a.s. Convergence and Convergence in Probability

Solution 1:

We only show that convergence in probability impies a.s. convergence since a.s. convergence always implies convergence in probability.

Let $\{\omega_{n} : n \in \mathbb{N}\}$ be the set of elements whose singletons, which are always measurable by the hypothesis $\mathcal{A} = 2^{\Omega}$, have positive probability. It suffices to show that if $X_{n} \to X$ in probability, then $X_{n}(\omega_{i}) \to X(\omega_{i})$ for each $i \in \mathbb{N}$.

So fix $i \in \mathbb{N}$, $\epsilon > 0$ and assume that $X_{n} \to X$ in probability. Then there is $N$ such that $$\mathbb{P}(\{|X_n - X| \geq \epsilon \}) < \mathbb{P}(\omega_{i})$$

whenever $n \geq N$. This means, in particular, that if $n \geq N$, then $|X_{n}(\omega_{i}) - X(\omega_{i})| < \epsilon$. Done.