Let $(f_n)$ in $L_p(\Omega)$ $1\leq p< \infty$ and $(g_n)$ bounded in $L_{\infty}(\Omega)$ assume that $f_n \rightarrow f$ in $L_p(\Omega)$ and $g_n \rightarrow g$ a.e. Prove that $$f_ng_n\rightarrow fg$$ in $L_p(\Omega)$.
I have done the following:
Note that $f_ng_n-fg=(f_n-f)g_n+f(g_n-g)$ therefore $$||f_ng_n-fg||_p=||(f_n-f)g_n+f(g_n-g)||_p\leq||f_n-f||_p||g_n||_{\infty}+||f(g_n-g)||_p$$ I know the first term is less than $\epsilon$ (because $f_n$ converges in $L_p(\Omega)$ to $f$) however there is a hint in the book that says $f(g_n-g)\rightarrow 0$ in $L_p(\Omega)$ by the dominated convergence theorem. Is the first part of my reasoning correct? how can I see what the hint is telling me? Thanks in advance.
The only remaining thing to is show that $\int_\Omega h_n\to 0$, where $h_n(x)=\left\lvert f(x)\left(g_n(x)-g(x)\right)\right\rvert^p$.
Since $f$ belongs to $\mathbb L^p$, the quantity $\left\lvert f(x) \right\rvert^p$ is finite for almost every $c$ hence the fact that $h_n(x)\to 0$ follows from the implication (if $c\geqslant 0$ and $a_n\to 0$) then $c\cdot a_n\to 0$.
For the domination condition, let $M:=\sup_{n\geqslant 1}\left\lVert g_n\right\rVert_\infty$. Then $\left\lvert g(x)\right\rvert\leqslant M$ for almost every $x$ hence $$\left\lvert h_n(x) \right\rvert\leqslant \left(2M\right)^p\left\lvert f(x) \right\rvert^p \mbox{ a.e.}$$