For a counter-example to reverse Fatou lemma without the domination hypothesis, take $f_n:=\chi_{(n,n+1)}$, with $X$ the real line, Borel $\sigma$-algebra and Lebesgue measure. We have $\limsup_{n\to +\infty}f_n(x)=0$ for all $x$ but $\int f_nd\mu=1$.
Since OP asks for an alternative way, I give a constructive approach. I prefer this since we see the truth along the proof. Moreover, the skill of converting an uncountable union into a countable one is often re-used.
For any fixed $\epsilon' > 0$,
\begin{align}
& \{x \in X : f_n(x) \not\to f_n(x) \} \\
=& \{x \in X : \exists \epsilon > 0, \forall N \in \Bbb{N}, \exists n \ge N, |f_n(x) - f(x)| > \epsilon \} \\
=& \bigcup_{\epsilon > 0} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon \}\\
=& \bigcup_{k \in \Bbb{N}} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \}.
\end{align}
Show that the last equality is true:
- $\subseteq$: take $k$ large enough so that $\epsilon > \epsilon'/2^k$
- $\supseteq$: for any $k \in \Bbb{N}$, choose $\epsilon$ sufficiently small so that $\epsilon'/2^k > \epsilon$
For each $k \in \Bbb{N}$, invoke the given condition (with $\epsilon = \epsilon'/2^k$) to find $N_k \in \Bbb{N}$ so that
$$\mu\left(\bigcup_{n = N_k}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \}\right) < \epsilon'/2^k.$$
It's not hard to check that
$$\bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \} \subseteq \bigcup_{n = N_k}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \},$$
from which we get our desired conclusion
\begin{align}
& \mu\left(\bigcup_{\epsilon > 0} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon \}\right) \\
=& \mu\left(\bigcup_{k \in \Bbb{N}} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \}\right) \\
\le& \sum_{k \in \Bbb{N}} \mu\left( \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \} \right) \\
\le& \sum_{k \in \Bbb{N}} \mu\left( \bigcup_{n = N_k}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \} \right) \\
\le& \sum_{k \in \Bbb{N}} \epsilon'/2^k = \epsilon'.
\end{align}
Since $\epsilon' > 0$ is arbitrary, we conclude that $\mu(\{f_n \not\to f\}) = 0$, i.e. $f_n \to f$ a.e.
Best Answer
$|f_k| \leq g$ for all $k$ so $|f| \leq g$ almost everywhere. Hence $|f_k-f| \leq 2g$ and $2g$ is integrable. Hence $\int |f_k-f| \to 0$ by DCT.