The converse does not hold. Take $X = [0,\infty)$ with Lebesgue measure. Let $a_n$ be a sequence of positive numbers such that $a_n \to 0$ but $\sum_n a_n = \infty$. (For instance, $a_n = 1/n$ would work.) Let $b_n = \sum_{i=0}^n a_k$. Then let $f_n = 1_{[b_{n-1}, b_{n})}$. Then $f_n \to 0$ a.e. and $\int f_n dm = a_n \to 0$ also. But if $f$ is a dominating function, we must have $f \ge \sup_n f_n = 1$ which is not integrable.
For a necessary and sufficient condition for convergence, look up the Vitali convergence theorem.
I think there is a point of reviving this post because I think this is also a nice proof pointed out to me when I posted this question by Chris Janjigian.
I have seen this question posted numerous times on this site, so I think there is a point of writing this out.
Let us consider the sequence $\int |f_n-f|$. then consider the following, take a subsequence $\int |f_{n_j}-f|$
For $f_{n_j}$, there must exist a sub-subsequence $f_{n_{j_k}}$ such that $f_{n_{j_k}}$ converges to $f(x)$ almost everywhere. (Since $f_n$, hence $f_{n_j}$ converges to $f(x)$ in measure)
It must also be the case $|f_{n_{j_k}}-f| \leq 2g$, we now apply dominated convergence to see that $\int|f_{n_{j_k}}-f| \rightarrow 0$
What we have shown is that, for every subsequence of $\int |f_n-f|$, we have a further subsequence, which converges to 0. Now using the lemma:
If for every subsequence of $x_n$, there exists a sub-subsequence which converges to 0, then $x_n$ converges to 0.
We are done.
The proof of the last lemma can be found Sufficient condition for convergence of a real sequence
Best Answer
Here is another proof:
I'll prove that $\displaystyle a_n:=\int_\Omega f_nd\mu\longrightarrow l:=\int_\Omega fd\mu$ as $n$ tends to $\infty$.
By a very useful fact in Analysis, it's enough to prove that for each subsequence of $\{a_n\}$ like $\{a_{n_k}\}$, there is a subsequence of $\{a_{n_k}\}$ like $\{a_{n_{k_l}}\}$ which converges to $l$.
Now, given a subsequence $\{a_{n_k}\}$, we have $$f_{n_k}\xrightarrow[]{\;\mu\;}f.\tag{I}$$
By this fact, there exists a subsequence of $\{f_{n_k}\}$ like $\{f_{n_{k_l}}\}$ which $$f_{n_{k_l}}\xrightarrow{\;a.e\;}f.\tag{II}$$ (Note that for $\text{(II)}$ we should have assumed that $\Omega$ is of finite measure, however we can get rid of it as @Leo Lerena pointed in the comments.)
Since $\{f_{n_{k_l}}\}$ is dominated by function $g\in\mathcal{L}^1(\Omega,\mu)$, by the original version of $LDCT$, $$\int_\Omega f_{n_{k_l}}\xrightarrow{\;a.e\;}\int_\Omega f.\tag{III}$$ That is : $$a_{n_{k_l}}\xrightarrow{n\rightarrow\infty}l \tag*{$\square$.}$$
Note
The technique of using subsequences, rather than sequences, is one of the most powerful tools of proof !