Weak Convergence of Product

Let $\Omega\subset \mathbb R^d$ be open (and possibly bounded), let $u_n,u,v_n,v\colon \Omega \to \mathbb R$ be measurable functions and suppose

• $u_n \to u$ a.e. in $\Omega$ and $\|u_n\|_{L^\infty(\Omega)} \leq C < \infty$,
• $v_n \rightharpoonup v$ in $L^1(\Omega)$ (where $\rightharpoonup$ denotes the weak convergence).

Is it true that $u_nv_n \rightharpoonup uv$ in $L^1(\Omega)$? What, if $L^1(\Omega)$ is replaced by $L^p(\Omega)$?

For $p> 1$, it holds. Let $g\in L^{p'}(\Omega)$ be some function. We have to prove that $$\int_\Omega gu_nv_n - guv$$ converges to $0$ as $n\to\infty$. To this end, note that $$\left|\int_\Omega gu_nv_n - guv\right| \leq \left|\int_\Omega gu_nv_n - guv_n \right|+ \left|\int_\Omega guv_n - guv\right|.$$ The first term on the right-hand side can be estimated by $$\left(\int_\Omega |gu_n - gu|^{p'}\right)^{1/p'} \|v_n\|_{L^p(\Omega)} ,$$ which converges to $0$, due to Lebesgue's dominated convergence theorem and the boundedness of $(v_n)$ in $L^p(\Omega)$. The second term converges to $0$, since $gu\in L^{p'}(\Omega)$ and $v_n\rightharpoonup v$ in $L^p(\Omega)$.
But for $p=1$, we have no convergence $gu_n \to gu$ in $L^\infty(\Omega)$, so I don't know how to procede.