TITLE: How do you prove the almost sure convergence is not (in general) metrizable?
QUESTION [9 upvotes]: How do you prove the almost sure convergence is not (in general) metrizable?
Many thanks for your help.

REPLY [18 votes]: Here is another answer I like. It is easy to prove that, in a metric space, a sequence $(x_n)$ converges to some $x$ if and only if, from every subsequence $(x_{n'})$, one can extract a sub(sub)sequence $(x_{n''})$ that converges to $x$.
Now assume that almost sure convergence is metrizable and consider a sequence $(X_n)$ that converges to some $X$ in probability. Any subsequence $(X_{n'})$ will also converge to $X$ in probability, so by a well-known result, a subsubsequence $(X_{n''})$ that converges almost surely to $X$ can be extracted. Therefore, the whole sequence $(X_n)$ converges almost surely to $X$.
So, if almost sure convergence was metrizable, then it would be equivalent to convergence in probability.