Suppose $X_n\to X$ in $L^1$ and $V_n\to V$ in $L^1$ and $(V_n)$ is a bounded sequence. I'm trying to show that then $\mathbb{E}X_nV_n\to \mathbb{E}XV$.
One has for all $N\in\mathbb{N}$ $$|\mathbb{E}X_nV_n-\mathbb{E}XV_n+\mathbb{E}XV_n-\mathbb{E}XV|\leq \mathbb{E}|V_n||X_n-X|+N\mathbb{E}(V_n-V)1_{\{|X|\leq N, V_n\geq V\}}+ N\mathbb{E}(V-V_n)1_{\{|X|\leq N, V\geq V_n\}}+|\mathbb{E}X(V_n-V)1_{\{|X|>N\}}|.$$ Now the first three members on the right converge to $0$ as $n\to\infty$. How does one go about the last one? Or maybe it should be done some other way?
For the last term, use $$ |\mathbb{E} X(V_n-V) 1_{|X|>N}| \leq C |\mathbb{E} X 1_{\{|X|>N\}}|, $$ where $C$ is our bound on $V_n$. So using your argument above you get a bound like $$ \lim_{n\to\infty} |\mathbb{E}X_nV_n - \mathbb{E}XV| \leq C |\mathbb{E} X1_{\{|X|>N\}}|, $$ which holds for all $N$. But $X\in L^1$, so as $N\to\infty$ the right hand side goes to $0$. Thus the limit on the left is $0$, so $\mathbb{E}X_nV_n\to \mathbb{E}XV$.