# Is showing almost surely convergence equivalent to lim sup = lim inf on a set with probability 1?

I know there are a lot of questions and answers concerning a.s. convergence on StackExchange, but I didn't find any addressing this in particular. What I am wondering is if you are given a problem of the variety that defines some sequence of random variables, say $X_n$ and you are asked to show that it converges almost surely, is this equivalent to showing that $P( \{ \omega \in \Omega: \lim \sup X_n(\omega) = \lim \inf X_n(\omega) \}) = 1$? ie, we are not given some $X$ and asked to show $X_n$ converges to $X$ almost surely, so is the best approach to show that the limit exists on a set with probability 1?

• Looks good. Note also that $\limsup X_n(\omega) = \liminf X_n(\omega) \iff \lim X_n(\omega)$ exists. So the event $\{\limsup X_n = \liminf X_n\}$ is the same as the event $\{\lim X_n \text{ exists}\}$. – Tom Nov 6 '13 at 16:27
• Provided the limsup and liminf are not infinite. – Did Nov 7 '13 at 9:43

As commenters said, the equality of $\liminf$ and $\limsup$ indeed implies the existence of limit, with the caveat about infinite limits: having $\lim=+\infty$ is usually considered diverging to $\infty$.
Is this the best approach when $X$ is not known? Depends. If you are using tools that naturally fit the language of $\liminf$ and $\limsup$ (Fatou's lemma), then probably yes. But often, what you are really doing is showing that $\{X_n(\omega)\}$ is a Cauchy sequence for a.e. $\omega$. Then the talk of upper/lower limits is not necessary, and would be a distraction.
Also consider that when $X$ is vector-valued, we don't have upper and lower limits at all.