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

Transcription

1 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 , India Current address: Department of Mathematics, Indian Institute of Technology, Powai, Mumbai , India address:

2 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: or, An Introduction to Measure and Integration, Narosa Publishers, Delhi, 1997 An Introduction to Measure and Integration,Second Edition, Graduate Text in Mathematics, Volume 45, American Mathematical Society, May, 2004 Mumbai Inder K. Rana

3 Contents Chapter 1. Classes of sets Semi-algebra and algebra of sets Sigma algebra and monotone class 5 Chapter 2. Measure Set functions Countably additive set functions on intervals Set functions on algebras Uniqueness problem for measures 17 Chapter 3. Construction of measures Extension from semi-algebra to the generated algebra Extension from algebra to the generated σ-algebra Choosing nice sets: Measurable sets Completion of a measure space The Lebesgue measure 27 Chapter 4. Integration Integral of nonnegative simple measurable functions Integral of nonnegative measurable functions Intrinsic characterization of nonnegative measurable functions Integrable functions The Lebesgue integral and its relation with the Riemann integral 55 vii

4 viii Contents 4.6. L 1 [a, b] as the completion of R[a, b] 59 Chapter 5. Measure and integration on product spaces Introduction Product of measure spaces Integration on product spaces: Fubini s theorems Lebesgue measure on R 2 and its properties 75 Chapter 6. L p -spaces Integration of complex-valued functions L p -spaces L (X, S, µ) L 2 (X, S, µ) L 2 -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

5 Chapter 1 Classes of sets 1.1. Semi-algebra and algebra of sets Concepts and examples: 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 sets C 1, C 2,...,C n C such that C i C j = for i j and A c = n i=1 C i 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) A c F, whenever A F 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

6 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 Ĩ of all intervals of the form (a, b], (, b], (a, ), (, + ). We call Ĩ the collection of all left-open, right-closed intervals of R. It is easy to check that Ĩ is also a semi-algebra of subsets of R. (iii) The collection { F(I) := E R E = } n I k, I k I, I k I l = for k l, n N. k=1 is an algebra of subsets of R. So is the class } n {E F(Ĩ) := R E = I k, I k Ĩ, I k I l = for k l, n N. 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) A c 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 E 1, E 2,...,E n F then F 1, F 2,...,F n F such that F i E i for each i, F i F j = for i j and n i=1 E i = n j=1 F j. The next set of exercise describes some methods of constructing algebras and semi-algebras.

7 1.1. Semi-algebra and algebra of sets 3 (1.3) Let X be a nonempty set. Let 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 α } α I be a family of algebras of subsets of a set X. Let F := α I F α. Show that F is also an algebra of subsets of X. (1.7) Let {F n } n 1 be a sequence of algebras of subsets of a set X. Under what conditions on F n can you conclude that F := n=1 F n 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 exist sets C i C, i = 1, 2,..., such that C i C j = for i j and i=1 C i = A. Prove the following: (i) For any finite number of sets C, C 1, C 2,...,C n in C, C\( n i=1 C i) is a finite union of pairwise disjoint sets from C and hence is a σ-set. (ii) For any sequence {C n } n 1 of sets in C, n=1 C n 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 {E R E = n k=1 I k, I k I, I k I l = if 1 k l n}. (1.11) Let C be any semi-algebra of subsets of a set X. Show that F(C), the algebra generated by C, is given by {E X E = n i=1 C i, C i C and C i C j = for i j, n N}.

8 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. (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 = {x n } n 1 x n Y, n = 1, 2,...}. For any positive integer k let A Y k, the k-fold Cartesian product of Y with itself, and let i 1 < i 2 < < i k be positive integers. Let C(i 1, i 2,...,i k ; A) := {x = (x n ) n 1 X (x i1,...,x ik ) A}. We call C(i 1, i 2,...,i k ; 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 Then the following hold: (a) (b) Let C E := {C E C C}. C E F(C) E := {A E A F(C)}. Deduce that F(C E) F(C) E. 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).

9 1.2. Sigma algebra and monotone class Sigma algebra and monotone class Concepts and examples: 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) A c S whenever A S. (iii) i=1 A i S whenever A i S, i = 1, 2,.... Such a class S is called a sigma algebra (written as σ-algebra) of subsets of X Examples: (i) Let X be any set. Then {, X} and P(X) are obvious examples of σ- algebras of subsets of X. (ii) Let X be any uncountable set and let S = {A X A ora c is countable}. Then S is a σ-algebra of subsets of X. (iii) Let X be any set and let C be any class of subsets of X. Let S(C) := S, where the intersection is taken over all σ-algebras S of subsets of X 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 Definition: Let X be a nonempty set and M be a class of subsets of X. We say M is a monotone class if (i) n=1 A n M, whenever A n M and A n A n+1 for n = 1, 2,..., (ii) n=1 A n M, whenever A n M and A n A n+1 for n = 1, 2, Examples: (i) Clearly, every σ-algebra is also a monotone class.

10 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) is a 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 C 1, C 2,... in C such that E S({C 1, C 2,...}). 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

11 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 B X. (ii) Let X = R. Let I be the class of all intervals and Ĩ the class of all left-open right-closed intervals. Show that I S(U), I S(Ĩ), Ĩ S(I) and hence deduce that S(I) = S(Ĩ) = B R. (1.21) Prove the following statements: (i) Let I r denote the class of all open intervals of R with rational endpoints. Show that S(I r ) = B R. (ii) Let I d denote the class of all subintervals of [0, 1] with dyadic endpoints (i.e., points of the form m/2 n for some integers m and n). Show that S(I d ) = B R [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.

12 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).

13 Chapter 2 Measure 2.1. Set functions Concepts and examples: 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 µ A i = µ(a i ). i=1 whenever A 1, A 2,...,A n C are such that A i A j = for i j and n i=1 A i C. (iii) µ is said to be countably additive if ( ) µ A n = µ(a n ) n=1 i=1 n=1 whenever A 1, A 2,... in C with A i A j = for i j and n=1 A n C. 9

14 10 2. Measure (iv) µ is said to be countably subadditive if µ(a) µ(a n ). n=1 whenever A C, A = n=1 A n with A n 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: Example: Let X be any infinite set and let x n X, n = 1, 2,.... Let {p n } n 1 be a sequence of nonnegative real numbers. For any A X, define µ(a) := p i. {i x i 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 i at x i. The measure µ is finite (i.e., µ(x) < + ) iff i=1 p i < +. If i=1 p i = 1, the measure µ is called a discrete probability measure/distribution. Note that µ({x i }) = p i i and µ({x}) = 0 if x x i. So, one can regard µ as a set function defined on the subsets of the set Y := {x n : n 1}. Some of the special cases when X = {0, 1, 2,...} are: R. (a) Binomial distribution: Y := {0, 1, 2,...,n} and, for 0 < p < 1, ( ) n p k = p k (1 p) n k, 0 k n. k (b) Poisson distribution: Y := {0, 1, 2,...} and p k := λ k e λ /k! for k = 0, 1, 2,..., where λ > 0. (c) Uniform distribution: Y := {1, 2,...,n}, p k := 1/k k. An important example of a set function on the collection of intervals in 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

15 2.1. Set functions 11 (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). Property (3): Let I I be such that I = n i=1 J i, where J i J j = for i j. Then n λ(i) = λ(j i ). i=1 This property of λ is called the finite additivity of λ, or one says that λ is finitely additive. Property (4): Let I I be a finite interval such that I i=1 I i, where I i I. Then λ(i) λ(i i ). i=1 Property (5): Let I I be a finite interval such that I = n=1 I n, where I n I and I n I m = for n m. Then λ(i) = λ(i n ). n=1 Property (6): Let I I be any interval. Then λ(i) = λ(i [n, n + 1)). n= Property (7): Let I I be any interval such that I = n=1 I n, I n I and I n I m = for n m. Then λ(i) = λ(i n ). n=1

16 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 n=1 I n, I n I. Then λ(i) λ(i n ). 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 A c is finite}. Let µ : F(C) [0, ) be defined by { 0 if A is finite, µ(a) := 1 if A c 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, µ n (A) := #{m : 1 m n, m A}. n Show that µ n is countably additive for every n on P(X). In a sense, µ n is the proportion of integers between 1 to n which are in A. Let C = {A X lim n µ n(a) exists}. Show that C is closed under taking complements, finite disjoint unions and proper differences. Is it an algebra?

17 2.1. Set functions 13 (2.3) Let µ : Ĩ (0, 1] [0, ] be defined by { b a if a 0, 0 < a < b 1, µ(a, b] := + otherwise. (Recall that Ĩ (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) 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 A c 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 U. Show that µ is finitely additive. (2.5) Let A be an algebra of subsets of a set X. (i) Let µ 1, µ 2 be measures on A, and let α and β be nonnegative real numbers. Show that αµ 1 + βµ 2 is also a measure on A. (ii) For any two measures µ 1, µ 2 on A, we say µ 1 µ 2 if µ 1 (E) µ 2 (E), E A. Let {µ n } n 1 be a sequence of measures on A such that Define E A, µ n µ n+1, n 1. µ(e) := lim n µ n(e). Show that µ is also a measure on A and E B, µ(e) = sup {µ n (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.

18 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: Proposition: Let F : R R be a monotonically increasing function. Let µ F : Ĩ [0, ] be defined by µ F (a, b ] := F(b) F(a), µ F (, b ] := lim [F(b) F( x)], x µ F (a, ) := lim [F(x) F(a)], x µ F (, ) := lim [F(x) F( x)]. x Then, µ F is a well-defined finitely additive set function on Ĩ. Further, µ F is countably additive if F is right continuous. One calls µ F the set function induced by F. The converse of proposition is also true Proposition: Let µ : Ĩ [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, F(x) := 0 if x = 0, µ(x,0] if x < 0.) Remarks: (i) In case µ(r) < +, a more canonical choice for the required function F in proposition is given by F(x) := µ(, x], x R.

19 2.3. Set functions on algebras 15 (ii) Propositions and 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 µ F 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 µ F. (2.8) α R. Show that F 1 := F + α is also a distribution function and µ F = µ F1. 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 semi-algebra, then µ is countably subadditive iff A C with A i=1 A i, A i C implies µ(a) µ(a i ). i= 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 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.

20 16 2. Measure Another characterization of countable additivity of set functions defined on algebras is given in the next theorem 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 then the following hold: (i) For any A A, if A = n=1 A n, where A n A and A n A n+1 n, then Conversely, µ(a) = lim n µ(a n). This is called the continuity from below of µ at A. (ii) For any A A, if A = n=1 A n, where A n A with A n A n+1 n and µ(a n ) < + for some n, then lim µ(a n) = µ(a). n This is called the continuity from above of µ at A. (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(a k) = 0, whenever {A k } k 1 is a sequence in A with k A k A k+1 k, and k=1 A k =. (b) µ is countably additive. (2.11 ) Extend the claim of theorem 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 {E n } n 1 in A, show that

21 2.4. Uniqueness problem for measures 17 (i) µ(lim inf n E n ) lim inf n µ(e n ). (ii) µ(lim sup n E n ) lim sup n µ(e n ). (Hint: For a sequence {E n } n 1 of subsets of a set X, lim inf n E n := n=1 k=n E k lim supe n := n n=1 k=n E k. ) 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 µ 1 and µ 2 be two measures on S(A), the σ-algebra generated by the algebra A, such that µ 1 (A) = µ 2 (A) A A. Is µ 1 = µ 2? 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 sets X n C, n = 1, 2,..., such that µ(x n ) < + for every n and X = n=1 X n Examples: (i) The length function λ on the class of intervals is σ-finite. (ii) Let A = A(Ĩ), the algebra generated by left-open right-closed intervals in R. For A A, let µ(a) = + if A 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 A ξ denote the algebra of subsets of R generated by A and {ξ}. Define for A A ξ, µ ξ (A) := { + if A \ {ξ}, 0 if either A = or A = {ξ}. It is easy to check that µ ξ is also a measure on A ξ 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 µ c (E) = c times the number of points in E. Show that µ c is a measure on X S(A) = S(X A) and is not σ-finite.

22 18 2. Measure Proposition: Let µ 1 and µ 2 be totally-finite measures on a σ-algebra S. Then the class M = {E S µ 1 (E) = µ 2 (E)} has the following properties: (i) M is a monotone class. (ii) If S = S(A), and µ 1 (A) = µ 2 (A) A A, then µ 1 (A) = µ 2 (A) A S(A). Exercises: (2.13 ) Let A be an algebra of subsets of a set X. Let µ 1 and µ 2 be σ-finite measures on a σ-algebra S(A) such that µ 1 (A) = µ 2 (A) A A. Then, µ 1 (A) = µ 2 (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) < +.

23 Chapter 3 Construction of measures Concepts and examples: 3.1. Extension from semi-algebra to the generated algebra Definition: Let C i, i = 1, 2 be classes of subsets of a set X, with C 1 C 2. Let function µ 1 : C 1 [0, + ] and µ 2 : C 2 [0, + ] be set functions. The set function µ 2 is called an extension of µ 1 if µ 1 (E) = µ 2 (E) for every E Examples: In example 2.4.2(ii), each µ ξ is an extension of the measure µ. Similarly, in example 2.4.2(iii) each µ c 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 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

24 20 3. Construction of measures [Hint: For E F(C), with E = n i=1 E i for pairwise disjoint sets E 1,...,E n C, define µ(e) := µ(e i ).] 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 with A i=1 A i, A i C implies µ(a) µ(a i ). 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 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 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

25 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, + ), and inf(a) := + otherwise 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 { } µ (E) := inf µ(a i ) A i A, A i E. i=1 The set function µ is called the outer measure induced by µ 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., i=1 µ (A) µ (B) whenever A B X. (iii) µ is countably subadditive, i.e., µ (A) µ (A i ) whenever A = A i. i=1 (iv) µ is an extension of µ, i.e., µ (A) = µ(a) if A A 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 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 let I 1, I 2,...,I n be any finite collection of open intervals such that E n i=1 I i. Then it is easy to see that n i=1 λ(i i) 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. i=1

26 22 3. Construction of measures Example: Let A := {A R Either A or A c 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 A c 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 that µ is countably subadditive on P(R). If A R is countable, then clearly A A, and hence µ (A) = µ(a) = 0. Further, µ (A) = 1 iff A is uncountable. Since, R = (, 0] (0, ) and µ (R) = 1 < 2 = µ(, 0] + µ(0, ). This shows that µ need not be even finitely additive on all subsets. Exercises: (3.4) Show that µ (E), as in definition 3.2.1, is well-defined. (3.5) The set function µ (E) can take the value + for some sets E. (3.6) Show that { µ (E) = inf µ(a i ) A i A, A i A j = for i j and i=1 } A i E. (3.7) Let X be any nonempty set and let A be any algebra of subsets of X. Let x 0 X be fixed. For A A, define { 0 if x0 A, µ(a) := 1 if x 0 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 x 0 A. Can you conclude that µ (A) = 1 implies x 0 A? Show that this is possible if {x 0 } A Choosing nice sets: Measurable sets i=1 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.

27 3.3. Choosing nice sets: Measurable sets 23 But the problem is how to pick these nice sets? This motivates our next definition 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 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 ) < +, (iv) For every A A, µ (Y ) µ (Y E) + µ (Y E c ). µ(a) µ (A E) + µ (A E c ). We give an equivalent definition of measurable sets when µ(x) < 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 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.

28 24 3. Construction of measures (iii) If A n S, n = 1, 2,..., then n=1 A n S and µ restricted to S is countably additive. (iv) Let N := {E X µ (E) = 0}. Then N S. This together with proposition gives us the following: 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 Remark: Theorems (iv) and 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 (3.9) 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. (3.10) Let µ F be the measure on the algebra A(Ĩ) as given in proposition Let µ F itself denote the unique extension of µ F to L F, the σ-algebra of µ F-measurable sets, as given by theorem Show that (i) B R L F. (ii) µ F ({x}) = F(x) lim F(y). Deduce that the function F is y x continuous at x iff µ F ({x}) = 0. (iii) Let F be differentiable with bounded derivative. If A R is a null set, then µ F (A) = 0. The measure µ F is called the Lebesgue-Stieltjes measure induced by the distribution function F Completion of a measure space Concepts and examples: Theorem 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 σ- algebra 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).

29 3.4. Completion of a measure space 25 We first give an equivalent ways of describing µ (E) for any set E X, µ being the outer measure induced by µ. Let A σ denote the collection of sets of the form i=1 A i, A i A Proposition: For every set E X, µ (E) = inf {µ (A) A A σ, E A} = inf {µ (A) A S(A), E A} = inf {µ (A) A S, E A} 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 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 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 Definition: Let (X, S, µ) be a measure space and let N := {E X E N for some

30 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 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 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 F ǫ A such that µ (E F ǫ ) < ǫ 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 G 1, G 2 be two measurable covers of E. Show that µ (G 1 G 2 ) = 0. (3.12) Let E 1 E 2 E 3... be subsets of X. Then µ ( n=1 E n ) = lim n µ (E n ). (3.13 ) Let E X, and let K 1, K 2 be two measurable kernels of E. Show that µ (K 1 K 2 ) = 0.

31 3.5. The Lebesgue measure 27 (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 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, { } λ (E) := inf λ(i i ) I i I i, I i I j = for i j and E I i. 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 L R, or simply by L. The σ-algebra S(I) = S(A) := B R, generated by all intervals, is called the σ-algebra of Borel subsets of R. We denote the restriction of λ to L or B R by λ itself. The measure space (R, L, λ) is called the Lebesgue measure space and λ is called 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 B R is unique. It is natural to ask the question: What is the relation between the classes B R, L and P(R)? Thus, As a special case of theorem 3.4.6, we have L = B R N, where The question arises: N := {N R N E B R, λ(e) = 0}. B R L P(R). Is B R 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 i=1

32 28 3. Construction of measures is at least 2 c (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 2 c. On the other hand, B R is the σ-algebra generated by all open intervals of R with rational endpoints. One can show that the σ- algebra B R 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 2 c 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 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 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 B R includes all topologically nice subsets of R, such as open sets, closed sets and compact sets. Also, for E B R, 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 B R. For this, note that the map y 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

33 3.5. The Lebesgue measure 29 techniques) that this is true for all sets E B R. 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 Theorem: Let E R and λ (E) < +. Then, given ǫ > 0, there exists a set F ǫ which is a finite disjoint union of open intervals and is such that λ (E F ǫ ) < ǫ. We give next some more characterizations of Lebesgue measurable sets 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 G ǫ such that E G ǫ and λ (G ǫ \ E) < ǫ. (iii) For every ǫ > 0, there exists a closed set F ǫ such that F ǫ E and λ (E \ F ǫ ) < ǫ. (iv) There exists a G δ -set G such that E G and λ (G \ E) = 0. (v) There exists an F σ -set F such that F E and λ (E \ F) = 0. [Hint: Prove the following implications: and (i) = (ii) = (iv) = (i) (i) = (iii) = (v) = (i).] Note: Theorem 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}.

34 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 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 B R L and λ(k) < +. It is natural to ask the question: can one obtain λ(e) for a set E B R, if λ(e) < +, from the knowledge of λ(k), K compact in R? The answer is given by the next proposition 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 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: 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 B R, the σ-algebra of Borel subsets of R. This gave us a measure λ on B R 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 B R, 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 B R and x R, E + x B R and λ(e + x) = λ(e). Thus the Lebesgue measure is a translation invariant σ-finite regular measure on B R. The question arises: are there other σ-finite measures on B R with these properties? Obviously, if c > 0 then cλ defined by (cλ)(e) := cλ(e), E B R, is also a σ-finite measure and is translation invariant. In fact the following hold:

35 3.5. The Lebesgue measure Theorem: Let µ be a measure on B R 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 B R and x R. Then there exists a positive real number c such that µ(e) = cλ(e) E B R 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) 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) g.h and G G, g g 1 are continuous with respect to the product topology on G G, one calls G a topological group. Given a topological group, let B G 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 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 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 B R\{0}, µ(e) := is a Haar measure on R \ {0}. E 1 dλ(x) (3.2) x 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 Ĩ 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 function µ F on the semi-algebra Ĩ, as in example Using theorem

36 32 3. Construction of measures 3.4.6, we can construct a complete measure µ F on a σ-algebra of subsets of R which includes B R. This measure µ F is called the Lebesgue-Stieltjes measure induced by the function F. Note that µ F has the property that µ F (a, b] < + a, b R, a < b. Conversely, given a measure µ on B R such that µ(a, b] < + a, b R, a < b, we can restrict it to Ĩ and, using proposition 2.2.2, define a monotonically increasing right continuous function F : R R such that the unique Lebesgue-Stieltjes measure µ F induced by F is nothing but µ (by the uniqueness of the extension). Thus measures µ on B R which have the property that µ(a, b] < + a < b can be looked upon as a Lebesgue-Stieltjes measure µ F for some F. We point out that it is possible to find different F 1, F 2 : R R such that both are monotonically increasing and right continuous and µ F1 = µ F2. 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 = µ F. 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 = µ F (R) = 1, is called a probability distribution function. Exercises: (3.15) Let I 0 denote the collection of all open intervals of R. For E X, show that { } λ (E) = inf λ(i i ) I i I 0 i, for i j and E I i. i=1 (3.16) Let E R and let ǫ > 0 be arbitrary. Show that there exists an open set U ǫ E such that λ(u ǫ ) λ (E)+ǫ. Can you also conclude that λ(u ǫ \ 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 {I n } n 1 of intervals such that E n=1 and λ ( n=1 \ E) < ǫ. Such sets are called Lebesgue null sets.prove 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 = {x 1, x 2, x 3,...} is a null set. (iii) Q, the set of rational numbers, is a null subset of R. i=1