Analysis Part 1. 1 Chapter Q1(q) 1.2 Q1(r) Book: Measure and Integral by Wheeden and Zygmund

Transcription

1 Analysis Part 1 Book: Measure and Integral by Wheeden and Zygmund 1 Chapter Q1(q) Let {T x k } be a sequence of points of T E. Since E is compact, {x k } has a subsequence {x kl } that converges to some x E. By continuity of T, lim T x k l = T x T E. l Thus {T x kl } is a subsequence of {T x k } that converges to T x T E, so T E is compact. 1.2 Q1(r) We are using notation in the book (pg 13), so R Γ = N k=1 f(ξ k)v(i k ), U Γ = N k=1 [sup x I k f(x)]v(i k ), etc. ( = ) Assume A = (R) f(x) dx exists. Given ɛ > 0, there exists δ > 0 I such that A R Γ < ɛ 2 for any Γ and any chosen {ξ k }, provided that Γ < δ. 1

2 By definition of supremum, there exists ξ k I k such that Thus Hence R Γ = > f(ξ k ) > sup f(x) ɛ x I k 2v(I). N f(ξ k )v(i k ) k=1 N N ɛ sup f(x)v(i k ) ( v(i k )) x I k 2v(I) k=1 = U Γ ɛ 2. k=1 U Γ A = U Γ R Γ + R Γ A < ɛ 2 + ɛ 2 = ɛ. Note that inf Γ <δ U Γ = inf Γ U Γ. This is clear once we note that any partition Γ has a refinement Γ with Γ < δ, and that U Γ U Γ for any refinement Γ of Γ. This implies A inf Γ U Γ ɛ. Since ɛ > 0 is arbitrary, we have A inf Γ U Γ. Similarly, we can show A sup Γ L Γ. Since L Γ U Γ, we have sup Γ L Γ inf Γ U Γ. Combined with the previous inequalities, this gives inf Γ U Γ = sup Γ L Γ = A. ( = ) Assume inf Γ U Γ = sup Γ L Γ = A. By definition of inf and sup, there exists a partition P 1 such that U P1 < A + ɛ/4, and a partition P 2 such that L P2 > A ɛ/4. Letting P to be the common refinement of P 1, P 2, we get a partition P such that U P L P < ɛ/2. Denote P = {I k } N k=1. We want to find a δ > 0 such that A R Γ < ɛ whenever Γ < δ. Note that it suffices to show that U Γ L Γ < ɛ instead. 2

3 Choose δ = ɛ 4M I k, where I k denotes the total area of the union of common boundaries of the intervals I k. For instance in dimension n = 1, I k = N 1. Let Γ = {J k } be a partition with Γ < δ. We split the sum according to whether J k I j for some j. U Γ L Γ = J k I j I j for some j (sup (sup f inf I j J k f inf f)v(j k ) + J k J k I j j I j f)v(i j ) + U P L P + 2Mδ I k < ɛ/2 + ɛ/2. The second last inequality J k I j J k I j j 2Mv(J k ) (sup J k f inf J k f)v(j k ) j 2Mv(J k) 2Mδ I k is due to the fact that the intervals J k I j (for all j) are precisely the intervals J k that intersect the boundary of some I j so the total volume J k I j j v(j k) is bounded above by δ I k. 1.3 Q4(a) First we show sup k j ( a k ) = inf k j (a k ) for all j. By definition of inf, inf k j (a k ) a k for all k j. Thus inf k j (a k ) a k. Since sup is the least upper bound, sup k j ( a k ) inf k j (a k ). Similarly, by noting that sup k j ( a k ) a k for all k j, we can show sup k j ( a k ) inf k j (a k ). Taking limits as j, we get lim sup( a k ) = lim sup j k j ( a k ) = lim inf (a j k) = lim inf (a k). k j 3

4 Q4(b) If the right hand side has the form + or ( ) + ( ), the inequality clearly holds. Assume that the expression on the right does not have the form + ( ) or +. gives Q4(c) Then, sup k j (a k +b k ) sup k j (a k )+sup k j (b k ). Taking limits as j lim sup(a k + b k ) lim sup(a k ) + lim sup(b k ). For k j, we have 0 a k sup k j a k < and 0 b k sup k j b k <. Thus, a k b k (sup k j a k )(sup k j b k ) for all k j. It follows that sup k j (a k b k ) (sup k j a k )(sup b k ). k j Note that lim j (sup k j a k ) exists since sup k j a k is a decreasing and bounded sequence. Similarly, lim j (sup k j b k ) exists. Taking limits as j, we get lim sup (a k b k ) (lim sup a k )(lim sup b k ). Q4(d) Consider a k = {0, 1, 0, 1,... }, b k = {1, 0, 1, 0,... }. Then lim sup (a k + b k ) = 1, and lim sup a k +lim sup b k = 1+1 = 2. Also, lim sup (a k b k ) = 0, and (lim sup a k )(lim sup b k ) = (1)(1) = 1. WLOG, suppose {a k } converges to a. Let b = lim sup b k. Using Theorem 1.4, there is a subsequence {b kj } of {b k } that converges to b. Then, a kj + b kj a + b as j. Let ɛ > 0. There exists K 1 such that b k < b + ɛ/2 for K K 1. There exists K 2 such that a k < a + ɛ/2 for k K 2. Thus if K = max{k 1, K 2 }, a k + b k < a + b + ɛ 4

5 for k K. By Theorem 1.4, this means lim sup a k + lim sup b k = a + b = lim sup(a k + b k ). For given ɛ > 0, there exists N N such that for all k N, a ɛ < a k < a + ɛ. For k j N, sup k j (a k + b k ) (a ɛ) sup k j b k. Taking limits as j gives lim sup(a k b k ) (a ɛ) lim sup b k. Since ɛ > 0 is arbitrary and lim sup b k is bounded, lim sup (a k b k ) (lim sup Along with part (c), this gives equality. a k )(lim sup b k ). 1.4 Q15 ( = ) Assume that bounded f is Riemann integrable on I, and A = (R) I f(x) dx. By Exercise 1(r), inf Γ U Γ = sup Γ = A. There exists a partition Γ 1 such that U Γ1 < A + ɛ/2, and a partition Γ 2 such that L Γ2 > A ɛ/2. Let Γ be the common refinement of Γ 1 and Γ 2. Note that U Γ U Γ1 and L Γ L Γ2. Hence 0 U Γ L Γ U Γ1 L Γ2 < A + ɛ/2 (A ɛ/2) = ɛ. ( = ) Assume that 0 U Γ L Γ < ɛ for some partition Γ of I. Then inf Γ U Γ sup L Γ U Γ L Γ < ɛ. Γ Since ɛ > 0 is arbitrary, thus inf Γ U Γ = sup Γ L Γ. By Exercise 1(r), this means that f is integrable. 5

6 1.5 Q16 By uniform convergence of {f k }, there exists N N such that whenever k N, f k (x) f(x) < ɛ 4v(I) for all x I. This also implies that f is bounded. By Exercise 15, for each k there is a partition Γ k = {I j } N k j=1 of I such that 0 U Γk (f k ) L Γk (f k ) < ɛ 2, where U Γk (f k ) := N k j=1 [sup x I j f k (x)]v(i j ), L Γk (f k ) := N k j=1 [inf x I j f k (x)]v(i j ). When k N, N k U Γk (f) L Γk (f) gives j=1 [sup x I j f k (x) + ɛ 4v(I) ]v(i j) = (U Γk (f k ) + ɛ 4 ) (L Γ k (f k ) ɛ 4 ) < ɛ 2 + ɛ 2. N k j=1 Thus by Exercise 15, f is Riemann integrable on I. For k N, and any partition Γ = {I j }, R Γ (f k ) R Γ (f) = < [ inf x I j f k (x) N [f k (ξ j ) f(ξ j )]v(i j ) j=1 N f k (ξ j ) f(ξ j ) v(i j ) j=1 N ɛ ( 4v(I) )v(i j) j=1 = ɛ 4. ɛ 4v(I) ]v(i j) Since f k and f are both Riemann integrable, taking limits as Γ 0 (R) f k (x) dx (R) f(x) dx ɛ I I 4. 6

7 This shows that (R) f I k(x) dx (R) f(x) dx. I 1.6 Q18 F c is open, hence it can be written as a countable union of disjoint open intervals. Write F c = (a i, b i ), where a i < b i. We may assume (a i, b i ) (, ), as that means F =, for which the statement is trivially true. Define f(x), if x F b i x b g(x) = i a i f(a i ) + x a i b i a i f(b i ), if x / F and x (a i, b i ), with a i, b i finite f(b i ), if x (, b i ) f(a i ), if x (a i, ). In short, g = f on F, and is the straight line segment connecting f(a i ) and f(b i ) on (a i, b i ). Thus g is clearly continuous on (, + ). If f(x) M for x F, then our constructed g also satisfies g(x) M for < x < +. 2 Chapter Q (i) Let V [f; a, b] = sup Γ S Γ M. Let x (a, b). Consider the partition Γ = {a, x, b}. Then f(x) f(a) f(x) f(a) + f(b) f(x) = S Γ sup S Γ M. Γ This implies f(a) M f(x) f(a) + M so f is bounded. 7

8 2.1.2 (ii) For any partition Γ = {x 0, x 1,..., x m }, S Γ (cf) = cf(x i ) cf(x i 1 ) = c S Γ (f). Taking sup over all partitions Γ, we get V (cf) = c V (f) <. S Γ (f + g) = f(x i ) + g(x i ) f(x i 1 ) g(x i 1 ) f(x i ) f(x i 1 ) + = S Γ (f) + S Γ (g). g(x i ) g(x i 1 ) Thus, V (f + g) V (f) + V (g) <. By part (i), let f(x) M f, g(x) M g on [a, b]. S Γ (fg) = f(x i )g(x i ) f(x i 1 )g(x i ) + f(x i 1 )g(x i ) f(x i 1 )g(x i 1 ) g(x i ) f(x i ) f(x i 1 ) + M g S Γ (f) + M f S Γ (g). f(x i 1 ) g(x i ) g(x i 1 ) Thus, V (fg) M g V (f) + M f V (g) <. 8

9 Assume there exists ɛ > 0 such that g(x) ɛ for x [a, b]. Note that 1 1. g(x) ɛ S Γ ( f m g ) = f(x i ) g(x i ) f(x i 1) g(x i 1 ) f(x i )g(x i 1 ) f(x i 1 )g(x i ) f(x i 1 )g(x i 1 ) + f(x i 1 )g(x i 1 ) g(x i )g(x i 1 ) 1 ɛ 2 Hence 2.2 Q5 m g(x i 1 ) f(x i ) f(x i 1 ) + 1 ɛ 2 M g ɛ 2 S Γ(f) + M f ɛ 2 S Γ(g). m V ( f g ) M g ɛ V (f) + M f V (g) <. 2 ɛ2 f(x i 1 ) g(x i ) g(x i 1 ) For any x (a, b], f is of bounded variation on (x, b]. By considering Γ = {x, b}, we see that f(b) f(x) M, which implies f(x) f(b) + M. Thus Next, let Γ = {a = x 0, x 1,..., x n = b} be any partition of [a, b]. Then S Γ [a, b] = f(x 1 ) f(a) + n f(x i ) f(x i 1 ) i=2 f(x 1 ) + f(a) + V [x 1, b] f(a) + f(b) + 2M. V [a, b] = sup[a, b] f(a) + f(b) + 2M <. Γ No, it is not necessary that V [f; a, b] M. Consider 0, x = a f(x) = 1, a < x b. 9

10 Note that V [f; a + ɛ, b] = 0 1 for any 0 < ɛ < b a. However 2 V [f; a, b] = 1 > 1. 2 The additional assumption that f is right continuous at a will ensure V [f; a, b] M. By right continuity, for any given γ > 0, there exists δ > 0 such that for all x < a + δ, f(x) f(a) < γ. For any partition Γ = {a = x 0, x 1,..., x n = b} with x 1 < a + δ, S Γ = f(x 1 ) f(a) + n f(x i ) f(x i 1 ) < γ + V [x 1, b] γ + M. i=2 For partitions Γ that do not satisfy x 1 < a+δ, we can always refine it (e.g. by adding the point a + δ/2) so that the refinement Γ satisfies x 1 < a + δ. Then S Γ S Γ γ + M. Thus V [f; a, b] = sup S Γ γ + M. Γ Since γ > 0 is arbitrary, V [f; a, b] M. 10