---- The following result gives one condition under which [[function limit|limits]] and [[integral|Lebesgue integration]] commute. > [!theorem] Theorem. ([[monotone convergence theorem for nonnegative measurable functions]]) > Suppose $(X, \Sigma, \mu)$ is a [[measure|measure space]] and $0 \leq f_{1} \leq f_{2} \leq \dots$ is a pointwise-increasing sequence of [[measurable function|measurable functions]] $X \to [0, \infty]$ . Then $\lim_{k \to \infty} \int f_{k} \, d\mu =\int f \, d\mu ,$ where $f:X \to [0, \infty]$ is the [[pointwise converge|pointwise limit]] of the $f_{k}$, i.e., $f(x):=\lim_{k \to \infty}f_{k}(x)$. ^theorem > [!proof]- Proof. ([[monotone convergence theorem for nonnegative measurable functions]]) > Note the function $f$ is [[measurable function|measurable]] by the result in [[measurable function]]. > > Since $f_{k}(x) \leq f(x)$ for each $k \in \mathbb{N}$ and $x \in X$, $\int f_{k} \, d\mu \leq \int f \, d\mu$ for each $k \in\mathbb{N}$, and so $\lim_{k \to \infty}\int f_{k} \, d\mu \leq \int f \, d\mu$ follows immediately from monotonicity of the [[integral]]. > > To show the reverse inequality, we will utilize the [[integral#^equivalence|equivalent]] definition of Lebesgue integration: suppose $A_{1},\dots,A_{m}$ are disjoint sets in $\Sigma$ and $s=\sum_{j=1}^{m} c_{j} \chi_{A_{j}}$ is a nonnegative [[simple function]] ($c_{1},\dots,c_{m} \in [0, \infty)$) with $s \leq f$. We have to show $\sum_{j=1}^{m} c_{j} \mu(A_{j}) \leq \lim_{k \to \infty} \int f_{k} \, d\mu$ . > > Let $t \in (0,1)$. For $k \in \mathbb{N}$, let $E_{k}:=\left\{ x \in X : f_{k}(x) \geq t\sum_{j=1}^{m} c_{j} \chi_{A_{j}}(x) \right\}$ > Then $E_{1} \subset E_{2} \subset \dots$ is an increasing sequence of sets in $\Sigma$ whose union equals $X$. Thus, for each $j \in [m]$, $A_{j} \cap E_{1} \subset A_{j} \cap E_{2} \subset \dots$ is an increasing union of sets in $\Sigma$ whose union equals $A_{j}$. Using the [[measure|continuity from below]] property of [[measure]], we therefore have $\mu(A_{j})=\lim_{k \to \infty} \mu(A_{j} \cap E_{k})$ for each $j \in [m]$. By construction, $f_{k}(x) \geq \sum_{j=1}^{m}tc_{j} \chi_{A_{j} \cap E_{k}}(x) \text{ for all }x \in X.$ > Integrating both sides, we see $\int f_{k} \, d\mu \geq t \sum_{j=1}^{m} c_{j} \mu(A_{j} \cap E_{k}).$ > Then taking the [[function limit]] as $k \to \infty$: $\lim_{k \to \infty} \int f_{k} \, d\mu \geq t \sum_{j=1}^{m} c_{j} \mu(A_{j}).$Now taking the limit as $t$ increases to $1$ recovers the desired result. > > ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```