b If E 1 F 1 E 2 F 2 M and E 1 F 1 E 2 F 2 where E i M and F i is a subset of a

# B if e 1 f 1 e 2 f 2 m and e 1 f 1 e 2 f 2 where e i

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(b) If E 1 F 1 , E 2 F 2 M and E 1 F 1 = E 2 F 2 where E i ∈ M and F i is a subset of a set N i of measure 0, then by monotonicity and subadditivity μ ( E 1 F 1 ) = μ ( E 2 F 2 ) μ ( E 2 N 2 ) μ ( E 2 )+ μ ( N 2 ) = μ ( E 2 )+0. Hence μ ( E 2 F 2 ) μ ( E 2 ). Monotonicity gives us the reverse inequality, so μ ( E 1 F 1 ) = μ ( E 2 F 2 ) = μ ( E 2 ). By a symmetric argument μ ( E 1 F 1 ) = μ ( E 2 F 2 ) = μ ( E 1 ). (c) We must show that μ ( ) = 0 and that if { E i F i } i =1 is a sequence of dis- joint sets in M , then μ ( S i =1 E i ) = i =1 μ ( E i ). Since ∅ ∈ M we know that μ ( ) = μ ( ) = 0. It follows immediately from part (b) that μ satisfies the rules for monotonicity and subadditivity. Applying the subadditivity of μ , part (b), additivity of μ , and monotonicity of μ (in that order) we get μ [ i =1 ( E i F i ) ! X i =1 μ ( E i F i ) = X i =1 μ ( E i ) = μ [ i =1 E i ! μ [ i =1 ( E i F i ) ! . Therefore, μ ( S i =1 E i ) = i =1 μ ( E i ), so μ is a measure. To see that μ is complete we note that if μ ( E F ) = 0, then μ ( E ) = 0 = μ ( F ). If E 0 F 0 E F where E 0 E and F 0 F , then each of the sets ( E F ) ( E F 0 ) ( E 0 F 0 ) is in M by definition. 3.Let(fn)be a sequence of measurable functionsfn:X. Prove that: (a) The set L = { x X | lim n →∞ f n ( x ) exists } belongs to M . (b) If ( X, M , μ ) is complete and f n f μ -a.e., then f is measurable. (a) We will break up the proof into two claims that combine to prove that L is measurable. Claim 1. { x X | lim n →∞ f n ( x ) = ∞} and { x X | lim n →∞ f n ( x ) = -∞} are measurable sets. Claim 2. If we assume that each function f n is finite valued, then L is measurable. Proof of Claim 1. Both sets { x X | lim n →∞ f n ( x ) = ∞} = T N =1 S n =1 { x | f n ( x ) > N } , and { x X | lim n →∞ f n ( x ) = -∞} = T N =1 S n =1 { x | f n ( x ) < - N } are count- able intersections of countable unions of measurable sets. Hence they are mea- surable. Page 3 of 5
Serge Ballif MATH 501 Homework 2 October 3, 2007 Proof of Claim 2. We note first that the real valued sequence lim n →∞ f n ( x ) exists if and only if f n ( x ) is a Cauchy sequence. That is, for > 0 there exists N

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• Fall '08
• WYSOCKI,KRZYSZTOF
• Math, measure, Lebesgue measure, Lebesgue integration, lim sup Ej, Hj Ej