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Unformatted text preview: 36 SPRING 2012 14. Measurability Criteria Theorem 14.1. L is a algebra of subsets of Rn , and is a measure on L. Proof. L0 , so L. If A L, then Ac M = M \A = M \(AM ) L0 for each M L0 , so Ac L. Finally, if Ai L for each i N and A = i Ai , then (A M ) (M ) < for each M L0 . Moreover, A M = (Ai M ),
i=1 so A M L0 by Theorem 13.6, whence A L. This shows that L is a algebra. Now () = 0 and (A) 0 for all A L by definition. If {Ai } is a pairwise disjoint collection of sets in L, then for every M L0 we have Ai M = (Ai M ) (Ai M ) (Ai ), since each Ai M L0 . Taking the sup over all M L0 gives (i Ai ) i (Ai ). For the reverse inequality, fix N N. Now take any Mi L0 , put M = N Mi , so M L0 . i (This is where N is important.) Thus: N N N (Ai Mi ) = (Ai Mi ) (Ai M )
i=1 i=1 i=1 i=1 i=1 i=1 i=1 Taking the sup over Mi L0 , we have i (Ai ) (i Ai ). i=1 (Ai M ) = N
i (Ai ) (i Ai ), and then since N was arbitrary, i=1 Ai M i=1 Ai . Exercise 14.1. For A Rn and x Rn , prove that A + x L if and only if A L, in which case (A + x) = (A). Exercise 14.2. For A Rn and c > 0, let cA = {cx  x A}. Prove that cA L if and only if A L, in which case (cA) = cn (A).
n At this point we know that (Rn , L, ) is a measure space: L is a algebra of subsets of R , and is a measure on L. We also know that L contains all open sets, and all closed sets, in Rn . A set N L is called Lebesguenull (or null, or just null ) if (N ) = 0 For Lebesgue measure, every subset of a null set is also measurable (and hence null), since if A N and N is null, then 0 (A) (A) (N ) = (N ) = 0, which forces (A) = (A) = 0. (We could have observed this a while ago.) Definition 14.2. Let F be the collection of all algebras of subsets of Rn which contain all the open sets. The Borel sets in Rn are the sets in the algebra B= A.
AF Date: February 21, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. MAT 473 37 In other words, B is the smallest algebra of subsets of Rn which contains the open sets. A subset of Rn is called an "F " set if it is a countable union of closed sets. A set which is a countable intersection of open sets in Rn is called a "G " set. Evidently, the F and G sets are Borel sets, and every Borel set is Lebesgue measurable. As we'll see later, there do exist Lebesguemeasurable sets which are not Borel (there are also Borel sets which are not F or G ). But the next theorem shows that the difference lies entirely in sets of measure zero. Theorem 14.3 (Approximation Property for L). A set B Rn is measurable if and only if for each > 0 there exist a closed set F and an open set G such that F B G and (G \ F ) < . Proof. Suppose the approximation property holds; so for each n N we may choose Fn closed and Gn open with Fn B Gn and (Gn \ Fn ) < 1/n. Put A= Fn ;
n=1 then A B, and B \ A Gn \ Fn for each n, so that for each n, which implies that (B \ A) = 0. Thus B \ A is measurable, and since A is measurable (in fact F ), the set B = A (B \ A) is measurable too. Conversely, suppose B is measurable, and fix > 0. For each n, let then {Dn  n N} is a partition of Rn by elements of L0 . Thus B Dn is in L0 for each n, so by the Approximation Property for L0 , we can choose Kn compact and Gn open with Kn B Dn Gn and (Gn \ Kn ) = (Gn \ Kn ) < /2n . Now let F = Kn and G= Gn ;
n=1 n=1 (B \ A) (Gn \ Fn ) = (Gn \ Fn ) < 1/n Dn = {x  n  1 x < n}; clearly G is open and F B G. Moreover, G\F Gn \ F n , n=1 n so (G \ F ) n (Gn \ Fn ) < n 1/2 = . To see that F is closed (which is the only thing which remains to be checked), suppose x F . Then x k Dn for some k, which is an open set. Thus any sequence in F which n=1 converges to x is eventually in F (k Dn ) = k Kn . But this latter set is closed, and n=1 n=1 therefore contains x, so we must have x F . Date: February 21, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. 38 SPRING 2012 Corollary 14.4. For A Rn , the following are equivalent: (i) A is Lebesgue measurable. (ii) There exists a F set B and a G set C such that B A C and (C \ B) = 0. (iii) There exists a Borel set B and a Lebesguenull set N such that A = B N . Exercise 14.3. Prove Corollary 14.4. Date: February 21, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. ...
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 Spring '09
 Algebra, Sets

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