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Unformatted text preview: Math 8601: REAL ANALYSIS. Fall 2010 Homework #5. Problems and Solutions. #1. Let E be a Lebesgue measurable set in R 1 with Lebesgue measure m ( E ) > 0. Show that for any < 1, there is an open interval I = ( a,b ) such that m ( E I ) > m ( I ). Proof. Suppose that this property is not true, i.e. m ( E I ) m ( I ) for any interval I = ( a,b ) , where = const < 1 . Fix a small > 0, such that (1+ ) < 1. By Theorem 1.18, there is an open set U E such that m ( U ) < (1 + ) m ( E ). By Proposition 0.21, the open set U is represented as a finite or countable union of disjoint open intervals I j . Therefore, m ( U ) = m ( E U ) = m E [ j I j = m [ j ( E I j ) = X j m ( E I j ) X j m ( I j ) = m ( U ) < (1 + ) m ( E ) < m ( E ) . This contradiction proves the desired property. #2. Let ( X, M ) be a measurable space, and let f n : X R 1 be a sequence of Mmeasurable functions on X . Show that the set E := { x X : lim...
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 Fall '08
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 Math

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