# Measurability of a set in the definition of almost sure convergence

Many books define almost sure convergence as follows:

The sequence of random variables ${(X_n)}_{n \in \mathbb{N}}$ defined on the probability space $(\Omega, \mathcal{F}, P)$ converges almost surely to a random variable $X$ defined on the same probability space, if $$P(\{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}) = 1.$$

In connection to this question, I wonder if the set $A := \{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}$ is implicitly assumed to be measurable or whether it is actually a priori measurable. If the latter is true, how can one show this?

Assuming $X$ is real-valued, $$\{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\} = ((X-\limsup_n X_n)=0)\cap ((X-\liminf_n X_n)=0)$$
Since $\limsup_n X_n$, $\liminf_n X_n$ and $X$ are measurable, $((X-\limsup_n X_n)=0)\cap ((X-\liminf_n X_n)=0)$ is measurable.
• By the way, if you consider $A=\{\omega\,:\,\limsup_{n\to\infty} \lvert X_n(\omega)-X(\omega)\rvert=0\}$, it works for complex (and vector-valued) random variables as well. – user228113 Nov 2 '17 at 11:56
• What to do if $X :\Omega \rightarrow \overline{\mathbb{R}}$? – Holden Nov 2 '17 at 13:04
• @Holden: Break the event up into several pieces according to whether the random variables in question are finite or equal $+\infty$ or $-\infty$. – Nate Eldredge Nov 2 '17 at 13:44
$$A=\{\omega\in\Omega\mid\text{ for all }\epsilon>0\text{ there is } N\text{ such that } |X_n(\omega)-X(\omega)|<\epsilon\text{ for all }n\geq N\}$$ $$=\{\omega\in\Omega\mid\text{ for all }m\text{ there is } N\text{ such that } |X_n(\omega)-X(\omega)|<1/m\text{ for all }n\geq N\}$$ $$=\bigcap_{m=1}^\infty\bigcup_{N=1}^\infty\bigcap_{n=N}^\infty\{\omega\in\Omega\mid|X_n(\omega)-X(\omega)|<1/m\}.$$