$\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.

Yes, it's true. By the very definition of $\liminf$, there exists a subsequence such that

$$\liminf_{n \to \infty} \mathbb{E}(X_n ) = \lim_{k \to \infty} \mathbb{E}(X_{n_k}). \tag{1}$$

Since, by assumption, $X_{n} \to X$ in probability (hence in particular $X_{n_k} \to X$ in probability), we can choose a further subsequence of $(X_{n_k})_{k \in \mathbb{N}}$ which converges almost surely to $X$. Now the claim follows directly from Fatou's lemma and $(1)$.

## Best Answer

Let $F=\{\lim_{n\to\infty}X_n=X\}$ and suppose that $\mathsf{P}(F)<1$. Since $F^c$ is countable, there exists $\omega'\in F^c$ such that $\mathsf{P}(\{\omega'\})=p>0$. Thus, there exists a subsequence $n_k$ such that $|X_{n_k}(\omega')-X(\omega')|>\epsilon>0$. However, this implies $$ \mathsf{P}(|X_{n_k}-X|>\epsilon)\ge \mathsf{P}(\{\omega'\})=p\not\to 0. $$