Editing Monotone convergence theorem (section)

===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.