> [!theoremb] Theorem > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], and $\seq{f_j} \in L^+$ be a [[Sequence|sequence]] of $\cm$-[[Measurable Function|measurable functions]]. If $f_j \le f_{j+1}$ and $f_j \to f$ [[Pointwise Convergence|pointwise]], then the [[Integral|integral]] can be interchanged with the [[Limit|limit]]: > $ > \limv{j}\int f_j d\mu = \int \limv{j}f_j d\mu > $ > > *Proof[^1]*. See [[Upper and Lower Approximations]]. Let $f \in L^+$, then take the strict approximation > $ > \Sigma_f = \bracs{\phi \text{ simple}:0 \le \phi \le f, \phi(x) = f(x) \Leftrightarrow f(x) = 0} > $ > as the collection of [[Simple Function|simple functions]] strictly below $f$, and $\int f = \sup_{\phi \in \Sigma_f}\int \phi$. > > ### Weak MCT > > Let $\psi \in \Sigma_f$, and suppose that $A_n \upto X$ and $\chi_{A_n} \cdot \psi \in \Sigma_{f_n}$, then the [[Limit Superior and Inferior|lower limit]] can be controlled: > $ > \int \psi \le \liminf_{n \to \infty} \int f_n > $ > *Proof*. As the integral on simple functions is a [[Measure Space|measure]], we can use continuity from below to obtain $\int \chi_{A_n} \cdot \psi \upto \int \psi$. Since each $\chi_{A_n} \cdot \psi \in \Sigma_{f_n}$, > $ > \begin{align*} > \int \chi_{A_n} \cdot \psi &\le \int f_n \\ > \int \phi &\le \liminf \int f_n > \end{align*} > $ > by controlling the lower limit of the sequence. > > ### Full MCT > > To relate $\int f$ to $\int f_n$, we use a chain reaction of sequences, where one controls the next until we reach the end. > > Let $\phi \in \Sigma_f$ and let $E_{n} = \bracs{x \in X: \phi(x) \le f_n(x)}$. Then $\chi_{E_{n}} \cdot \phi \le f_n(x)$ and $\chi_{E_{n}} \cdot \phi \in \Sigma_{f_n}$. > > Moreover, if $f(x) < \infty$, > $ > \exists n \in \nat: f(x) - f_n(x) \le f(x) - \phi(x) \Rightarrow \phi(x) \le f_n(x) > $ > and if $f(x) = \infty$, > $ > \exists n \in \nat: \phi(x) \le f_n(x) > $ > since $f_n(x) \upto f(x)$. Therefore $E_{n} \upto X$. > > Using the weak MCT, we can obtain > $ > \int \phi \le \liminf \int f_n > $ > > Since $\liminf \int f_n \ge \int \phi \forall \phi \in \Sigma_f$, > $ > \int f \le \liminf_{n \to \infty} \int f_n = \lim_{n \to \infty}\int f_n > $ > and we can collapse the lower limit to obtain > $ > \int \limv{n}f_n = \limv{n}\int f_n > $ > [!theorem] > > Let $\seq{f_j} \subset L^+$ be an increasing sequence such that $\limv{j}f_j = f$ a.e., then $\int f = \limv{j}\int f_j$. > > *Proof*. $f - \limv{j}f_j$ is positive and equal to $0$ a.e. Therefore > $ > \int f = \int \paren{f - \limv{j}f_j} + \int \limv{j}f_j = > \limv{j}\int f_j > $ > by monotone convergence theorem. > [!theorem] > > Let $\bracs{f_n} \subset L^+$ be a finite or infinite [[Sequence|sequence]] in $L^+$ and $f = \sum_{n}f_n$, then the sum can be interchanged with the integral > $ > \int f = \sum_{n}\int f_n > $ > > *Proof*. First deal with finite sums by induction on $n$. Let $f_1, f_2 \in L^+$, then there exists increasing sequences $\seq{\phi_j}, \seq{\psi_j}$ such that $f_1 = \limv{j}\phi_j$ and $f_2 = \limv{j}\psi_j$. By [[Monotone Convergence Theorem|monotone convergence theorem]], $\limv{j}\int \phi_j = \int f_1$ and $\limv{j}\int \psi_j = \int f_2$. Since $\limv{j}(\phi_j + \psi_j) = f_1 + f_2$, > $ > \int(f_1 + f_2) = \limv{j}\int \phi_j + \limv{j}\int \psi_j = \int f_1 + f_2 > $ > Suppose that $\int\sum_{j = 1}^{n}f_n = \sum_{j = 1}^{n}\int f_n$, then > $ > \begin{align*} > \int \sum_{j = 1}^{n + 1}f_j &= \int \sum_{j = 1}^{n}f_j + \int f_{j + 1} \\ > &= \sum_{j = 1}^{n}\int f_j + \int f_{j + 1} \\ > &= \sum_{j = 1}^{n + 1}\int f_{j} > \end{align*} > $ > and the proposition holds for all $n \ge 2$. > > Now suppose that the sequence is infinite, then $\seq{\sum_{j = 1}^{n}f_j}$ is an increasing sequence of functions with $\limv{n}\sum_{j = 1}^{n}f_j = \sum_{n \in \nat}f_j$. By the [[Monotone Convergence Theorem|monotone convergence theorem]], > $ > \int f = \int \limv{n}\sum_{j = 1}^{n}f_j = \limv{n}\int\sum_{j = 1}^{n}f_j = \limv{n}\sum_{j = 1}^{n}\int f_j = \sum_{n \in \nat}\int f_n > $ [^1]: Alternate route that starts from the Uniform Convergence assumption used for Riemann integrals: [[From Uniform Convergence to Monotone Convergence]].