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solns4 - Mathematics 105 Spring 2004 — Problem Set IV...

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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 right-hand side are pairwise disjoint. Since B is a σ-algebra, all three of these sets are measurable. Therefore | A ∪ B | = | A \ B | + | A ∩ B | + | B \ A | and consequently the left-hand 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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solns4 - Mathematics 105 Spring 2004 — Problem Set IV...

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