Theorem 16.3.6 (Dominated Convergence Theorem [Theorem 2.24, Fol99]). Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n}\subset \mathcal{L}^{1}(X)$, and $f:X \to \complex$ such that
$f_{n} \to f$ pointwise.
There exists $g \in \mathcal{L}^{1}(X)$ such that $\abs{f_n}\le \abs{g}$ for all $n \in \natp$.
then $\int fd\mu = \limv{n}\int f_{n} d\mu$.
Proof. By (1) and (2), $f$ is measurable with $\int \abs{f}d\mu \le \int \abs{g}d\mu < \infty$, so $f \in \mathcal{L}^{1}(X)$. Now, since $g + f, g - f \ge 0$, by Fatou’s lemma and Proposition 16.3.4,
\begin{align*}\int g d\mu + \int f d\mu&\le \liminf_{n \to \infty}\int g + f_{n} d\mu = \int g d\mu + \liminf_{n \to \infty}f_{n} d\mu \\ \int g d\mu - \int f d\mu&\le \liminf_{n \to \infty}\int g - \int f_{n} d\mu = \int g d\mu - \limsup_{n \to \infty}\int f_{n} d\mu\end{align*}
so
\[\limsup_{n \to \infty}\int f_{n} d\mu \le \int f d\mu \le \liminf_{n \to \infty}\int f_{n} d\mu\]
and $\int f d\mu = \limv{n}\int f_{n} d\mu$.$\square$