NnbljG-13

NnbljG-13 - MAT 473 33 13. Lebesgue measure: inner and...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAT 473 33 13. Lebesgue measure: inner and outer measures Proposition 13.1. Let Ai be subsets of RN . (i) If A2 , then (A1 ) (A2 ) and (A1 ) (A2 ). A1 (ii) i=1 Ai i=1 (Ai ). (iii) If the Ai are disjoint, then Ai (Ai ). i=1 i=1 Proof. (i) is routine. For (ii), let > 0 and for each i choose an open set Gi containing Ai with (Gi ) < (Ai ) + /2i . Then i Gi is an open set containing i Ai , so Ai Gi (Gi ) < ( (Ai ) + /2i ) = (Ai ) + . i=1 i=1 i=1 Since > 0 was arbitrary, we have (i Ai ) i (Ai ). Similarly, for (iii), fix N N and for each i let Ki be a compact subset of Ai . Then N Ki i is a compact subset of i Ai , and since the union of Ki 's is finite and disjoint we have i=1 N N Ai Ki = (Ki ). i=1 i=1 i=1 i=1 Taking the sup over all such Ki gives (i Ai ) we have (i Ai ) i (Ai ). Definition 13.2. Let For A L0 we set (A) = (A). N i (Ai ), and now since N was arbitrary L0 = {A Rn | (A) = (A) < }. By Exercise 12.4, all the compact sets and all the open sets with finite measure are in L0 , and the definition of (A) just given agrees with the existing definitions when A is compact or open. Lemma 13.3. Let A and B be disjoint members of L0 . Then A B L0 , and (A B) = (A) + (B). Proof. (AB) (A)+ (B) by Proposition 13.1 (ii), so AB has finite outer measure. Furthermore, (using Proposition 13.1 (iii) and disjointness), so equality holds throughout. (A) + (B) = (A) + (B) (A B) (A B) Theorem 13.4 (Approximation Property for L0 ). Suppose A Rn with (A) < . Then A L0 if and only if for every > 0 there exists a compact set K and an open set G such that K A G and (G \ K) < . Proof. Suppose A L0 , and let > 0. Since (A) < , by the definition of outer measure and the approximation property of infima, there exists an open set G containing A such that (G) < (A) + /2. Similarly, there exists a compact subset K of A such that Date: February 16, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. 34 SPRING 2012 (K) > (A) - /2. But G = (G \ K) K is a disjoint union, and G \ K is in L0 because it is open; thus Lemma 13.3 implies that Conversely, suppose the approximation property holds. Then for every > 0 there exist K A G such that (G \ K) < ; it follows (again using Lemma 13.3) that Since > 0 was arbitrary, this gives (A) (A), so equality holds, so A L0 . (A) (G) = (K) + (G \ K) < (A) + . (G \ K) = (G) - (K) < ( (A) + /2) - ( (A) - /2) = . Exercise 13.1. Fix a < b in R, and suppose f : [a, b] [0, ) is (bounded and) Riemann integrable. Prove that the set A = {(x, y) | a x b and 0 y f (x)} R2 b is in L0 , with (A) = a f (x) dx. Hint: Use the upper-sum-lower-sum characterization of Riemann integrability. Corollary 13.5. L0 is closed under finite unions, finite intersections, and set differences, and is monotonic on L0 . Proof. We have disjoint unions, so by the algebra of sets it suffices to show that L0 is closed under set differences. So suppose A, B L0 , and fix > 0. Now choose compact sets KA A and KB B, and open sets GA A and GB B such that (GA \ KA ) < /2 and similarly for B. Then K = KA \ GB is a compact subset of A \ B, G = GA \ KB is an open set containing A \ B, and G \ K (GA \ KA ) (GB \ KB ), so that It follows from Theorem 13.4 that A \ B L0 . For monotonicity, if A B in L0 , then (B) = (A)+(B \A) (A), using Lemma 13.3. Theorem 13.6. If Ai L0 for each i N, A = i Ai , and (A) < , then A L0 and (i Ai ) i (Ai ). If in addition the Ai are disjoint, then equality holds. Proof. For the disjoint case, Ai (Ai ) = (Ai ) Ai Ai , i=1 i=1 i=1 i=1 i=1 (G \ K) (GA \ KA ) + (GB \ KB ) < /2 + /2 = . so equality holds, so i Ai L0 . Otherwise, set B1 = A1 , let B2 = A2 \ A1 , and in general put Bk = Ak \ (k-1 Ai ). Then i=1 the Bi are disjoint, they're all in L0 , and i Bi = i Ai , so i Ai L0 by the above. Moreover, Bi Ai for each i, so Ai = Bi = (Bi ) (Ai ). i=1 i=1 i=1 i=1 Date: February 16, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. MAT 473 35 Definition 13.7. Let We call the elements of L the Lebesgue measurable sets in Rn . For A L, we define the Lebesgue measure of A to be (A) = sup{(A M ) | M L0 }. By the "intersection" part of Corollary 13.5, we have L0 L. Moreover, by the "monotonic" part of Corollary 13.5, is (still) well-defined on L0 . Exercise 13.2. For each k N, let Sk = [-k, k] [-k, k] [-k, k] Rn . Prove that a set A Rn is in L if and only if A Sk L0 for all k. Also prove that for A L we have (A) = lim (A Sk ). k L = {A Rn | A M L0 for all M L0 }. Date: February 16, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. ...
View Full Document

This document was uploaded on 03/11/2012.

Ask a homework question - tutors are online