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# hw6 - Math 245A Homework 6 Brett Hemenway 46 Let X = Y =[0...

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Math 245A Homework 6 Brett Hemenway November 30, 2005 46. Let X = Y = [0 , 1], M = N = B [0 , 1] , μ = Lebesgue measure, and ν = counting measure. Let D = { ( x, x ) : x [0 , 1] } ⊂ X × Y . Then D x = { ( x, x ) } and D y = { ( y, y ) } , thus χ D dμdν = 0 = 0 and χ D dνdμ = 1 = 1 Let { A i × B i } be a sequence of rectangles in M × N such that D i =1 A i × B i . For each x [0 , 1], x A j × B j for some j , then x A j B j . So [0 , 1] i =1 ( A i B i ). Since μ ([0 , 1]) = 1, we have that μ ( A j B j ) > 0 for some j . Then μ ( A j ) > 0, and μ ( B j ) > 0, so ν ( B j ) = . Thus i =1 μ × ν ( A i × B i ) = for any covering set { A i × B i } . Thus μ × ν ( D ) = , so have χ D d ( μ × ν ) = μ × ν ( D ) = 49. a. Let E ∈ M × N and μ × ν ( E ) = 0. Then by Fubini’s theorem 0 = μ × ν ( E ) = χ E × ν = χ E x = χ E y So ν ( E x ) = μ ( E y ) = 0. b. Let f be L measurable and f = 0 λ -a.e. Let A = { ( x, y ) X × Y : f ( x, y ) = 0 } 1

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We have λ ( A ) = 0, so there exists a set E ∈ M × N with A E and μ × ν ( E ) = 0. Then by Fubini’s theorem 0 = fdμ × ν = f x = f y Thus ν ( E x ) = μ ( E y ) = 0. But A x E x and A y E y , so since μ and ν are complete, A x , A y are measurable and have measure zero. To prove Theorem 2.39, Let f be a L -measurable function. Then by Theorem 2.12, there is a μ × ν measurable function g such that f = g λ -a.e. If (i) f 0, or (ii) f L 1 ( λ ), then applying lemma (b) to f - g we have that f x is N -measurable for a.e. x and f y is M -measurable for a.e. y . then by Fubini’s theorem x g x and y f y are measurable, and by lemma (b) g x = f x for a.e. x and g y = f y for a.e. y . So x f x and x f y are measurable. If we are in case (ii), then
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hw6 - Math 245A Homework 6 Brett Hemenway 46 Let X = Y =[0...

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