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Unformatted text preview: Mathematics 105, Spring 2004 — Problem Set IV Solutions 1 IV.A Let { I n } be any finite set of open intervals that covers [0 , 1] ∩ Q . Show that ∑ n  I n  ≥ 1. Explain why this does not prove that  [0 , 1] ∩ Q  e ≥ 1. Solution. This does not prove that  [0 , 1] ∩ Q  e ≥ 1, because in the definition of outer measure, infinite covers are also allowed. This problem shows that there is a dramatic difference between outer measure and the variant that would be obtained by allowing only finite covers. Suppose that { I n : 1 ≤ n ≤ N } is some open cover of [0 , 1] ∩ Q by finitely many intervals. We claim more generally that any finite cover of ( A, B ) ∩ Q by intervals I n must satisfy ∑ n  I n  ≥ B A . We prove this by induction on N , the case N = 1 being obvious. Clearly there must exist some index k such that either A ∈ I k , or A is the left endpoint of I k . Let b be the right endpoint of I k . If b > B then already  I k  ≥ B A , and we’re done. If b < B then { I n : n 6 = k } is a cover of ( b, B ) ∩ Q by N 1 intervals, so by induction ∑ n 6 = k  I n  ≥ 1 b . Thus ∑ n  I n  ≥ ( B b ) +  I k  ≥ ( B b ) + ( b A ) = B A . IV.B Show that for any Lebesgue measurable sets A, B ⊂ R n ,  A ∪ B  +  A ∩ B  =  A  +  B  . Solution. Decompose A ∪ B = ( A \ B ) ∪ ( A ∩ B ) ∪ ( B \ A ); the three sets on the righthand side are pairwise disjoint. Since B is a σalgebra, all three of these sets are measurable. Therefore  A ∪ B  =  A \ B  +  A ∩ B  +  B \ A  and consequently the lefthand side of our inequality is  A ∪ B  =  A \ B  + 2  A ∩ B  +  B \ A  . Now A = ( A \ B ) ∪ ( A ∩ B ) and the two sets on the right are disjoint, so  A  =  A \ B  +  A ∩ B  . Likewise  B  =  B \ A  +  A ∩ B  . Thus the inequality holds....
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 Spring '04
 MichaelChrist
 Math, #, Cantor, 2K, Lebesgue measure, Stroock, Michael Christ

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