Let $(X_{n,m})_{n\geq 1,m\geq1}$ be a double sequence of random variables such that $X_{n,m}\Rightarrow X_m$ (weak convergence) as $n\rightarrow\infty$, $X_{n,m}\rightarrow X_n$ (almost sure convergence) as $m\rightarrow\infty$, and $X_m\rightarrow X$ (almost sure convergence) as $m\rightarrow\infty$. I am trying to determine whether these conditions are sufficient to establish that $X_n \Rightarrow X$ as $n\rightarrow\infty$, or if not, what additional assumptions would have to be made in order to ensure the result. This is what I have done so far. Let $f$ be continuous and bounded. Using continuity of $f$ and then its boundedness (to apply dominated convergence theorem) we have: $$\mathbb{E}(f(X))=\lim_{m\rightarrow\infty}\mathbb{E}(f(X_m))=\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}\mathbb{E}(f(X_{n,m})).$$ If I could interchange the two above limits, I would be able to conclude, since: $$\mathbb{E}(f(X))=\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}\mathbb{E}(f(X_{n,m}))=\lim_{n\rightarrow\infty}\mathbb{E}(f(X_{n})),$$ also by continuity of $f$ and dominated convergence. However, I am having difficulties proving that we can effectively interchange these two limits (to do so, one of the two limits needs to be uniform), and wondering if the assumptions I currently have are sufficient? Any comments or ideas would be greatly appreciated.
update: I think it's probably better to consider $f$ to be continuous with compact support instead, as we can therefore use the uniform continuity of $f$.
