Editing Monotone convergence theorem (section)

==Beppo Levi's lemma==
The following result is a generalisation of the monotone convergence of non negative sums theorem above to the measure theoretic setting. It is a cornerstone of measure and integration theory with many applications and has [[Fatou's lemma]] and the [[dominated convergence theorem]] as direct consequence. It is due to [[Beppo Levi]], who proved a slight generalization in 1906 of an earlier result by [[Henri Lebesgue]].
<ref name="BigRudin">{{cite book 
  |last1=Rudin 
  |first1=Walter 
  |title=Real and Complex Analysis 
  |date=1974 
  |publisher=Mc Craw-Hill 
  |page=22 
  |edition=TMH}}
</ref>
<ref>{{Citation
  | last1 = Schappacher
  | first1 = Norbert
  | author-link1 = Norbert Schappacher
  | last2 = Schoof
  | first2 = René
  | author-link2 = René Schoof
  | title = Beppo Levi and the arithmetic of elliptic curves
  | journal = [[The Mathematical Intelligencer]]
  | volume = 18
  | issue = 1
  | year = 1996
  | doi = 10.1007/bf03024818
  | mr = 1381581
  | zbl = 0849.01036
  | url = http://irma.math.unistra.fr/~schappa/NSch/Publications_files/1996_RSchNSch.pdf
  | page = 60
| s2cid = 125072148
 }}</ref>

Let <math>\operatorname{\mathcal B}_{\bar\R_{\geq 0}}</math> denotes the <math>\sigma</math>-algebra of Borel sets on the upper extended non negative real numbers <math>[0,+\infty]</math>.  By definition, <math>\operatorname{\mathcal B}_{\bar\R_{\geq 0}}</math> contains the set <math>\{+\infty\}</math> and all Borel subsets of <math>\R_{\geq 0}.</math>

===Theorem (monotone convergence theorem for non-negative measurable functions)===
Let <math>(\Omega,\Sigma,\mu)</math> be a [[measure (mathematics)|measure space]], and <math>X\in\Sigma</math> a measurable set. Let <math>\{f_k\}^\infty_{k=1}</math> be a pointwise non-decreasing sequence of <math>(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})</math>-[[Measurable function|measurable]] non-negative functions,  i.e. each function <math>f_k:X\to [0,+\infty]</math> is <math>(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})</math>-measurable and for every <math>{k\geq 1}</math> and every <math>{x\in X}</math>,

:<math> 0 \leq \ldots\le f_k(x) \leq f_{k+1}(x)\leq\ldots\leq \infty. </math>

Then the pointwise supremum
:<math> \sup_k f_k : x \mapsto \sup_k f_k(x) </math>
is a <math>(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})</math>-measurable function and

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

'''Remark 1.''' The integrals and the suprema may be finite or infinite, but the left-hand side is finite if and only if the right-hand side is.

'''Remark 2.''' Under the assumptions of the theorem,
{{ordered list|type=lower-alpha
| <math>\textstyle \lim_{k \to \infty} f_k(x) = \sup_k f_k(x) =  \limsup_{k \to \infty} f_k(x) = \liminf_{k \to \infty} f_k(x) </math>
| <math>\textstyle \lim_{k \to \infty} \int_X f_k \,d\mu = \sup_k \int_X f_k \,d\mu = \liminf_{k \to \infty} \int_X f_k \,d\mu =  \limsup_{k \to \infty} \int_X f_k \,d\mu.  </math>
}}

Note that the second chain of equalities follows from monoticity of the integral (lemma 2 below). Thus we can also write the conclusion of the theorem as

:<math> \lim_{k \to \infty} \int_X f_k(x) \, d\mu(x) = \int_X \lim_{k\to \infty} f_k(x) \, d\mu(x)
</math>
with the tacit understanding that the limits are allowed to be infinite.

'''Remark 3.''' The theorem remains true if its assumptions hold <math>\mu</math>-almost everywhere. In other words, it is enough that there is a [[null set]] <math>N</math> such that the sequence <math>\{f_n(x)\}</math> non-decreases for every <math>{x\in X\setminus N}.</math> To see why this is true, we start with an observation that allowing the sequence <math>\{ f_n \}</math> to pointwise non-decrease almost everywhere causes its pointwise limit <math>f</math> to be undefined on some null set <math>N</math>. On that null set, <math>f</math> may then be defined arbitrarily, e.g. as zero, or in any other way that preserves measurability. To see why this will not affect the outcome of the theorem, note that since <math>{\mu(N)=0},</math> we have, for every <math>k,</math>

:<math> \int_X f_k \,d\mu = \int_{X \setminus N} f_k \,d\mu</math> and <math>\int_X f \,d\mu = \int_{X \setminus N} f \,d\mu, </math>

provided that <math>f</math> is <math>(\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}})</math>-measurable.<ref name="SCHECHTER1997">See for instance {{cite book |first=Erik |last=Schechter |title=Handbook of Analysis and Its Foundations |location=San Diego |publisher=Academic Press |year=1997 |isbn=0-12-622760-8 }}</ref>{{rp|at=section 21.38}} (These equalities follow directly from the definition of the Lebesgue integral for a non-negative function).

'''Remark 4.''' The proof below does not use any properties of the Lebesgue integral except those established here. The theorem, thus, can be used to prove other basic properties, such as linearity, pertaining to Lebesgue integration.

===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>.