$g \circ X$ continuous a.s., $X \to X_{n}$ a.s., then $g(X_{n}) \to g(X)$ a.s.?

Question

Let $\{X_{n}\}$ be a sequence of random variables converging almost surely to a random variable $X$ (i.e. $X_{n} \overset{a.s.}{\to} X$).

Let $g: \mathbb{R} \to \mathbb{R}$ be a measurable function and let $D_{g} = \{x : x \text{ is discontinuity point of $g$}\}$. Suppose that $$\mathbb{P}(X \in D_{g}) = 0$$

I am wondering if it is true that $g(X_{n})\overset{a.s.}{\to} g(X)$?

My intuition is yes, but I am have difficulty pinning down an exact proof. I am looking for a proof of this if it is true, or a counterexample if it is not.

Answer 1

Yes, it is true. ("$g$ continuous a.s." and "$P(X\in D_g)=0$" are two different things. You might want to edit the title of your question.)

Since $X_n\to X$ a.s., there exists a null set $N\subset\Omega$ such that $X_n\to X$ pointwise in $N^c$. Now denote $D=\{X\in D_g\}$ and let $A=N^c\setminus D$. In the measurable set $A$, we have the following properties

• $X_n(\omega)\to X(\omega)$ for $\omega\in A$;
• $g$ is continuous at $X(\omega)$ for $\omega\in A$;
• $P(A)=1$.