If $X_n$ converges to X in probability then $X_n$ converges to X a.s. We know that if $X_n$ converges to X a.s. then $X_n$ converges to X in probability in the space ( Ω, A, P) . But the converse is not true in general . We can construct  such examples on any probability space (Ω, A, P) which has a “non-atomic part”, meaning that, there is an $Ω_0$ ∈ A with P($Ω_0$)>0, such that, for any A ∈ A, A ⊂ $Ω_0$, with P(A)>0 and any 0< δ <1, one has B ∈A, B ⊂A with P(B)= δP(A).
But when the probability space is “purely atomic”, not only do we have no counter-example as above, but indeed, convergence in probability implies a.s. convergence. My question is how can I show that if (Ω, A, P) be a discrete probability space, that is, Ω is a countable set and A consists of all subsets of Ω. Then for random variables $X_n$, n ≥ 0 and X on (Ω, A, P), $X_n$ converges to X in probability then $X_n$ converges to X a.s.
 A: Let's review the definitions.  To say that $X_n$ converges to $X$ in probability means that for any $\epsilon > 0$,
$$\lim\limits_{n \to \infty} P(|X_n - X| \geq \epsilon) = 0.$$
To say that $X_n$ converges to $X$ almost surely means that
$$P\Big( \{ s \in \Omega : \lim\limits_{n \to \infty} X_n(s) = X(s)\}\Big) = 1.$$
Proposition: For a discrete probability space $(\Omega, A, P)$, and random variables $X, X_n : n\in \mathbb N$, the following are equivalent:
(i): $X_n$ converges to $X$ in probability.
(ii): $X_n$ converges to $X$ almost surely.
(iii): $X_n$ converges to $X$ pointwise at all points of nonzero measure, which is to say that for every $s \in \Omega$ such that $P(s) > 0$, we have $\lim\limits_{n \to \infty} X_n(s) = X(s)$.
(i) $\Rightarrow$ (iii): Suppose that there exists an $s_0 \in \Omega$ with $\eta := P(s_0) > 0$ for which we do not have $\lim\limits_{n \to \infty} X_n(s_0) = X(s_0)$.  Then there exists an $\epsilon > 0$ such that $|X_n(s_0) - X(s_0)| \geq \epsilon$ for infinitely many $n$.  If we set
$$E_n = \{ s \in \Omega : |X_n(s) - X(s)| \geq \epsilon\}$$
then we have $s_0 \in E_n$ for infinitely many $n$.  In particular, $P(E_n) \geq P(s_0) = \eta$ for infinitely many $n$.  Therefore,
$$\lim\limits_{n \to \infty} P(E_n) = \lim\limits_{n \to \infty} P(|X_n - X| \geq \epsilon)$$
cannot be zero.  This shows that $X_n$ does not converge to $X$ in probability.
(iii) $\Rightarrow$ (ii): Suppose that $X_n$ does not converge to $X$ almost surely.  Then
$$P \Big( \{s \in \Omega : \lim\limits_{n \to \infty} X_n(s) = X(s)\} \Big) < 1.$$
Since $\Omega$ is a discrete probability space, the only way this can happen is if there exists an $s_0 \in \Omega$, with $P(s_0) > 0$, for which $\lim\limits_{n \to \infty} X_n(s) \neq X(s)$.
(ii) $\Rightarrow$ (i): This is well known.
A: We can do this by contrapositive. Assume that $X_n\not\to X$ a.s., then there exists a set $B$ with $\mathbb{P}(B)>0$ where $B$ is the set of all elements $i$ where $X_n(i)\not\to X(i)$. Since $\Omega$ is assumed to be countable, then $B$ is a countable set. Now the $\sigma-$algebra $\mathcal{A}=\mathcal{P}(\Omega)$ by the problem, so we can write $B$ in the following manner.
$$B= \bigcup_{i\in B}\{i\} $$
These singleton sets are disjoint and we have countable number of disjoint sets, thus
$$\mathbb{P}(B)=\sum_{i\in B}\mathbb{P}(\{i\}) $$
Since $\mathbb{P}(B)>0$ and the measure is positive, then there must exist some $j\in B$ such that $\mathbb{P}(\{j\})>0$. Otherwise you would have that $\mathbb{P}(B)=0$.
Since $j\in B$, then $X_n(j)\not\to X(j)$. This implies that for any $\epsilon > 0$, for any $N\in\mathbb{N}$ we can find $n\geq N$ such that $|X_n(j)-X(j)|>\epsilon$. So by monotonicity of $\mathbb{P}$ we have that for any $\epsilon>0$ that
$$\lim_{n\to\infty}\mathbb{P}(\{i: |X_n(i)-X(i)|>\epsilon\}) \geq \mathbb{P}(\{j\})>0$$
Hence $X_n$ does not converge to $X$ in probability.
A: Let $(\Omega, A, \mathbf{P})$ be a discrete probability space, $\{\omega_n: n \in \mathbb{N}\} \subseteq \Omega$ with $\sum_{n=1}^\infty \mathbf{P}(\{\omega_n\}) = 1$. Assume $(X_n)_n, X$ are random variables on $\Omega$ such that $X_n$ does not converge almost sure to $X$. We show that $X_n$ cannot converge to $X$ in probability neiter.
As $X_n \not \rightarrow X$ a.s. there is some $\omega := \omega_n \in \Omega$ such that $X_n(\omega) \not \rightarrow X(\omega)$ for $(n \rightarrow \infty)$ and $\mathbb{P}(\omega) =: \delta > 0$. Note that here we use that $\Omega$ is discrete. In particular it follows that for every $\epsilon > 0$ there are infintely many $n \in \mathbb{N}$ with $$|X_n(\omega) - X(\omega)| > \epsilon. $$ Hence we obtain $$\mathbf{P}(|X_n - X| > \epsilon) \geq \mathbb{P}(\{\omega\}) = \delta > 0 $$ for arbitrarily large $n \in \mathbb{N}$. Thus $X_n$ cannot converge to $X$ in probability.
