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Unformatted text preview: Math 245A Homework 3 Brett Hemenway October 31, 2005 Real Analysis Folland Chapter 1 18 Let A be an algebra, and A the collection of countable unions of sets in A . Let be a premeasure on A , and * the induced outer measure. a. Let E X then * ( E ) = inf X j =1 ( A j ) : A j A , E [ j =1 A j So for any > 0, we can find sets A j A with E S j =1 A j and X j =1 ( A j ) * ( E ) + Then letting A = S j =1 A j A , we have * ( A ) X j =1 * ( A j ) = X j =1 ( A j ) * ( E ) + 1 b. Suppose * ( E ) < , then if E is *measurable, we have, for each n , A n E , A n A with * ( E ) * ( A n ) * ( E ) + 1 n Letting B = T n =1 A n we have B A , E B and * ( E ) * ( B ) * ( E ) Using the measurability of E , we have * ( B ) = * ( B E ) + * ( B E c ) = * ( E ) + * ( B \ E ) so, subtracting, we conclude that * ( B \ E ) = 0. To show the converse, assume we have an A set B with E B , and * ( E ) = * ( B ). Then for F X * ( F ) * ( E F ) + * ( E c F ) by the subadditivity of * . We know that B is *measurable because B A which is contained in the algebra generated by A , and all sets in the algebra generated by A are *measurable. This gives * ( F ) = * ( B F ) + * ( B c F ) * ( E F ) + * ( B c F ) = * ( E F ) + * ( B c F ) + * ( B \ E F ) * ( E F ) + * (( B c B \ E ) F ) = * ( E F ) + * ( E c F ) So E is *measurable....
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.
 Spring '10
 tao/analysis
 Algebra, Sets

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