# hw2 - Math 245A Homework 2 Brett Hemenway October 19, 2005...

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Math 245A Homework 2 Brett Hemenway October 19, 2005 Real Analysis Gerald B. Folland Chapter 1. 13. Every σ -ﬁnite measure is semiﬁnite. Let μ be a σ -ﬁnite measure on X, M . Since μ is σ -ﬁnite, there exist a sequence of sets { E n } n =1 in M , with μ ( E n ) < and S n =1 E n = X . Let F ∈ M with μ ( F ) = . Deﬁne a new sequence A n = n [ i =1 E n F This is an increasing sequence in M , and S n =1 = F . The continuity from below of μ gives us that lim n →∞ μ ( A n ) = μ ( lim n →∞ A n ) = μ ( F ) = So for any C > 0 there exists an N such that μ ( A N ) > C . But μ ( A N ) < because μ ( A n ) N n =1 μ ( E n ) and μ ( E n ) is ﬁnite for each n . A N F , so we conclude that F is semiﬁnite. 17. Let μ * be an outer measure on X , and { A j } j =1 a sequence of disjoint μ * measurable sets. μ * ( E ( [ j =1 A j )) = μ * ( [ j =1 E A j ) X j =1 μ * ( E A j ) Since μ * is subadditive. To prove the opposite inequality, we slightly modify the argument on 1

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page 30. Let B n = S n j =1 A j . Then the B n form an increasing sequence of μ * - measurable sets. Let B = S j =1 A j = S n =1 B n . Then μ * ( E B n ) = μ * ( E B n A n ) + μ * ( E B n A c n ) = μ * ( E A n ) + μ * ( E B n - 1 ) So by induction μ * ( E B n ) = n j =1 μ * ( E A j ). This gives μ * ( E ) = μ * ( E B n ) + μ * ( E B c n ) μ * ( E B n ) + μ * ( E B c ) = n X j =1 μ * ( E A j ) + μ * ( E B c ) Then taking the limit as
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hw2 - Math 245A Homework 2 Brett Hemenway October 19, 2005...

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