Let $(\Omega, \mathcal A,\mathbb P)$ be such that $\Omega$ is countable and $\mathcal A = 2^{\Omega}$. Prove that almost sure convergence and convergence in probability are the same on this probability space. fa

  • $\begingroup$ Am I correct in assuming that you mean "Let $(\Omega, \mathcal{A}, \mathbb{P})$ be such that $\Omega$ is countable and..."? $\endgroup$ Mar 18, 2013 at 4:56
  • $\begingroup$ Of course, thanks for catching that and sorry for the typo. $\endgroup$
    – user66322
    Mar 18, 2013 at 4:59

1 Answer 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .