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Unformatted text preview: Math 245C Homework 4 Brett Hemenway June 1, 2007 Folland Chapter 7 4. Let X be an LCH space. (a) Suppose f C c ( X, [0 , )). Then f 1 ([( a, )) is closed since f is continuous, and f 1 ([ a, )) is compact since it is a closed set contained in the support of f which is compact. Now, f 1 ([ a, )) = f 1 n =1 ( a 1 n , ) ! = n =1 f 1 (( a 1 n , )) . But f 1 (( a 1 n , )) is open for each n since f is continuous, thus f 1 ([ a, )) is the countable intersection of open sets. (b) Let K be a compact G set, i.e. K = T n =1 U n , where the U n are open. Then by the Locally Compact version of Urysohns Lemma (Lemma 4.32), We can find an f n such that 0 f n 1, and f n  K = 1, and f n = 0 outside a compact subset of U n . Let f = X n =1 1 2 n f n . Then 0 f 1 and if x K , f ( x ) = X n =1 1 2 n f n ( x ) = X n =1 1 2 n = 1 . On the other hand, if x 6 K , then there exists an n such that x 6 U n , thus f n ( x ) = 0, which means f ( x ) 1 1 2 n < 1. So f 1 ( { 1 } ) = K . 1 11. Suppose is a Radon measure on X such that ( { x } ) = 0 for all x X , and A B X is such that 0 < ( A ) < . Since is Radon, for any < ( A ), we can find a compact set K A such that < ( K ), by possibly replacing A with K , we may assume that A is compact. For each x A , we have ( { x } ) = 0. Since is outer regular, for any > 0, we can find an open neighborhood E x of x such that ( E x ) < . Now, S x A E x covers A , and since A is compact, it has a finite subcover E x 1 ,...,E x n . Now, for any < ( A ), we can find an n such that  n [ i =1 E x i ! To find a set with measure exactly , we repeat this construction. Using the preceding construction, find E 1 such that  ( E 1 ) . Now, find E 2 E c 1 such that 2 ( E 2 )  ( E 1 ), continuing, we can find E n ( E 1 E n 1 ) c such that 2 n 1 ( E n )  n 1 [ i =1 E i !...
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 Spring '10
 tao/analysis
 Math

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