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How do you prove the almost sure convergence is not (in general) metrizable?

Many thanks for your help.

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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.

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  • $\begingroup$ This is very neat! Thanks a lot. $\endgroup$ – Frank Apr 8 '15 at 14:09
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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 $[0,1]$.

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