Real Variables: Solutions to Homework 3

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1 Real Variables: Solutions to Homework 3 September 3, 011 Exercise 0.1. Chapter 3, # : Show that the cantor set C consists of all x such that x has some triadic expansion for which every is either 0 or. Proof. We are being asked to prove that x C if and only if x = for c k = {0, }. First suppose that x C. We proceed by induction on n in that we will show that the left endpoint of the intervals in C n are x which have triadic expansions with is either 0 or. Consider first C 1 = [0, 1] [, 1]. Then either x = 0 or x =, which are both the correct triadic expansions as claimed. Now suppose this is true for C n 1, that is x = n 1 with = {0, } is a left endpoint of C n 1. Now for C n, let x = n with c k = {0, }. If a n = 0 the x is the left endpoint of the left subinterval and if x = then x is the left endpoint of the right subinterval. This is the construction. Now, suppose that x = with c k = {0, }. Then we have the inequalities 3 x k 3 + k = n. This shows us that if I n C n and n is the left endpoint of the I n, then x C n. This holds for all n so therefore x C. Say that x C 1, then c 1 = 0 if x is in the left interval and c 1 = if x is in the right interval. By the same construction suppose that x C n 1 and that a i have been defined for i = 1,,..., n 1. If x I n 1, an interval in C n 1 then it is written a k I n 1. Either x in the left or right interval of I n 1. Again, if it is the left say x = n 1 that a n = 0 and that a n = if it is in the right. But then we see that so x = x 3 k, with = {0, }. 1

2 Exercise 0.. Chapter 3, # 7: If {I k } N is a finite collection of non-overlapping, I k is measurable I k = Ik. Proof. From the book, we know that I k is measurable follows from Now, to complete the proof, let I k denote the interior of sets I k. These interiors are disjoint so the enjoy some nice properties like additivity in measure. Then I k Ik by subadditivity. We seek to show the other direction for the inequality. Consider I k Ik = Ik = Ik. Thus Ik = Ik. Exercise 0.3. Chapter 3, # 8: Show that the Borel σ-algebra B in R n is the smallest σ-algebra containing the closed sets in R n. Proof. By definition, B is the smallest σ-algebra containing the open sets. In particular, B is a field and is therefore closed under set complementation. Because B contains all the open sets in R n, it also contains the compliments of all such set, which are exactly all closed sets in R n. To see that it is the smallest such σ-algebra, imagine that there is one smaller that contains all the closed sets, call it A. Again, because A is a σ-algebra it is a field and is closed under complementation so in particular A contains all the open sets in R n. However this is a contradiction because by definition, B is the smallest σ-algebra containing the open sets in R n and we are done. Exercise 0.4. Chapter 3, # 9: If {E n } n=1 is a collection of sets with n=1 E n e <, show that lim sup E and lim inf E have measure zero. Proof. Now, since E n e <, there must be some m such that Then, lim sup E n = E n e < ɛ. E n j=1 n=j Proceed to calculate, lim sup E n e = E n E n = j=1 n=j e e E n. E n e < ɛ.

3 Also lim inf E n = E n E n, j=1 n=j and the same argument follows. Thus lim sup E n e = lim inf E n e = 0. Exercise 0.5. Chapter 3, # 10: If E 1 and E are measurable show that E 1 E + E 1 E = E 1 + E. Proof. We know since E 1 is measurable that E 1 E = E 1 E E 1 + E 1 E E c 1 = E 1 + E E c 1 Furthermore we know that so E 1 E + E E c 1 = E, E 1 E + E 1 E = E 1 + E. Exercise 0.6. Chapter 3, # 17: Our example will be the Cantor-Lebesgue function: F x = b k where b k k = a k if x = a k. We first must go through the tedious task of showing it is continuous, that F 0 = 0 and F 1 = 1, and that F is surjective. We begin: Proof. The ternary expansion for zero is a k = 0 for all k. Looking at the function, F 0 = k 0 1 k = 0 The ternary expansion for 1 is a k = for all k. F 1 = k 1 k = 1 Thus F 0 = 0 and F 1 = 1. Now we need to prove continuity. To do this, we need to show that for all ɛ > 0 there exists δ > 0 s.t. F a F b < ɛ when a b < δ for a, b [0, 1]. Let ɛ > 0 and pick 3

4 and n such that n < ɛ. Pick δ = 3 n+1. Pick a, b s.t. a b < δ. Let a have a ternary expansion with a k {0, } and let b have ternary expansion with b k {0, }. We need to show that a k = b k for k < n. To do so, pick the minimal k where a k b k, k. 3 b k k b k + k=k +1 3 b k k Again, we conclude that 1 b k + k=k +1 3 b k k giving us that k > n. So for the case where a k = b k, k < n and = a k /, d k = b k /, then ck F a F b = d k k k = ck d k k k k = n < ɛ This shows the function F x to be continuous on C. We need to prove F to be surjective. We start with the fact that every element x [0, 1] has a binary expansion. That is b n x =. n n=1 Define a k = b k. Because a k {0, } then by Exercise 1 in this homework, for any element c C, a n c = 3 n and F c = x. Therefore F is surjective. Finally, we can get to the example: n=1 Proof. Te A [0, 1] to be non measurable. Now, we know that F C = [0, 1]. Every subset of a set of measure zero is measurable so then in particular B = F 1 A C is measurable. However, is non measurable. F B = F F 1 A C = A 4

5 Exercise 0.7. Chapter 3, # 3: let Z be a subset of R 1 with measure zero. Show that the set Z = {x : x Z} also has measure zero. Proof. Either Z is bounded or it is infinite. Suppose first that Z is bounded, that is x M R for all x Z. Because Z has measure zero we can cover it will a collection of intervals I n B M 0 such that n I n < ɛ. Now, Z is covered by a union of intervals In. But clearly In diamb M 0 I n = M I n. So Z n I n M n I n < Mɛ. Thus, ting ɛ 0 shows that Z = 0. Now, in the unbounded case, define Z restricted to a bounded interval on the real line as Z M = Z B M 0 and Z M = Z B M 0. By construction Z = M Z M and Z = M Z M. Now, for any M, Z M = 0 this is simply the bounded case we did above. Using the property of subadditivity of the measure Z = Z M Z M = 0. M M We have shown that Z = {x : x Z} has measure zero if Z has measure zero. 5