MON - Monotone Convergence Theorem Theorem 1. If 0 fn ↑ f...

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Unformatted text preview: Monotone Convergence Theorem Theorem 1. If 0 fn ↑ f then fn dµ ↑ f dµ. Proof. As fn increases to f it is plain that limn fn dµ f dµ. In order to establish the reverse inequality we retrace the way in which the integral is defined, considering first f simple and then f arbitrary. The case f simple is conveniently handled in two stages. Suppose first that f = a1A for some a ∈ [0, ∞] and A ∈ F . If a = 0 then there is nothing to do; so let a > 0 and choose α ∈ (0, a): from f fn fn 1{fn α1{fn α} α} it follows by monotonicity of the integral that f dµ αµ({fn fn dµ whence from n ↑ ∞ ⇒ {fn α}) α} ↑ A it follows that f dµ lim fn dµ n αµ(A); as α ∈ (0, a) was arbitrary, we deduce that f dµ lim n fn dµ aµ(A) = Now let f be simple: say M f= am 1Am m=1 1 f dµ. with a1 , . . . , aM nonnegative and A1 , . . . , AM ∈ F disjoint. We may rewrite this decomposition as M f= f 1Am m=1 and decompose fn similarly: M fn = fn 1Am . m=1 As fn 1Am ↑ f 1Am = am 1Am the case just handled yields fn 1Am dµ ↑ am µ(Am ) whereupon summation over 1 m M yields M M fn 1Am dµ ↑ M here, m=1 fn 1Am dµ = member (because M=1 m am µ(Am ) = fn dµ is squeezed between left member and right M f ) so we concludee that m=1 and fn f dµ = lim n whenever f f dµ; m=1 m=1 fn dµ 0 is simple. Finally, let f 0 be arbitrary. Choose any simple s with 0 s f : as s ∧ fn ↑ s ∧ f = s it follows from the simple (limit) case just covered that sdµ = lim n (s ∧ fn ) dµ lim n fn dµ; this being the case for all such s ensures (taking supremum) that f dµ lim n as required to complete the proof. 2 fn dµ ...
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This note was uploaded on 07/08/2011 for the course MHF 3202 taught by Professor Larson during the Spring '09 term at University of Florida.

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MON - Monotone Convergence Theorem Theorem 1. If 0 fn ↑ f...

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