Product of two functions converging in $L^1(X,\mu)$ 
Let $f_n\to f$ in $L^1(X,\mu)$, $\mu(X)<\infty$, and let $\{g_n\}$ be a sequence of measurable functions such that $|g_n|\le M<\infty\ \forall n$ with some constant $M$, and $g_n\to g$ almost everywhere. Prove that $g_nf_n\to gf$ in $L^1(X,\mu)$.

This is a question from one of my past papers, but unfortunately there is no solution provided.
Here is as far as I have gotten:
$$\int|gf-g_nf_n|=\int|(f-f_n)g_n+(g-g_n)f|\le\int|f-f_n||g_n|+\int|g-g_n||f|$$
$$\le M\int|f-f_n|+\int|g-g_n||f|$$
We know that $f_n\to f$ in $L^1$, so $\int|f-f_n|\to 0$, and by Lebesgue's bounded convergence theorem it follows that $\int|g-g_n|\to 0$. But I am unsure whether this also implies $\int|g-g_n||f|\to0$.

Hint: Observe that $2M|f|$ is an integrable bound for $|g_n - g|\cdot |f|$ and the latter converges a. e. to $0$. Now apply the bounded convergence theorem.
