For the Dominated Convergence to apply, you must have that the $f_n$'s are dominated by a function $\varphi$ whose integral is finite.
Now I added one extra condition to the given because I don't see how it is possible without it. So all in all the conditions become
- $\lim_{n\to\infty}f_n=f$
- $\lim_{n\to\infty}g_n=g$
- $|f_n|\leq g_n$.
- $\lim_{n\to\infty}\int g_nd\mu=\int g d\mu$
- $\int g d\mu<\infty$ (extra condition)
Now notice that it is enough to show
$$\lim_{n\to\infty}\int|f_n-f|d\mu=0 $$
The result you want follows immediately from the triangle inequality.
To start the proof, define $$\varphi_n=g_n+g-|f_n-f| .$$
Then $\varphi_n$ is positive measurable since
$$|f_n|\leq g_n\implies|f|\leq g\implies\varphi_n\geq g_n+g-(|f_n|+|f|)\geq g_n+g-(g_n+g)=0.$$
Also notice that $\varphi_n\to 2g$ as $n\to\infty$ almost everywhere on $E$.
Therefore, we can use Fatou's Lemma to deduce
$$\int 2g d\mu=\int2\lim_{n\to\infty}g_nd\mu=
\int\lim_{n\to\infty}\varphi_nd\mu=
\int\liminf_{n\to\infty}\varphi_nd\mu\leq\liminf_{n\to\infty}\int \varphi_nd\mu,$$
and furthermore,
$$\varphi_n\leq g_n+g\implies\int \varphi_nd\mu\leq\int(g_n+g)d\mu\implies
\limsup_{n\to\infty}\int \varphi_nd\mu\leq\int 2gd\mu.$$
Hence
$$\lim_{n\to\infty}\int \varphi_nd\mu=\int 2g d\mu.$$
Since
$$\int (g_n+g) d\mu=\int\varphi_nd\mu+\int |f_n-f|d\mu,$$
then by letting $n\to\infty$ we see that
$$\int 2gd\mu=\int 2gd\mu+\lim_{n\to\infty}\int|f_n-f|d\mu.$$
Before you rush and do the cancellation, you need to make sure that $\int2gd\mu$ is finite which is why I needed the extra condition. You are done. Notice that this is very similar to the proof of the Dominated Convergence.
