15 - MAT Caratheodory's Theorem Nonmeasurable and Non-Borel...

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Unformatted text preview: MAT 473 39 15. Caratheodory's Theorem; Nonmeasurable and Non-Borel Sets Theorem 15.1 (Carathodory). A RN is measurable if and only if e for each E RN . (E) = (E A) + (E Ac ) Lemma 15.2. For every measurable set B and every A B, Exercise 15.1. Prove Lemma 15.2 by showing first that (B) (A) + (B \ A), and then that (B) (A) + (B \ A). (B) = (B \ A) + (A). Proof of Theorem 15.1. Suppose A is measurable, and fix E RN . By the subadditivity of outer measure, we have (E) (E A) + (E Ac ). On the other hand, by the monotonicity of outer measure, for every open set G which contains E we have Taking the inf over all such G gives (E) (E A) + (E Ac ), hence equality. Conversely, suppose the intersection property holds for every E. Then it certainly holds for every M L0 , which is to say On the other hand, the lemma gives (M ) = (M A) + (M Ac ). (G) = (G A) + (G Ac ) = (G A) + (G Ac ) (E A) + (E Ac ). It follows from these two displayed equations that (M A) = (M A), which is to say M A L0 for every M L0 , which is to say A is measurable. Define an equivalence relation on R by x y if and only if y - x Q, and let A R be a set consisting of exactly one element from each equivalence class in R/ . (Such a set exists by the Axiom of Choice.) Proposition 15.3. A is not measurable. Proof. Let (qn ) be an enumeration of Q. Then (15.1) R= (qn + A). n=1 (M ) = (M \ (M Ac )) + (M Ac ) = (M A) + (M Ac ). To see this, fix x R. Then x a for some a A by the definition of A, which is to say x - a = qn for some n. Thus x = qn + a qn + A, so x is in the union. Now, since outer measure is translation-invariant3, = (R) = (qn + A) (qn + A) = (A), n=1 n=1 n=1 proved in the homework that Lebesgue measure is translation invariant, and even if you didn't prove that outer measure is translation-invariant in the course of your proof, it follows from the result, since for every open set G containing A + x, G - x is an open set containing A, and (G) = (G - x). Date: March 7, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. 3You've 40 SPRING 2012 and it follows that (A) > 0. On the other hand, let K A be compact. Then if (rn ) is an enumeration of Q [0, 1], the set B= (rn + K) n=1 is bounded, for if K [-M, M ], then B [-M, M + 1]. Moreover, this union is disjoint: for if z (rn + K) (rm + K), then z = rn + a = rm + b for some a, b K A; but then b - a = rn - rm Q, so a b, so a = b by the definition of A. It follows that rn = rm , so n = m. Since Lebesgue measure is translation-invariant, we now have > (B) = (rn + K) = (rn + K) = (K), n=1 n=1 n=1 which forces (K) = 0. Since K was an arbitrary compact subset of A, it follows that (A) = 0. Thus (A) < (A). If (A) < , then A is by definition not measurable. If (A) = , then A fails the Approximation Property, since for every closed F A we have (F ) = (F ) (A) = 0, while for every open G A we have (G) = (G) (A) = . Thus for any such F and G, we have (G \ F ) = . Corollary 15.4. Every measurable subset of R which has positive measure contains a nonmeasurable subset. Proof. We retain the notation of the proof of Proposition 101.1. Suppose (B) > 0. Then since B= ((qn + A) B) , n=1 we have 0 < (B) = (B) so we must have ((qN + A) B) > 0 for some N . But (qN + A) B qN + A, whence4 ((qN + A) B) = 0. Thus (arguing as at the end of the proof of Proposition 101.1), (qN + A) B is a nonmeasurable subset of B. n=1 ((qn + A) B), Definition 15.5. Let F0 = [0, 1], and let F1 = [0, 1 ] [ 2 , 1]. Continue recursively, defining 3 3 Fn+1 to be the disjoint union of the 2n+1 closed intervals got by deleting the open middle third of each of the 2n closed intervals comprising Fn . The set C= Fn n=0 is called the Cantor set. 4Here I've implicitly used translation-invariance of inner measure, which follows from translationinvariance of Lebesgue measure because K - x is a compact subset of A with (K - x) = (K) whenever K is a compact subset of A + x. Date: March 7, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. MAT 473 41 Proposition 15.6. Let C be the Cantor set. (i) C is compact. (ii) C has empty interior ; equivalently, [0, 1] \ C is dense in [0, 1]. (iii) (C) = 0. Proof. C is a closed subset of the compact set [0, 1], so is compact. This gives (i). Now fix x C, and let > 0. Choose N such that 1/3N < ; then B (x) properly contains the subinterval [k/3N , (k + 1)/3N ] of FN which contains x, and hence contains points of [0, 1] \ C. Thus B (x) C, and since x C and > 0 were arbitrary, this shows that C has empty interior, proving (ii). For (iii), note that C Fn and (Fn ) = 2n /3n for each n, so (C) (2/3)n for each n, which implies (C) = 0. Definition 15.7. Let (n ) be a sequence of positive real numbers such that 0 = 1 and n+1 < n /2 for each n. Let F0 = [0, 1], and let F1 = [0, 1 ] [1 - 1 , 1]. Continue recursively, defining Fn+1 to be the disjoint union of the 2n+1 closed intervals of length n+1 got by deleting the open middles of each of the 2n closed intervals comprising Fn . The set C= Fn n=0 is called a fat Cantor set. Notice that the "fat" Cantor set corresponding to n = 1/3n is just the usual Cantor set. Proposition 15.8. Let C be the fat Cantor set corresponding to the sequence (n ). Then C is compact, has empty interior, and (C) = lim 2n n . n Exercise 15.2. Prove the statement about (C) in Proposition 15.8. Also prove that for any (0, 1), there exists a sequence (n ) as in Definition 15.7 such that the corresponding fat Cantor set C has (C) = . Let C be the Cantor set, and define f : [0, 1] \ C [0, 1] by 2k - 1 f (x) = x is in the k th piece (from the left) of Fn-1 \ Fn . n 2 After some reflection you should agree that |f (y) - f (x)| < 1/2n for any x, y [0, 1] \ C such that |y - x| < 1/3n . Thus f is uniformly continuous on [0, 1] \ C, and since [0, 1] \ C is dense in [0, 1], f has a unique uniformly continuous extension to [0, 1]. Definition 15.9. The unique uniformly continuous extension of f to [0, 1] is called the Lebesgue function for the Cantor set C. Proposition 15.10. There exists a measurable subset of R which is not Borel. Date: March 7, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. 42 SPRING 2012 Proof. Let f be the Lebesgue function for the (usual) Cantor set, and define g : [0, 1] [0, 2] by g(x) = f (x) + x. Then g is continuous, strictly increasing (hence one-to-one), and onto; thus g -1 is also continuous, since (g -1 )-1 ((a, b)) = (g(a), g(b)) for any open interval (a, b) [0, 1]. It follows that B [0, 2] is a Borel set if and only if g -1 (B) [0, 1] is. Now g is simply translation on each of the countably many disjoint open intervals In comprising [0, 1] \ C, whence g [0, 1]\C = g In = g(In ) = (g(In )) = (In ) = [0, 1]\C = 1. n=1 n=1 n=1 n=1 It follows that (g(C)) = ([0, 2] \ g([0, 1] \ C)) = 2 - 1 = 1. In particular, (g(C)) > 0, so we may choose a nonmeasurable set B g(C). But then A = g -1 (B) is not Borel, since B is not even measurable; but A C, so A is measurable, since (C) = 0. Thus A is a measurable non-Borel subset of R. Possibly more interesting than the proposition itself is the construction in the proof of a continuous function which takes a set of measure 0 to a set of measure 1. Date: March 7, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. ...
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This document was uploaded on 03/11/2012.

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