) can be written uniquely as f = g + h, where g ∈ M and h ∈ M ⊥ . Let us denote g by PM (f ), called the projection of f onto 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 the next theorem. 6.4.17. Theorem: PM : L2 (X, S, µ) → L2 (X, S, µ), as defined in note 6.4.16 above, has the following properties: ∀ f, g ∈ L2 (X, S, µ) (i) PM is linear and PM (f ) = f whenever f ∈ M . (ii) f − PM (f ) ∈ M ⊥ . (iii) hPM (f ), gi = hf, PM (g)i = hPM (f ), PM (g)i. (iv) PM (PM (f )) = f. (v) kf k22 = kPM (f )k22 + kf − PM (f )k22 . (vi) PM is continuous.
Exercises: (6.12) Let f ∈ L2 (X, S, µ) be such that f ⊥ g ∀ g ∈ L2 (X, S, µ). What can you conclude about f ? (6.13) Pythagoras identity: Let f, g ∈ L2 (X, S, µ) and f ⊥ g. Show that kf + gk22 = kf k22 + kgk22 . (6.14) (Bessel’s inequality):Let f1 , f2 , . . . , fn be elements of L2 (X, S, µ) such that kfi k2 = 1 ∀ i and fi ⊥ fj for i 6= j. Show that n X
|hf, fj i|2 ≤ kf k22 .
j=1
(Hint: Consider hz, zi, where z = f −
Pn
j=1 hf,
fj ifj .)
(6.15) Let {fn }n≥1 , {gn }n≥1 be sequences in L2 (X, S, µ) such that lim kfn → n→∞
f k2 = 0, lim kgn → gk2 = 0 for f, g ∈ L2 (X, S, µ). Show that n→∞
hfn , hi → hf, hi, hh, gn i → hh, gi, hfn , gn i → hf, gi for every h ∈ L2 (X, S, µ). In other words, the inner product map h· , ·i : L2 × L2 −→ C is continuous in each variable and jointly. (6.16) (Parallelogram identity): Let f, g ∈ L2 (X, S, µ). Then kf + gk22 + kf − gk22 = 2kf k22 + 2kgk22 .
6.5. L2 -convergence of Fourier series
93
6.5. L2 -convergence of Fourier series Concepts and examples: The problem of investigating the convergence of Fourier series motivated mathematicians to look for an extension of the integral concept. It is natural 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 one of the pioneering applications of the Lebesgue integral lies in the study of Fourier series. We shall not go into the general theory of Fourier series, but only prove an important result (the Riesz-Fischer theorem) which has applications in many branches of mathematics, physics and electrical engineering. For the general theory of Fourier series, one can consult Bhatia [4] and other references given there. 6.5.1. Definition: For f ∈ Lr1 [−π, π], let 1 an := π
Z
π
f (x) cos nx dλ(x), n = 0, 1, 2, . . . ,
−π
and 1 bn := π
Z
π
f (x) sin nx dλ(x), n = 1, 2, . . . .
−π
The scalars an , bn are called the Fourier coefficients of f , and the series ∞ X a0 + (an cos nx + bn sin nx) 2 n=1
is called the Fourier series of f. Let sn (x) :=
n X a0 + (ak cos kx + bn sin kx), x ∈ [−π, π]. 2 k=1
The function sn (x) is called the nth -partial sum of the Fourier series of f at x. 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 the pointwise convergence problem of Fourier series. For f ∈ L1 [−π, π] and x ∈ [−π, π], if lim sn (x) = f (x), we say that f has pointwise representation n→∞ by its Fourier series at x. The answer to the pointwise convergence problem is not easy, and, given the nature and scope of this text, we shall not go into it. For details one can consult any one of the texts listed above. We state below an important relation between functions and their Fourier coefficients.
94
6. Lp -spaces
6.5.2. Theorem (Uniqueness of the Fourier series): If f ∈ Lr1 [−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 ∈ Lr1 [−π, π] can have a pointwise representation one has to put extra conditions on f to get a pointwise representation. Secondly, f ∈ Lr1 [−π, π] need be defined only a.e. Thus pointwise representation makes sense only a.e. Finally, from the point of view of many applications, pointwise representations have very little utility. Such consideration motivated mathematicians to look for some other methods of analyzing the convergence problem. We consider below one such method, namely the convergence of Fourier series in the L2 -metric. For other methods we refer to the 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 previous section , if we write φk (x) := cos kx and ψk (x) := sin kx, then ak = hf, φk i/π and bk = hf, ψk i/π. 6.5.3. Lemma: For nonnegative integers n and m, the following relations hold: hψn , φm i = 0; hψn , ψm i = 0 if n 6= m and = π if n = m; hφn , φm i = 0 if n 6= m and = π if n = m.
6.5.4. Theorem (Bessel’s inequality): For f ∈ Lr2 [−π, π], ∞ X 1 |a0 |2 + (|an |2 + |bn |2 ) ≤ kf k22 . 2 π n=1
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 ∞ X |a0 |2 + (|an |2 + |bn |2 ) < +∞. 2 n=1
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 ∈ Lr2 [−π, π]. Then the Fourier series of f converges to f in the L2 norm, i.e., ksn − f k2 −→ 0 as n −→ ∞. Further, ∞ X 1 |a0 |2 (|an |2 + |bn |2 ). kf k22 = + π 2 n=1
6.5.7 Note: A careful observation of the above arguments will tell the reader that Fourier coefficients can be defined for Riemann integrable functions also (as was done historically first), and Bessel’s inequality remains valid. However, RieszFischer critically uses the fact that Lr2 [−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 ∞ X ℓ=0
1 π2 = . (2ℓ + 1)2 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 minus infinity. We extend the algebraic operations and the order relation of R to R∗ 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 = −∞ x(+∞) = (+∞)x = −∞ x(−∞) = (−∞)x = +∞
if x > 0,
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 is called the set of extended real numbers. The symbol +∞ is also denoted by ∞, 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 bounded above, 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 not bounded above, we say {xn }n≥1 is convergent to +∞ and write lim xn = +∞.
n→∞
Similarly, if {xn }n≥1 is a monotonically decreasing sequence which is not bounded below, we say {xn }n≥1 is convergent to −∞ and write lim xn = −∞.
n→∞
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 {inf k≥j xk }j≥1 always converge. We write lim sup xn := lim (sup xk ) n→∞
j→∞ k≥j
and lim inf xn := lim (inf xk ). n→∞
j→∞ k≥j
limn→∞ sup xn is called the limit superior of the sequence {xn }n≥1 and lim inf n→∞ xn is called the limit inferior of the sequence {xn }n≥1 . Note that lim inf xn ≤ lim sup xn . n→∞
n→∞
We say a sequence {xn }n≥1 is convergent to x ∈ R∗ if lim inf n→∞ xn = lim supn→∞ xn =: x, say. In that case we write limn→∞ xn := x. A.4. Series in R∗ P Let {xk }k≥1 be a sequence in R∗ suchPthat for every n ∈ N, sn := nk=1 xk is ∞ well-defined. We say that the series P 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
99
Extended real numbers
P∞ P the series ∞ k=1 xk k=1 xk . For example, if each xk ≥ 0, then the series is always convergent.
Let {xk }k≥1 be a sequence in R∗P , xn ≥ 0 for every n. Let σ : N → N be any bijective map. Then the series ∞ k=1 xσ(k) is called an arrangement P x . of the series ∞ k=1 k P∞ P • For every rearrangement σ, the series ∞ k=1 xk conk=1 xσ(k) and verge to the same sum. P Clearly, both the series are convergent in R∗ . Let α := ∞ k=1 xk and β := P∞ x . To prove α = β, in view of the symmetry, it is enough to prove k=1 σ(k) that β ≤ α. Now n m ∞ X X X β= xσ(k) ≤ xk ≤ xk = α, k=1
k=1
k=1
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 P∞ sequence of ∗ . For every fixed n, the series nonnegative elements of R m=1 xn,m is conP P∞ vergent in R∗ .PLet yn := ∞ x . Further, y is also convergent in m=1 n,m n=1 n P∞ ∞ ∗ R . Let y P := 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 the same sum, i.e., y = z, which is written as ! ! ∞ ∞ ∞ ∞ X X X X xn,m . xn,m = m=1
n=1
n=1
m=1
To see this, we note that ∀ r, s ∈ N, r X s s X ∞ ∞ X ∞ X X X xn,m ≤ xn,m ≤ xn,m . n=1 m=1
Hence
y :=
m=1 n=1
∞ ∞ X X
xn,m ≤
n=1 m=1
m=1 n=1
∞ ∞ X X
xn,m =: z.
m=1 n=1
The reverse inequality follows from symmetry.
A similar result holds for rearrangement of double-indexed series of nonnegative terms in R∗ : • Let σ : N → N×N be any bijective map, and let {xn,m }n≥1,m≥1 be a double-indexed sequence of nonnegative elements of R∗ . Then ∞ X r=1
xσ(r) =
∞ ∞ X X
n=1 m=0
xn,m =
∞ ∞ X X
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 and B, to be the empty set if either A or B or both are empty, and to be the set of 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 their Cartesian product to be the set A1 × · · · × An := {(x1 , x2 , . . . xn ) | xi ∈ Ai , i = 1, 2, . . . , n}. Obviously, A1 × · · · × An is a nonempty set. We S can think of A1 × · · · × An as the set of all functions f : {1, 2, . . . , n} −→ ni=1 Ai with f (i) = xi ∈ Ai for each i. One can copy this to define the Cartesian product of any arbitrary family of sets, say {Aα }α∈I , to be the set ) ( Y [ Aα := f : I → Aα f (α) ∈ Aα , ∀ α ∈ I . α∈I
α∈I
Q i.e.,we do not However, there is no surety that the set α∈I Aα is nonempty,S know that there always exists at least one function f : I → α∈I Aα such that f (α) ∈ Aα ∀ α ∈ I, although intuitively it seems obvious that such a function should always exist. However, this cannot be proved with the usual axioms of set theory. (For a short introduction to axiomatic set theory, see Rana [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 above dilemma 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 S each Aα is nonempty, then there exists a function f : I −→ α∈I Aα such that f (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 is the following: • If {Aα | α ∈ I} is a nonempty family of pairwise disjoint sets S such that 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 mathematics. (We have used it in section 4.6 to construct nonmeasurable subsets of R.)
Appendix C
Continuum hypothesis
Let X and Y be two sets. We say X and Y are equipotent iff there exists a bijection between them. We write this as X ≈ Y. In a sense, equipotent sets have same ‘number’ of elements. We say a set A is finite if A ≈ {1, 2, . . . , n} for some n ∈ N, and we say A is countable if A ≈ N, the set of natural numbers. A set which is not countable is called uncountable. For example, N, Z, Q are all countable sets while R is uncountable. (For a detailed discussion, see Rana [30].) The statement X ≈ Y means X and Y have the same ‘number of elements’, and can be made precise as follows (see Halmos [15] for details). Let C be a collection of sets such that any two members of C are equipotent to each 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 set A is also called the cardinality of A. For example, 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 any set 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 set X, 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-one map from A into B. We write card(A)
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 but A 6≈ N? The answer to this question is not known. The statement that the answer to the above question is in the negative is called the continuum hypothesis: • 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 of R is preceded by only countably many elements. We used this in section 3.4, to prove Ulam’s theorem. It is known that the continuum hypothesis is independent of the Zermelo-Fraenkel axioms of set theory.
References
[1]
Aliprantis, C.D. and Burkinshaw, O. Principles of Real Analysis (3rd Edition). Academic Press, Inc. New York, 1998.
[2]
Apostol, T.M. Mathematical Analysis. Narosa Publishing House, New Delhi (India), 1995.
[3]
Bhatia, Rajendra Fourier Series. Hindustan Book Agency, New Delhi (India), 1993.
[4]
Billingsley, Patrick Probability and Measure. 3rd Edition, John Wiley and Sons, New York, 1995.
[5]
Friedman, A. Foundations of Modern Analysis. Holt, Rinehart and Winston, 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. SpringerVerlag, Heidelberg, 1969.
[9]
Kolmogorov, A.N. Foundations of Probability Theory. Chelsea Publishing 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, Academic Press, 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, American 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 E y , 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 hf, gi, 88 R f dµ, 39, 80 R sdµ, 36 R f dµ, 40 RE E f dµ, 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
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
Index
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 H¨ older’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
109
Index
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
ultrafilter, 13 uncountable, 103 Uniform distribution, 10 Vitali’s example, 34
Index
