Editing Monotone convergence theorem (section)

==Proof of theorem==
Set <math> f(x) = \sup_k f_k(x)</math> for the pointwise supremum at <math>x \in X</math>.

'''Step 1.''' The function <math>f</math> is <math> (\Sigma, \operatorname{\mathcal B}_{\bar\R_{\geq 0}}) </math>–measurable, and the integral <math>\textstyle \int_X f \,d\mu </math> is well-defined (albeit possibly infinite)<ref name="SCHECHTER1997"/>{{rp|at=section 21.3}}

''proof:'' From <math>0 \le f_k(x) \le \infty</math> we get <math>0 \le f(x) \le \infty</math>. Hence we have to show that <math>f</math> is <math>(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})</math>-measurable. For this, it suffices to prove that <math>f^{-1}([0,t])</math> is <math>\Sigma </math>-measurable for all <math>0 \le t \le \infty</math>, because the intervals <math>[0,t]</math> generate the [[Borel sigma algebra]] on <math>[0,\infty]</math> by taking countable unions, complements and countable intersections.

Now since the <math>f_k(x)</math> is a non decreasing sequence, 
<math> f(x) = \sup_k f_k(x) \leq t</math> if and only if <math>f_k(x)\leq t</math> for all <math>k</math>. Hence
:<math>f^{-1}([0, t]) = \bigcap_{k =1}^\infty f_k^{-1}([0,t]).</math>
is a measurable set, being the countable intersection of the measurable sets <math>f_k^{-1}([0,t])</math>.

Since <math>f \ge 0</math> the integral is well defined (but possibly infinite) as 
:<math> \int_X f \,d\mu = \sup_{s \in SF(f)}\int_X s \, d\mu</math>.

'''Step 2.''' We have the inequality

:<math>\sup_k \int_X f_k \,d\mu \le \int_X f \,d\mu </math>

''proof:'' We have <math>f_k(x) \le f(x)</math> for all <math>x \in X</math>, so <math>\int_X f_k(x) \, d\mu \le \int_X f(x)\, d\mu</math>  by "monotonicity of the integral" (lemma 1). Then step 2 follows by the definition of suprememum.

'''step 3''' We have the reverse inequality

:<math> \int_X f \,d\mu  \le \sup_k \int_X f_k \,d\mu </math>.

''proof:'' 
Denote by <math>\operatorname{SF}(f)</math> the set of 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 f</math> on <math>X</math>.
We need an "epsilon of room" to manoeuvre. 
For <math>s\in\operatorname{SF}(f)</math> and <math>0 <\varepsilon < 1</math>, define
:<math>B^{s,\varepsilon}_k=\{x\in X\mid (1 - \varepsilon) s(x)\leq f_k(x)\}\subseteq X.</math>

We '''claim''' the sets <math>B^{s,\varepsilon}_k</math> have the following properties: 
#<math>B^{s,\varepsilon}_k</math> is <math>\Sigma</math>-measurable. 
# <math>B^{s,\varepsilon}_k\subseteq B^{s,\varepsilon}_{k+1}</math>
#<math> X=\bigcup_{k = 1}^\infty B^{s,\varepsilon}_k</math>

Assuming the claim, by the definition of <math>B^{s,\varepsilon}_k</math> and "monotonicity of the Lebesgue integral" (lemma 1) we have
:<math>\int_{B^{s,\varepsilon}_k}(1-\varepsilon) s\,d\mu\leq\int_{B^{s,\varepsilon}_k} f_k\,d\mu \leq\int_X f_k\,d\mu \le \sup_k \int_X f_k \, d\mu
</math> 
Hence by "Lebesgue integral of a simple function as measure" (lemma 2), and "continuity from below" (lemma 3) we get:
:<math> (1- \varepsilon)\int_X s \, d\mu = 
\int_X (1- \varepsilon)s \, d\mu  = \sup_k
\int_{B^{s,\varepsilon}_k} (1-\varepsilon)s\,d\mu  
\le \sup_k \int_X f_k \, d\mu. </math>
so
:<math>(1-\varepsilon)\int_X f d\mu \le \sup_k\int_X f_k d\mu</math>
which since <math>\varepsilon</math> is arbitrary, proves step 3.   
 
Ad 1: Write  <math>s=\sum_{1 \le i \le m}c_i\cdot{\mathbf 1}_{A_i}</math>, for  non-negative constants <math>c_i \in \R_{\geq 0}</math>, and measurable sets <math>A_i\in\Sigma</math>, which we may assume are pairwise disjoint and with union <math>\textstyle X=\coprod^m_{i=1}A_i</math>.  Then for <math> x\in A_i</math> we have <math>(1-\varepsilon)s(x)\leq f_k(x)</math> if and only if <math> f_k(x) \in [( 1- \varepsilon)c_i, \,\infty],</math> so 
:<math>B^{s,\varepsilon}_k=\coprod^m_{i=1}\Bigl(f^{-1}_k\Bigl([(1-\varepsilon)c_i,\infty]\Bigr)\cap A_i\Bigr)</math>
which is measurable since the <math>f_k</math> are measurable.

Ad 2: For <math> x \in B^{s,\varepsilon}_k</math> we have <math>(1 - \varepsilon)s(x) \le f_k(x)\le f_{k+1}(x)</math> so <math>x \in B^{s,\varepsilon}_{k + 1}.</math>

Ad 3: Fix <math>x \in X</math>. If <math>s(x) = 0</math> then <math>(1 - \varepsilon)s(x) = 0 \le f_1(x)</math>, hence <math>x \in B^{s,\varepsilon}_1</math>. Otherwise, <math>s(x) > 0</math> and <math>(1-\varepsilon)s(x) < s(x) \le f(x) = \sup_k f(x)</math> so <math>(1- \varepsilon)s(x) < f_{N_x}(x)</math> for <math>N_x</math>
sufficiently large, hence <math>x \in B^{s,\varepsilon}_{N_x}</math>.

The proof of the monotone convergence theorem is complete.

===Relaxing the monotonicity assumption===
Under similar hypotheses to Beppo Levi's theorem, it is possible to relax the hypothesis of monotonicity.<ref>coudy (https://mathoverflow.net/users/6129/coudy), Do you know important theorems that remain unknown?, URL (version: 2018-06-05): https://mathoverflow.net/q/296540</ref>  As before, let <math>(\Omega, \Sigma, \mu)</math> be a [[measure (mathematics)|measure space]] and <math>X \in \Sigma</math>.  Again, <math>\{f_k\}_{k=1}^\infty</math> will be a sequence of <math>(\Sigma, \mathcal{B}_{\R_{\geq 0}})</math>-[[Measurable function|measurable]] non-negative functions <math>f_k:X\to [0,+\infty]</math>.  However, we do not assume they are pointwise non-decreasing.  Instead, we assume that <math display="inline">\{f_k(x)\}_{k=1}^\infty</math> converges for almost every <math>x</math>, we define <math>f</math> to be the pointwise limit of <math>\{f_k\}_{k=1}^\infty</math>, and we assume additionally that <math>f_k \le f</math> pointwise almost everywhere for all <math>k</math>.  Then <math>f</math> is <math>(\Sigma, \mathcal{B}_{\R_{\geq 0}})</math>-measurable, and <math display="inline">\lim_{k\to\infty} \int_X f_k \,d\mu</math> exists, and <math display="block">\lim_{k\to\infty} \int_X f_k \,d\mu = \int_X f \,d\mu.</math>