Theorem 16.2.4 (Monotone Convergence Theorem, [Theorem 2.14, Fol99]). Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n}\subset \mathcal{L}^{+}(X, \cm)$, and $f \in \mathcal{L}^{+}(X, \cm)$ such that $f_{n} \upto f$ pointwise, then
\[\limv{n}\int f_{n} d\mu = \int f d\mu\]
Proof. By Definition 16.2.2, $\int f_{n} d\mu \le \int f d\mu$ for each $n \in \natp$.
Let $\phi \in \Sigma^{+}(X, \cm)$ with $\phi|_{\bracs{f > 0}}< f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $f_{n} \upto f$, $\bracs{f_n \ge \phi}\upto X$. In which case, since $A \mapsto \int \one_{A} \phi d\mu$ is a measure (Proposition 16.1.3),
\[\limv{n}\int f_{n} d\mu \ge \limv{n}\int \one_{\bracs{f_n \ge \phi}}\cdot \phi d\mu = \int \phi d\mu\]
by continuity from below (Proposition 14.1.5). Therefore $\limv{n}\int f_{n} d\mu \ge \int f$ by Lemma 16.2.3.$\square$