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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 σalgebra. 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 σalgebra. 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 σalgebra 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 = √ π 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 Tonelli’s Theorem we can integrate with respect to either variable....
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 Fall '08
 WYSOCKI,KRZYSZTOF
 Math, Sets

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