How do you prove the almost sure convergence is not (in general) metrizable? How do you prove the almost sure convergence is not (in general) metrizable?
Many thanks for your help.
 A: Suppose you had a metric $d$ on the bounded measurable functions on, say, the interval $[0,1]$ with Lebesgue measure, such that $f_n \to 0$ almost surely iff $d(f_n, 0) \to 0$.  For $t \in [0,1]$ and $r > 0$ let 
$$g_{t,r}(s) = \cases{1 & if $|s-t| < r$\cr 0 & otherwise\cr}$$
For each $t \in [0,1]$, $g_{t,1/n}$  converge a.s. to $0$ as $n \to \infty$, so there is $r(m,t)>0$ such
that $d( g_{t,r(m,t)}, 0) < 1/m$.  Now for each $m$ the intervals $\{(t-r(m,t), t+r(m,t)): t \in [0,1]\}$ cover $[0,1]$, so Heine-Borel says there is a finite subcover
$\{ (t_{j,m} - r(m,t_{j,m}), t_{j,m} + r(m,t_{j,m}): j = 1 \ldots N(m)\}$.  Consider the sequence of functions $G_k$ obtained by listing $g_{t_{j,m}, r(m,t_{j,m})}$, $j = 1 \ldots N(m)$, for each positive integer $m$.  We have
$d(G_k, 0) \to 0$.  However, for each $s\in [0,1]$ and each $m$, at least one $g_{t_{j,m},r(m,t_{j,m})}(s) = 1$, so this sequence does not converge to $0$ pointwise anywhere in $[0,1]$.
A: 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.
