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).
Best Answer
$\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,
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 $\cF$.
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.
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.