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Let $(\Omega, \mathcal F,\mathbb P)$ be such that $\Omega$ is countable. I'm trying to find a simple example of random variables $X_n$ which converge to $0$ in probability but not a.s.

If $\mathcal F = 2^{\Omega}$ (i.e., $\{\omega\} \in \mathcal F$ for each $\omega$ since $\Omega$ is countable), it is shown here that such random variables do not exist. But now we only assume (of course) that the $X_n$ are measurable.

This question is essentially the same (but the measure there need not be finite).

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  • $\begingroup$ @Pedro I believe there is a gap in the proof below: it assumes that $\mathcal F$ is separable in order to show that the atoms are measurable. Could someone please explain to me if and how this can be fixed? Thanks! $\endgroup$ – Aubrey Feb 11 '15 at 18:36
  • $\begingroup$ Separable means that there is a countable dense subset, did you mean to say that I'm assuming that $\mathcal{F}$ is countable? I don't think I'm assuming anything of the sort, see my comment below my answer. $\endgroup$ – Pedro M. Feb 11 '15 at 19:39
  • $\begingroup$ @Pedro Sorry, I meant separated, i.e. that for each pair of distinct points in $X$, there is a set in $\mathcal F$ that contains exactly one of them (see Cohn, Measure Theory, p. 288). $\endgroup$ – Aubrey Feb 11 '15 at 19:49
  • $\begingroup$ Update: Pedro explained to me that there is no gap. $\endgroup$ – Aubrey Feb 11 '15 at 20:35
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$\newcommand{\NN}{\mathbb{N}} \newcommand{\cF}{\mathcal{F}} $ Very good question; I believe I managed to prove that there is no such counter-example, i.e., that the hypothesis $\cF = 2^{\Omega}$ is not necessary. More precisely,

Proposition: Let $(\Omega, \cF, \mathbb{P})$ be a countable probability space. Then convergence in probability implies almost sure convergence.

Given a countable $\Omega$ and a $\sigma$-algebra $\cF$, define an equivalence relation $\sim$ on $\Omega$, as follows: we write $\omega_1 \sim \omega_2$ if, for every $A \in \cF$, $\omega_1 \in A \Leftrightarrow \omega_2 \in A$.

Let $A_\omega$ denote the equivalence class of $\omega \in \Omega$. We claim that $A_\omega \in \cF$ but no proper subset of $A_\omega$ is in $\cF$. To see this, for each $\alpha \notin A_\omega$ let $B_\alpha \in \cF$ be such that $\alpha \in B_\alpha$ but $\omega \notin B_\alpha$. Then $$A_\omega = \left(\bigcup_{\alpha \in \Omega \setminus A_\omega} B_\alpha\right)^c \in \cF,$$ because the union is countable. From the definition of $\sim$, it follows that no proper subset of $A_\omega$ is in $\cF$.

Lemma: Every $A \in \cF$ is a (finite or countable) disjoint union of equivalence classes.

Proof: Let $A \in \cF$. If $\omega \in A$, then by definition we have $A_\omega \subset A$, so that $\bigcup_{\omega \in A} A_\omega \subset A$, which implies the result.

Corollary: A random variable $X$ is constant on each equivalence class $A_\omega$.

Proof: For each value $k$, $X^{-1}(k) \in \cF$, and thus is a union of equivalence classes.

Now just adapt the proof given here as if each equivalence class were a singleton, and you get the result.

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  • $\begingroup$ Great! It should perhaps be noted that $\{A_{\omega}\}$ is countable, which is clear since the map sending each $\omega$ to $A_{\omega}$ is surjective. $\endgroup$ – Aubrey Feb 11 '15 at 16:27
  • $\begingroup$ Wait... I just read it again and it seems that the existence of $B_\alpha \in \cF$ is not obvious. How exactly do you prove that $A_\omega \in \cF$? $\endgroup$ – Aubrey Feb 11 '15 at 17:54
  • $\begingroup$ @Aubrey Well, if $\alpha \notin A_\omega$, then either there exists $B_{\alpha} \in \cF$ such that $\alpha \in B, \omega \notin B$ or there exists $C \in \cF$ such that $\omega \in C, \alpha \notin C$: but in this case we may set $B_{\alpha} = C^c \in \cF$. $\endgroup$ – Pedro M. Feb 11 '15 at 19:30
  • $\begingroup$ Why? This is precisely the additional assumption that $\mathcal F$ is separated. $\endgroup$ – Aubrey Feb 11 '15 at 19:52
  • $\begingroup$ @Aubrey I don't assume that $\mathcal{F}$ is separable. Since $\mathcal{F}$ is a $\sigma$-algebra, $\Omega \in \mathcal{F}$, so that there exists at least one set containing both $\alpha$ and $\omega$. On the other hand, if all sets of $\mathcal{F}$ contain both, then $\alpha \sim \omega$. Therefore, if $\alpha \not\sim \omega$, there must exist some set containing either only $\alpha$, or only $\omega$. Since $\mathcal{F}$ is closed by complements, we're done. $\endgroup$ – Pedro M. Feb 11 '15 at 19:54

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