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Unformatted text preview: Serge Ballif MATH 501 Homework 7 November 16, 2007 1. Let E be a Borel subset of R 2 . Prove that for every y R , the slice E y is a Borel set in R . Hint: Consider the collection C of subsets of R 2 having the property that E y is a Borel set in R for every y R . Claim 1. C is a algebra. Proof. C is nonempty because it contains the empty set. If E 1 ,E 2 ,... are subsets of C , then we know that ( E j ) y is Borel, and S j =1 ( E j ) y is Borel. We calculate [ j =1 E j ! y = ( x R ( x,y ) [ j =1 E j ) = [ j =1 { x R  ( x,y ) E j } = [ j =1 ( E j ) y . If E C , then we know E y is Borel and hence ( E y ) c is Borel. Therefore, ( E c ) y = { x R  ( x,y ) E c } = { x R  ( x,y ) E } c = ( E y ) c . Therefore C is a algebra. We note that if P and Q are closed subsets of R , then sets of the form P Q , are in C because ( P Q ) y = P if y Q and ( P Q ) y = otherwise. Every open set in R 2 is a countable union of sets of the form P Q . Therefore, C must contain all open sets. Thus, C contains the Borel algebra on R 2 . Thus, E is in C , i.e. E y is a Borel set for all y . 2. Let f ( x,y ) = xe x 2 (1+ y 2 ) for ( x,y ) I I where I = (0 , ) . Integrate f in two different ways and deduce that Z (0 , ) e x 2 dx = 2 . This is one of the most important integrals. The function f ( x,y ) is continuous and hence measurable. It is also positive, so by Tonellis Theorem we can integrate with respect to either variable....
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 Fall '08
 WYSOCKI,KRZYSZTOF
 Math, Sets

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