MATH501_HW7

# MATH501_HW7 - Serge Ballif MATH 501 Homework 7 1 Let E be a...

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Serge Ballif MATH 501 Homework 7 November 16, 2007 1. Let E be a Borel subset of 2 . Prove that for every y , the slice E y is a Borel set in . Hint: Consider the collection C of subsets of 2 having the property that E y is a Borel set in for every y . Claim 1. C is a σ -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 ( x, y ) [ j =1 E j ) = [ j =1 { x | ( 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 | ( x, y ) E c } = { x | ( x, y ) E } c = ( E y ) c . Therefore C is a σ -algebra. We note that if P and Q are closed subsets of , 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 2 is a countable union of sets of the form P × Q . Therefore, C must contain all open sets. Thus, C contains the Borel σ -algebra on 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. Z (0 , ) Z (0 , ) xe - x 2 (1+ y 2 ) d x d y = Z (0 , ) " - e - x 2 (1+ y 2 ) 2(1 + y 2 ) 0 # d y = Z (0 , ) - 1 2(1 + y 2 ) d y = 1 2 arctan( y ) | 0 = π 4 Set A = R (0 , ) e - x 2 dx = π 2 . Then squaring both sides and converting to polar

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