In fact, we can prove the following more generalized conclusion.
Theorem (The extended dominated convergence theorem or $E D C T) . \quad$ Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $f_{n}, g_{n}: \Omega \rightarrow \mathbb{R}$ be $\langle\mathcal{F}, \mathbb{R}\rangle$-measurable functions such that $\left|f_{n}\right| \leq g_{n}$ a.e. ( $\left.\mu\right)$ for all $n \geq 1$. Suppose that
(i) $g_{n} \rightarrow g$ a.e. $(\mu)$ and $f_{n} \rightarrow f$ a.e. $(\mu)$;
(ii) $g_{n}, g \in L^{1}(\Omega, \mathcal{F}, \mu)$ and $\int\left|g_{n}\right| d \mu \rightarrow \int|g| d \mu$ as $n \rightarrow \infty$. Then, $f \in$ $L^{1}(\Omega, \mathcal{F}, \mu)$,
$$
\lim _{n \rightarrow \infty} \int f_{n} d \mu=\int f d \mu \quad \text { and } \quad \lim _{n \rightarrow \infty} \int\left|f_{n}-f\right| d \mu=0
$$
Proof: By Fatou's lemma,
$$
\int|f| d \mu \leq \liminf _{n \rightarrow \infty} \int\left|f_{n}\right| d \mu \leq \liminf _{n \rightarrow \infty} \int\left|g_{n}\right| d \mu=\int|g| d \mu<\infty
$$
Hence, $f$ is integrable. For proving the second part, let $h_{n}=f_{n}+g_{n}$ and $\gamma_{n}=g_{n}-f_{n}, n \geq 1$. Then, $\left\{h_{n}\right\}_{n \geq 1}$ and $\left\{\gamma_{n}\right\}_{n \geq 1}$ are sequences of nonnegative integrable functions. By Fatou's lemma and (ii),
$$
\begin{aligned}
\int(f+g) d \mu &=\int \liminf _{n \rightarrow \infty} h_{n} d \mu \\
& \leq \liminf _{n \rightarrow \infty} \int h_{n} d \mu \\
&=\liminf _{n \rightarrow \infty}\left[\int g_{n} d \mu+\int f_{n} d \mu\right] \\
&=\int g d \mu+\liminf _{n \rightarrow \infty} \int f_{n} d \mu
\end{aligned}
$$
Similarly,
$$
\int(g-f) d \mu \leq \int g d \mu-\limsup _{n \rightarrow \infty} \int f_{n} d \mu .
$$
According to the linearity of integral,$\int(g \pm f) d \mu=\int g d \mu \pm \int f d \mu .$ Hence,
$$
\int f d \mu \leq \liminf _{n \rightarrow \infty} \int f_{n} d \mu
$$
and
$$
\limsup _{n \rightarrow \infty} \int f_{n} d \mu \leq \int f d \mu
$$
yielding that $\lim _{n \rightarrow \infty} \int f_{n} d \mu=\int f d \mu$. For the last part, apply the above argument to $f_{n}$ and $g_{n}$ replaced by $\tilde{f}_{n} \equiv\left|f-f_{n}\right|$ and $\tilde{g}_{n} \equiv g_{n}+|f|$, respectively.
This certificate is quoted from Measure Theory and Probability Theory. Krishna B. Athreya, Soumendra N. Lahiri.