TITLE: Measurability of a set in the definition of almost sure convergence
QUESTION [6 upvotes]: 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?

REPLY [10 votes]: $$A=\{\omega\in\Omega\mid\text{ for all }\epsilon>0\text{ there is } N\text{ such that } |X_n(\omega)-X(\omega)|<\epsilon\text{ for all }n\geq N\}$$
$$=\{\omega\in\Omega\mid\text{ for all }m\text{ there is } N\text{ such that } |X_n(\omega)-X(\omega)|<1/m\text{ for all }n\geq N\}$$
$$=\bigcap_{m=1}^\infty\bigcup_{N=1}^\infty\bigcap_{n=N}^\infty\{\omega\in\Omega\mid|X_n(\omega)-X(\omega)|<1/m\}.$$

REPLY [5 votes]: 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.