Problem set 1, Real Analysis I, Spring, 015. (1) Let f n : D R be a sequence of functions with domain D R n. Recall that f n f uniformly if and only if for all ɛ > 0, there is an N = N(ɛ) so that if n N and x D, then f n (x) f(x) < ɛ. Now let D = [0, 1], f n (x) = x n and { 0 for x [0, 1) f(x) = 1 for x = 1. Show by using the definition of uniform convergence that f n does not approach f uniformly. Solution: The negation of f n f uniformly is that there is an ɛ > 0 so that for all N, there are an n N and an x D so that f n (x) f(x) ɛ. So we take ɛ = 1, and let N be a positive integer. Now if x [0, 1), f(x) = 0, and so f n (x) f(x) = x n. So if we set x N = 1, this will suffice to find a counterexample. So choose x = 1 N [0, 1). This shows f n does not approach f uniformly. () Let N = {1,, 3,... } denote the set of natural numbers. For i, j N, assume that c ij 0. Prove that sup c ij = c ij, where the supremum is over all finite subsets B of N N. Solution: We first show. Since c ij 0, we have that c ij N c ij for all N. If B N N is a finite set, then B {1,..., N} {1,..., N} for some N. Therefore, N N N c ij c ij c ij c ij, where the last inequality follows since each c ij 0. Now to prove, first consider the case where for some i c ij =. In this case, for all M there is an N so that N c ij M. But then B = {(i, 1), (i, ),..., (1, N)} is a finite set so that c ij M. Thus in this case, both sides are infinite. 1
Now consider the case that each c ij = d i <. Let ɛ > 0. For each i, there is an N i so that N i c ij d i i 1 ɛ. Now if d i = L <, there is an M so that M d i L ɛ. Therefore, in this case M N i M c ij (d i ɛ ) > M ɛ d i+1 i = M d i+1 i ɛ L ɛ. So therefore, this finite sum is within ɛ of the sum of the series L, and so sup c ij c ij = L. The last case is L =. In this case, let P be a positive number. Then there is an M so that M d i P + ɛ. Then as above, we have M Ni c ij P. This show the supremum of the finite sums is. (4) Construct a closed set Ĉ so that at the kth stage of the construction, one removes k 1 centrally situated open intervals each of length l k, with l 1 + l + + k 1 l k < 1. (a) If the l k are chosen small enough, then k=1 k 1 l k < 1. In this case, show that m(ĉ) > 0, and in fact m(ĉ) = 1 k=1 k 1 l k. Solution: Here is an example of such a sequence l k. Let l k = 4 k. Then k=1 k 1 4 k = k=1 k 1 = 1 < 1. By construction, [0, 1] is the disjoint union of Ĉ and a countable collection of intervals, with k 1 intervals each of length l k. All of these sets are measurable (the intervals are open and Ĉ is closed), and so 1 = m([0, 1]) = m(ĉ) + k 1 l k. (b) Show that if x Ĉ, then there exists a sequence of points {x n } n=1 such that x n / Ĉ, yet x n x and x n I n, where I n is a sub-interval in the complement of Ĉ with I n 0. Solution: First of all, recall Ĉ = k=0 Ĉk, where Ĉk is a disjoint union of k closed intervals. Each of these closed intervals must then have length less than k. Then from k=1
each of these closed intervals, an open sub-interval of length l k < k is removed to form Ĉk+1. Now let x Ĉ. Then x C k for all k, and must be contained in a closed interval of length less than k. The center x k of the open subinterval of length I k = l k is then of distance less than k 1 to x, and I k = l k 0 as k. (c) Prove as a consequence that Ĉ is perfect, and contains no open interval. Solution: Recall that a closed set is perfect if it has no isolated points. Part (b) above shows that if x Ĉ, there is a sequence x n x, with x n I n an interval in the complement. Let y n be an endpoint of I n which is not equal to x. Then since x n x, and I n 0, we have y n x x n x + y n x n x n x + I n 0. So y n x. But the endpoints y n of the intervals I n are elements of Ĉ. Thus Ĉ is perfect. To show Ĉ contains no open interval, note part (b) shows Ĉ c is dense in [0, 1]. (d) Show also that Ĉ is uncountable. Solution: If Ĉ were countable, it would have measure 0. (5) Suppose E is a given set, and O n is the open set O n = {x : d(x, E) < 1/n}. Show: (a) If E is compact, then m(e) = lim n m(o n ). Solution: First of all, note that for E closed, n=1 O n = E. Proof: Recall d(x, E) = inf y E x y. The intersection n=1 O n = {x : d(x, E) = 0}. Let x have d(x, E) = 0. Then for all n, there is a y n E so that y n x 1/n. This shows y n x, and so x E, as E is closed. Now as E is compact, it is both closed and bounded. Each open set O n is also bounded, and so is measurable with finite measure. The previous paragraph then implies O n E. Then since m(o n ) <, Corollary 3.3(ii) implies m(e) = lim n m(o n ). (b) However, the conclusion in (a) may be false if E is closed and unbounded, or E open and bounded. Solution: For E closed and unbounded, let E be the line y = 0 in R. Then O n = {(x, y) : x < 1 } has infinite n measure for all n. On the other hand, E itself has measure 0: we may cover it by a union of closed rectangles 3
of measure at most ɛ. For simplicity, we only cover the ray [0, ) {0} by rectangles. The negative ray will be handled by symmetry across the x axis. So let ɛ > 0. Cover [0, 1] {0} with a rectangle of width 1 and height ɛ/4. In general, cover [n, n + 1] {0} by a rectangle of width 1 and height ɛ n. The sum of the areas of these rectangles is then ɛ/. So with the corresponding rectangles along the negative ray, we have E covered by an almost-disjoint collection of rectangles of total measure ɛ. So m(e) ɛ for all ɛ > 0, and m(e) = 0. For the case of E open and bounded, consider the open bounded set E = [0, 1] Ĉ from problem (4). This set is dense in [0, 1], and so each O n [0, 1], and m(o n ) 1. On the other hand, we showed m(e) < 1. This shows that m(o n ) does not approach m(e) in this case. (16) The Borel-Cantelli Lemma. Suppose {E k } k=1 is a countable family of measurable subsets of R d and that m(e k ) <. Let k=1 E = {x R d : x E k, for infinitely many k} = lim sup(e k ). k (a) Show that E is measurable. Solution: This is clear by the hint, since countable unions and countable intersections of measurable sets are measurable. To prove the hint, assume x is an element of E k for infinitely many k. Since {1,..., N} is finite for all N, this shows that for all N, there is an n > N so that x E n. Therefore x n=1 k>n E k. Conversely, if x n=1 k>n E k, then for all n, x k>n E k. If x were an element of only finitely many E k, then there would be a largest N so that x E N. This contradicts the fact that x k>n E k. Thus x is an element of E k for infinitely many k. This proves the hint. (b) Prove m(e) = 0. Solution: For all n, E k>n E k, and so by countable subadditivity, m(e) k=n+1 m(e k). But this is the tail of a convergent series, and so k=n+1 m(e k) 0 as n. So m(e) = 0. 4
[Hint: Write E = n=1 k>n E k.] (6) Suppose A E B, where A and B are measurable sets of finite measure. Prove that if m(a) = m(b), then E is measurable. Solution: In this case E = A (E A) a disjoint union, and E A B A. Now B = A (B A) is also a disjoint union, which shows m(b) = m(a)+m(b A). Since m(a) = m(b) <, we have m(b A) = 0. But now E A is a subset of a set of measure 0. This shows E A is measurable (by Property of Lebesgue-measurable sets; this is the fact that Lebesgue measure is complete). So E = A (E A) is measurable. 5