$\int \lvert f_n-f\rvert^p g\rightarrow 0$ implies convergence in measure?

Question

I was thinking about whether $\int_{\Omega} \lvert f_n-f\rvert g\rightarrow 0$ implies $f_ng\rightarrow fg$ in measure. Here $f_n$ is a sequence in $L^1(\Omega)$, $f\in L^1(\Omega)$, $g \in C_0(\Omega)$ and $\Omega \subseteq \mathbb{R}^n$ is open and nonempty.

So clearly, $L^p$ convergence implies convergence in measure. In this Case, $g$ might bei negative, so $L^p$ convergence of $f_n$ to $f$ might not be true, even on the support of $g$.

So could there be a counterexample?

Answer 1

Counter-example: $\Omega =(-1,1),f(x)=0, f_n(x)=x^{2} $ for all $n$ and let $g(x)$ be any odd function with compact support, not identically $0$.