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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online