Solutions to Tutorial 1 (Week 2)

Transcription

1 THE UNIVERSITY OF SYDNEY SCHOOL OF MATHEMATICS AND STATISTICS Solutions to Tutorial 1 (Week 2 MATH3969: Measure Theory and Fourier Analysis (Advanced Semester 2, 2017 Web Page: Lecturer: Daniel Daners Questions marked with * are more difficult questions. Material covered (1 review of basic set theory (2 definition and examples of σ-algebras (3 definition and properties of the Lebesgue outer measure Outcomes After completing this tutorial you should (1 be familiar with the basics of abstract measure and integration theory; (2 have a good idea on the construction of measures from outer measures with application to the Lebesgue measure and related measures Summary of essential material Set theory For your reference we collect some facts on set theory. We always look at a base set X and subsets thereof. Here are some precise definition: A X is a subset of X if x mplies x X. The collection of all subsets of X is called the power set of X. We denote the power set of X by (X. If A, B X we write A B if x B for every x A and A = B if A B and B A. Let I be an arbitrary index set, for instance I = {1, 2,, n finite, I = N countable or I = R uncountable. There is no restriction on the cardinality of I. Suppose that for every i I there is a set X. We define the union = {x X x for some i I and the intersection = {x X x for all i I. We define the complement A c of A X by For X A we also say X minus A. A c = X A = {x X x A. Copyright 2017 The University of Sydney 1

2 As in algebra, there are rules on how to do computations with sets. We state the most basic ones below. We assume that X for all i I and B X, where I is an arbitrary index set. Distributive laws: B = ( B and ( B = ( B De Morgan s laws: c = A c i and ( c = A c i Intersection of sets is also associative, that is, it does not matter how we bracket an expression. For instance A (B C = (A B C. Unions have a similar property. Associativity is true for arbitrary unions/intersections of sets. σ-algebras and measures Let X be a set and a collection of subsets of X. We call a σ-algebra if (i ; (ii A implies A c ; (iii A j for all j N implies j N A j. σ-algebras are the domains of definitions of measures. Let X be a set and a σ-algebra of subsets of X. A function μ [0, ] is called a measure if (i μ( = 0; (ii If A j for all j N are such that A k A j = for k j, then ( μ A j = μ(a j j N (countable additivity. To be able to measure the size of an arbitrary subset of X we also introduce the notion of an outer measure. We weaken the condition on the countable additivity. A function μ (X [0, ] is called an outer measure if (i μ ( = 0; (ii If A j X for all j N and A X with A A j, then μ (A μ (A j (countable sub-additivity. We emphasize that and outer measure is defined for all subsets of X, not just restricted to a σ-algebra. The most important outer measure for us is the Lebesgue outer measure m N on RN defined by { m N (A = inf vol(r k A R k, R k open rectangles, where vol(r k is the usual volume of a rectangle, namely the product of the side lengths. 2

3 Questions to complete during the tutorial 1. Let A, B, C, be subsets of a set X. Simplify the following expressions. (a (A B A c Solution: Using the laws from above and the fact that A A c = (b S (A B f A B =. (c (A B A c = (A A c (B A c = B A c. Solution: S (A B A = S ( (A A (B A = S A. (A c B c (A c B Solution: (A c B c (A c B = A c (B c B = A c X = A c, where we use the distributive law in the first equality. (d (A B (A B c Solution: (A B (A B c = A (B B c = A X = A. 2. Let μ [0, ] be a Borel measure measure on R with μ(r = 1. Define the function F (t = μ ( (, t]. The function F is called the distribution function of the measure μ and is often used in probability theory. Some of the questions below rely on monotone continuity properties of measures for nested sets: Let A n for all n N. A n A n+1 for all n N lim μ(a n = μ ( A n A n A n+1 for all n N and μ(a 0 < lim μ(a n = μ (a Show that F is increasing and right continuous. To prove the continuity use the sequential characterisation of continuity and the monotonicity properties of measures stated above. A n Solution: If s t, then (, s] (, t] and so by the monotonicity of measures F (s = μ ( (, s] μ ( (, t] = F (t, so F is increasing. For the right continuity fix t R and let t n be a decreasing sequence with t n t. Then (, t n ] (, t n+1 ] for all n N and (, t] = (, t n ]. Since μ((, t 0 ] μ(r = 1 < the monotonicity properties of measures implies lim F (t n = lim μ ( (, t n ] = μ (, t n ] = μ ( (, t] = F (t. Since this is true for all decreasing sequences with t n t it follows that F is right continuous. (b Show that lim t F (t = 0 and lim t F (t = 1. Solution: Again by the monotonicity of measures lim F (t n = lim μ ( (, t n ] = μ for every decreasing sequence t n and lim F (t n = lim μ ( (, t n ] = μ ( (, t n ] (, t n ] = μ( = 0. = μ(r = 1 for every increasing sequence t n. Hence F (t 0 if t and F (t 1 if t. 3

4 (c Show that μ({t = F (t lim s t F (s (The height of the jump of F at t. Solution: Since the function is bounded and monotone lim s t F (s exists at every point t R. Now for every increasing sequence t n with t n t we have from the monotonicity property as claimed. μ{t = μ (t n, t] = lim μ ( (t n, t] = F (t lim F (t n = F (t lim s t F (s (d Show that a bounded increasing function on R can have at most countably many discontinuities. Solution: Since F is monotone, left and right limits exist at every point. Let S n = {t R lim s 0+ F (s lim s 0 F (s > 1 n be the set of discontinuities with a jump larger than 1 n. This is a finite set for every n as otherwise the function is unbounded. The set of all discontinuities S n is therefore at most countable. 3. Suppose that X, Y are sets and that f X Y. For A Y consider the inverse image (or pre-image of A, given by f 1 [A] = {x X f(x A. The following parts ask you to prove equality of sets, say C and D. Often this is done by showing x C x D and x D x C. The first means C D, the second D C, hence C = D. (a Prove that ( f 1 [A] c = f 1 [A c ]. Solution: For every x X either f(x A or f(x A c, but not both. Hence X = f 1 [A] f 1 [A c ] is a disjoint union and so our claim follows. (b Suppose that Y for all i I, where I is an arbitrary index set. Prove that f 1 [ ] = f 1[ ] and f 1 [ ] = f 1[ ]. [ Solution: We prove the first identity. Let x f 1 ] A j. Then f(x A j, and therefore f(x A j for some j. But this means that x f 1 (A j, and so x f 1 [A j ], [ showing that f 1 ] A j f 1 [A j ]. If on the other hand x f 1 [A j ], then x f 1 [A j ] for some j N, and so f(x A j [ A j. Hence x f 1 ] A j and so [ f 1 [A j ] f 1 ] A j as required. The other identity follows in a similar way, replacing some j I by all j I in the above proof. 4. Suppose that X, Y are sets and f X Y a function. To do the following questions you need to verify all conditions given in the definition of a σ-algebra. (a If is a σ-algebra in Y, show that 0 = {f 1 [A] A is a σ-algebra in X. Solution: (i Clearly = f 1 [ ] 0. (ii By Question 3 we have (f 1 [A] c = f 1 [A c ]. Since A implies A c, we get (f 1 [A] c 0 for every A. (iii Let B j 0, j N. Then there exist A j with B j = f 1 [A j ]. Hence from Question 3 B j = f 1 [A j ] = f 1[ ] A j 0. 4

5 (b If is a σ-algebra in X, show that 1 = {A Y f 1 [A] is a σ-algebra in Y. Solution: (i Clearly = f 1 [ ], so 1. ( (ii Suppose that A Y is such that f 1 [A]. Since f 1 [A] implies f 1 [A c ] = f 1 [A] c by Question 3, we get A c 1. (iii Let A j Y with f 1 [A j ] for all j N. Then there exist A j with B j = f 1 [A j ]. Hence by Question 3 f 1[ ] A j = f 1 [A j ], so A j Let A R. We call A bounded from below if there exists m R such that m a for all a A. If such m exists, we call m a lower bound for A. There is a special lower bound, namely the infimum or greatest lower bound. By definition we call M the infimum of A and write inf f M is a lower bound for A; and for every lower bound m of A we have m M. (a Let M = inf A. Show that for every ε > 0 there exists a A such that M a < M + ε. Solution: We prove the contrapositive. If ε > 0 is such that M + ε a for all a A, then M + ε is a lower bound, and hence M is not the infimum. (b Suppose that M is a lower bound for A, and that for every ε > 0 there exits a A such that a < M + ε. Show that M = inf A. Solution: Assume that m > M and set ε = m M. Then ε > 0 and m = M +ε. By assumption there exits a A such that a < M + ε = m. Hence m is not a lower bound for A. Therefore, if m is a lower bound for A, then m M. As M is a lower bound of M, it is the infimum of A. 6. Let J = [a, b] R be a closed bounded interval. Let m be the Lebesgue outer measure on R. (a Use the definition of m to show that m (J b a. *(b Solution: Clearly (a ε, b + ε [a, b] for all ε > 0. Hence the family (a ε, b + ε,,, covers [a, b]. By definition of m (J as an infimum over all possible covers by open intervals we have m (J (b + ε (a ε = b a + 2ε for all ε > 0. Hence m (J b a. Suppose that I k = (α k, β k, k N, is such that J k N I k. Prove that and hence deduce that m (J b a. b a (β k α k Solution: By assumption J is closed and bounded and therefore by the Heine-Borel theorem it is compact. (This will not mean much for those who have not done MATH3961, you need to accept the result here. Hence there are finitely many intervals I kj, j = 1,, m, such that J m j=1 I k j. By renumbering the intervals we can assume that J m j=1 I k. We order the set of endpoints α 0, α 1, and β 0, β 1, in increasing order a = ξ 0 < ξ 1 < < ξ n = b. Then for every i = 1,, n we have (ξ i 1, ξ i I k for at least one k = 1,, n. Hence b a = n (ξ i ξ i 1 i=0 m (β k α k (β k α k. As the above is true for every covering with intervals I k we conclude that m (J b a by definition of an infimum. 5

6 Extra questions for further practice 7. Denote by m the Lebesgue outer measure in R N as defined in lectures. (a Show that {x is measurable and that m ({x = 0 for every x R N. Solution: Fix ε > 0 and let R 0 = (x 1 ε, x 1 + ε (x 2 ε, x 2 + ε (x N ε, x N + ε and R k = for all k 1. Then {x R k and m ({x vol(r k = (2ε N Since the above is true for all ε > 0 we get m ({x = 0. From lectures we know that sets of outer measure zero are measurable. (b Let C R N be a countable set. Show that m (C = 0. Solution: Since C is countable we can enumerate its elements by x 0, x 1, x 2,. By the countable subadditivity of outer measures and the previous part 0 m (C 8. Consider the following set functions of A R N : m ({x k = 0. { m (A = inf vol(r k A R k, R k open rectangles, { m (A = inf vol(r k A R k, R k open rectangles. The first one is the definition of the Lebesgue outer measure. The questions below show that we can replace open by closed rectangles in the definition of the Lebesgue outer measure. As a consequence, one can take any rectangles, not necessarily open or closed. (Why? [Note: This question as asked originally did not quite work as intended. Part (a was the more difficult one and should have been starred, part (b as was easier. We present them interchanged here in the solutions. Thanks to Ting-Ying Chang for carefully reading the solutions and pointing that out.] (a Prove that m (A m (A for all A R N. *(b Solution: We use the fact that the infimum over a larger set is smaller than or equal to the infimum over a smaller set. The sets of interest are { C = vol(r k A R k, R k open rectangles R, { D = vol(r k A R k, R k open rectangles R. We show that C D. Assume that α C. This means that there exist a family (R k k N of open rectangles such that A R k and such that α = vol(r k. As R k R k we also have A R k, so α D as well by definition of D. Hence C D and therefore m (A = inf D inf C = m (A. Prove that m (A m (A for all A R N and hence that m (A = m (A. Solution: Let the sets C, D be defined as in the previous part and fix ε > 0. Since m (A = inf D there exists α D such that α < m (A + ε 2 (here we use 5(a. By definition of D there 6

7 exists a family (R k k N of open rectangles such that A R k and α = vol(r k. By enlarging R k slightly on each side we can construct open rectangles R εk such that R k R εk and vol(r εk vol(r k + ε 2 k+2. Then clearly A R εk and m ε (A vol(r εk vol(r k + 2 = α + ε 2 < k+2 m (A + ε 2 + ε 2 = m (A + ε. Since the above works for every choice of ε > 0 we conclude that m (A m (A. Hence equality follows. Note that instead of closed rectangles we could take any rectangle between open and closed. The above arguments work in the same fashion and show equality of definitions. 9. (a Prove the distributive laws for unions and intersections of sets. Solution: For the proof we extensively use the definition of the intersection and the union of sets without saying so every time. ( (i Assume that x B. Then x or x B. In the first case x for every i I, and so x B for all i I. In the second case x B, so x B for all i I again. Hence in both cases x ( B. This shows that B ( B. To prove the opposite inclusion assume that x ( B. Then x B for all i I. If x B, then x for all i I, so x (. In the other case x B. In either case x B by definition of a union of sets. Hence B ( B. Since we have proved both inclusions, equality follows. ( (ii Suppose that x B. Then x and x B. Hence there exists j I ( with x A j. Therefore x A j B and so x B. Hence ( B ( B. to prove the opposite inclusion assume that x ( B. Then there exists j I such that x A j B and therefore x B and x A j. Hence x and so ( x B. This proves the opposite inclusion ( B ( B. (b Prove de Morgan s laws for the complement of unions and intersections of sets. Solution: ( c. (i Suppose that x A i Then x Ai, so there exists j I such that x A j, that is, x A c j. Hence x Ac j and so c A c i. 7

8 (ii To prove the opposite inclusion let x Ac i. Then there exists j I such that x Ac j, that is, x A j. Therefore x ( c. and so x i A Hence and equality follows. c A c i We prove the second law by using the first one by using that (B c c = B for every set: ( c c ( A c i = = (A c i c c ( c. = A c i 10. Let X be a set and a σ-algebra of subsets of X. Let S be fixed. (a Show that S = {A S A is a σ-algebra of subsets of S. Solution: (i = S S. (ii If B S, then there exists A with B = A S. Then S B = A c S S as well because A c. (iii Finally, if B k = A k S S for A k, then k N B k = k N (A k S = ( k N A k S S since k N A k. (b If μ [0, ] is a measure, show that the restriction μ S S [0, ] defined by μ S (A = μ(s s a measure as well. Solution: This follows from the properties of measures since all sets S A are measurable if S and A are measurable. Challenge questions (optional The idea below is to generalise the Lebesgue outer measure on R, and to show how it relates to a class of increasing functions. *11. Let F R R be a right continuous increasing function, that is, F (t = lim s t+ F (s for all t R. For a b define ν F ((a, b] = F (b F (a. If A R let { μ F (A = inf ν F (I k I k = (a k, b k ], A I k. (a Show that μ is an outer measure (called the Lebesgue-Stieltjes outer measure F Solution: We first prove that μ F ( = 0. For that we choose I k = (0, 0] = for all k N. Since k N (0, 0] = the definition of μ F implies 0 μ F ( ( ν F ((0, 0] F (0 F (0 = 0. Hence μ F ( = 0. We next show the countable sub-additivity. Let A, A k R, k N and A A k. Let ε > 0 and choose intervals I jk such that A j j=1 I jk and μ F (A k +ε 2k+1 > ν F (I jk. Then A and hence I jk j=1 μ F (A ( ν F (I jk μ F (A k + ε = 2 k+1 μ F (A k + ε. Since ε > 0 was arbitrary we get μ F (A μ F (A k. This completes the proof. 8

9 (b Show that μ F ((a, b] ν F ((a, b] = F (b F (a. Solution: Fix an interval (a, b]. We can produce a cover by I 0 = (a, b] and the intervals I k = (0, 0] = for k 1. Then (a, b] I k and μ F ((a, b] ( ν F (I k F (b F (a + F (0 F (0 = F (b F (a = νf ((a, b] as claimed. *(c Show that μ F ((a, b] ν F ((a, b] = F (b F (a and hence from (b μ F ((a, b] ν F ((a, b]. Solution: The opposite inequality is a bit more tricky. We have to show that for every possible cover (a, b] I k with intervals of the form I k = (a k, b k ] we have F (b F (a ν F (I k. (1 We distinguish two cases. (i First assume that finitely many of the I k s cover (a, b]. Renumbering we can assume that (a, b] n I k. Now order all endpoints of these k intervals in (a, b] and denote them by a = ξ 0 < ξ 1 < < ξ m = b. Then for each i = 1,, m the interval J i = (ξ i 1, ξ i ] I k for some k. Hence ν F (J i ν F (I k and so n ν F (I k ν F (I k m ν F (J i = i=1 m (F (ξ i F (ξ i 1 = F (b F (a. i=1 (ii If there is no finite sub-cover, the proof is more difficult. Fix ε > 0 and choose δ k > 0 such that F (b k + δ k F (b k < ε 2 k+1. Set I εk = (a k, b k + δ k ]. Then we have (a, b] I εk and ( ν F (I εk = F (bk + δ k F (a k ( = F (bk + δ k F (b k ( + F (bk F (a k ε < ν F (I k + 2 < k+1 ν F (I k + ε. (2 By the right continuity of F we can also choose δ > 0 such that F (a + δ F (a < ε. Now clearly by construction, the open intervals (a k, b k + ε cover [a + δ, b]. Since [a + δ, b] is compact the Heine-Borel theorem implies that there is a finite sub-cover. Renumbering we can assume these intervals are I εk, k = 0,, n. Then (a+δ, b] n I εk. The argument used in (i above implies F (b F (a + δ Using (2 and F (a + δ F (a < ε we get n ν F (I εk ν F (I εk. F (b F (a ( F (b F (a + δ + ( F (a + δ F (a < ε + ν F (I εk < 2ε + ν F (I k. Since the above procedure works for every choice of ε > 0 we finally get (1. 9