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Monotone convergence theorem
(section)
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===Proof=== This proof does ''not'' rely on [[Fatou's lemma]]; however, we do explain how that lemma might be used. Those not interested in this independency of the proof may skip the intermediate results below. ====Intermediate results==== We need three basic lemmas. In the proof below, we apply the monotonic property of the Lebesgue integral to non-negative functions only. Specifically (see Remark 4), ====Monotonicity of the Lebesgue integral==== '''lemma 1.''' let the functions <math>f,g : X \to [0,+\infty]</math> be <math>(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})</math>-measurable. *If <math>f \leq g</math> everywhere on <math>X,</math> then :<math>\int_X f\,d\mu \leq \int_X g\,d\mu.</math> *If <math> X_1,X_2 \in \Sigma </math> and <math>X_1 \subseteq X_2, </math> then :<math>\int_{X_1} f\,d\mu \leq \int_{X_2} f\,d\mu.</math> '''Proof.''' Denote by <math>\operatorname{SF}(h)</math> the set of [[simple function|simple]] <math>(\Sigma, \operatorname{\mathcal B}_{\R_{\geq 0}})</math>-measurable functions <math>s:X\to [0,\infty)</math> such that <math>0\leq s\leq h</math> everywhere on <math>X.</math> '''1.''' Since <math>f \leq g,</math> we have <math> \operatorname{SF}(f) \subseteq \operatorname{SF}(g), </math> hence :<math>\int_X f\,d\mu = \sup_{s\in {\rm SF}(f)}\int_X s\,d\mu \leq \sup_{s\in {\rm SF}(g)}\int_X s\,d\mu = \int_X g\,d\mu.</math> '''2.''' The functions <math>f\cdot {\mathbf 1}_{X_1}, f\cdot {\mathbf 1}_{X_2},</math> where <math>{\mathbf 1}_{X_i}</math> is the indicator function of <math>X_i</math>, are easily seen to be measurable and <math>f\cdot{\mathbf 1}_{X_1}\le f\cdot{\mathbf 1}_{X_2}</math>. Now apply '''1'''. =====Lebesgue integral as measure===== '''Lemma 2.''' Let <math>(\Omega,\Sigma,\mu)</math> be a measurable space, and let <math>s:\Omega\to{\mathbb R_{\geq 0}}</math> . be a simple <math>(\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}})</math>-measurable non-negative function. For a measurable subset <math>A \in \Sigma</math>, define :<math>\nu_s(A)=\int_A s(x)\,d\mu.</math> Then <math>\nu_s</math> is a measure on <math>(\Omega, \Sigma)</math>. ====Proof (lemma 2)==== Write <math>s=\sum^n_{k=1}c_k\cdot {\mathbf 1}_{A_k},</math> with <math>c_k\in{\mathbb R}_{\geq 0}</math> and measurable sets <math>A_k\in\Sigma</math>. Then :<math>\nu_s(A)=\sum_{k =1}^n c_k \mu(A\cap A_k).</math> Since finite positive linear combinations of countably additive set functions are countably additive, to prove countable additivity of <math>\nu_s</math> it suffices to prove that, the set function defined by <math>\nu_B(A) = \mu(B \cap A)</math> is countably additive for all <math>A \in \Sigma</math>. But this follows directly from the countable additivity of <math>\mu</math>. =====Continuity from below===== '''Lemma 3.''' Let <math>\mu</math> be a measure, and <math>A = \bigcup^\infty_{i=1}A_i</math>, where :<math> A_1\subseteq\cdots\subseteq A_i\subseteq A_{i+1}\subseteq\cdots\subseteq A </math> is a non-decreasing chain with all its sets <math>\mu</math>-measurable. Then :<math>\mu(A)=\sup_i\mu(A_i).</math> ====proof (lemma 3)==== Set <math>A_0 = \emptyset</math>, then we decompose <math> A = \coprod_{1 \le i } A_i \setminus A_{i -1} </math> as a countable disjoint union of measurable sets and likewise <math>A_k = \coprod_{1\le i \le k } A_i \setminus A_{i -1} </math> as a finite disjoint union. Therefore <math>\mu(A_k) = \sum_{i=1}^k \mu (A_i \setminus A_{i -1})</math>, and <math>\mu(A) = \sum_{i = 1}^\infty \mu(A_i \setminus A_{i-1})</math> so <math>\mu(A) = \sup_k \mu(A_k)</math>.
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