The answer is negative even if $f,g_n$ are supported in $[0,1]$ and $g_n \to 0$ uniformly. The reason is that $f \in L^1$ does not suffice for the Hardy-Littlewood maximal function to be in $L^1$, so choosing $g_n$ to capture that maximal function  will yield a counterexample. It is still interesting to see an explicit counterexample. I will add that next.