Editing Monotone convergence theorem (section)

===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. Consider a simple <math>(\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}})</math>-measurable non-negative function <math>s:\Omega\to{\mathbb R_{\geq 0}}</math>. For a measurable subset <math>S \in \Sigma</math>, define
:<math>\nu_s(S)=\int_Ss\,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(S)=\sum_{k =1}^n c_k \mu(S\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_A(S) = \mu(A \cap S)</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>S = \bigcup^\infty_{i=1}S_i</math>, where
:<math>
S_1\subseteq\cdots\subseteq S_i\subseteq S_{i+1}\subseteq\cdots\subseteq S
</math>
is a non-decreasing chain with all its sets <math>\mu</math>-measurable. Then
:<math>\mu(S)=\sup_i\mu(S_i).</math>

====proof (lemma 3)====
Set <math>S_0 = \emptyset</math>, then 
we decompose <math> S = \coprod_{1 \le i } S_i \setminus S_{i -1} </math>  as a countable disjoint union of measurable sets and likewise <math>S_k = \coprod_{1\le i \le k } S_i \setminus S_{i -1} </math> as a finite disjoint union. Therefore
<math>\mu(S_k) = \sum_{i=1}^k \mu (S_i \setminus S_{i -1})</math>, and <math>\mu(S) = \sum_{i = 1}^\infty \mu(S_i \setminus S_{i-1})</math> so <math>\mu(S) = \sup_k \mu(S_k)</math>.