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

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Monotone Convergence Theorem Theorem 1. If 0 6 f n f then R f n d μ R f d μ . Proof. As f n increases to f it is plain that lim n R f n d μ 6 R f d μ . 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 = a 1 A 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 > f n > f n 1 { f n > α } > α 1 { f n > α } it follows by monotonicity of the integral that Z f d μ > Z f n d μ > αμ ( { f n > α } ) whence from n ↑ ∞ ⇒ { f n > α } ↑ A it follows that Z f d μ > lim n Z f n d μ > αμ ( A ); as α (0 , a ) was arbitrary, we deduce that Z f d μ > lim n Z f n d μ > ( A ) = Z f d μ. Now let f be simple: say f = M X m =1 a m 1 A m 1
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with a 1 , . . . , a M nonnegative and A 1 , . . . , A M ∈ F disjoint. We may rewrite this decomposition as f = M X m =1 f 1 A m and decompose f n similarly: f n = M X m =1 f n 1 A m . As f n 1 A m f 1 A m = a m 1 A m the case just handled yields Z f n 1 A m d μ a m μ ( A m
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