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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 uppersumlowersum 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 \ (k1 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) welldefined 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. ...
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