Editing Monotone convergence theorem (section)

====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>.