Almost dominated convergence from Tao's measure theory book

I am trying to solve the following exercise from Tao's measure theory book (Exercise 1.4.46).

Let $(X, B, \mu)$ be a measure space, and let $f_1, f_2, \ldots: X \to C$ be a sequence of measurable function that converge pointwise $\mu$-almost everywhere to a measurable limit $f: X \to C$. Suppose that there is an unsigned absolutely integrable function $G, g_1, g_2, \ldots: X \to [0, +\infty]$ such that the $|f_n|$ are pointwise $\mu$-almost everywhere bounded by $G + g_n$, and that $\int_{X} g_n d\mu \to 0$ as $n \to \infty$. Show that $$\lim_{n \to \infty} \int_{X} f_n d\mu = \int_{X} f d\mu$$

Note: $f$ is an absolutely integrable function if $\int_{X} |f| d\mu < \infty$.

Solution:

Without loss of generality, assume $f_n$ converge pointwise to $f$ and $|f_n|$ are pointwise bounded by $G+g_n$. Since $G+g_n$ is absolutely integrable and $|f_n| \le G + g_n$, then $f_n$ are absolutely integrable. The integrability of $f$ follows from the Fatou's lemma$$ \int_{X} |f| d\mu = \int_{X} \liminf_{n \to \infty}|f_n| d\mu \le \liminf_{n \to \infty} \int_{X} |f_n| d\mu \le \liminf_{n \to \infty} \int_{X} (G+g_n) d\mu = \int_{X} G d\mu < \infty$$

Now, by the triangle $|f_n - f| \le G + g_n + |f|$, hence$$ G + g_n + |f| - |f_n-f| \ge 0.$$Since $g_n$ are all non-negative, then $\liminf_{n \to \infty} g_n \ge 0$. Thus, $G(x) + |f(x)| \le G(x) + |f(x)| + \liminf_{n \to \infty} g_n(x)$ for all $x \in X$.By Fatou's lemma, we have\begin{align} \int_{X} (G + |f|) d\mu &\le \int_{X} (G+|f|+\liminf_{n \to \infty} g_n) d\mu \newline&= \int_{X} \liminf_{n \to \infty}(G+|f|+ g_n - |f_n-f|) d\mu \newline&\le \liminf_{n \to \infty} \int_{X}(G+|f|+ g_n - |f_n-f|) d\mu \newline&=\int_{X}(G+|f|) d\mu - \limsup_{n \to \infty} \int_{X}|f_n-f| d\mu,\end{align}because $\int_{X} g_n d\mu \to 0$ as $n \to \infty$. Since $\int_{X}(G+|f|) d\mu$ is finite, we have$$\limsup_{n \to \infty} \int_{X}|f_n-f| d\mu \le 0,$$thus $\int_{X}|f_n-f| d\mu \to 0$.Then,$$ \left|\int_{X} f_n d\mu - \int_{X} f d\mu \right| = \left|\int_{X} (f_n-f) d\mu \right| \le \int_{X} \left| f_n-f \right| d\mu \to 0.$$Therefore, $\lim_{n \to \infty} \int_{X} f_n d\mu = \int_{X} f d\mu$.

I am aware of solutions in the previous posts (A strong version of the Dominated Convergence Theorem and Almost Dominated Convergence Exercise (not Generalized Dominated Convergence Theorem)), but the proposed solutions used the relationship between different modes of convergence. This exercise is given before the discussion of modes of convergence, so I think I need not use such results.

I would like to know whether my solution is correct or not. The step I doubt most is where I bound $G+|f|$ using $\liminf_{n \to \infty} g_n$. I think it is correct, but if it is wrong, please correct me! Thanks.