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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.
 Spring '09
 LARSON

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