measure theory: $g$ continuous a.e., $f_n \to f$ a.e., does $g \circ f_n \to g \circ f$ a.e.?

Question

Question: Let $(f_n)_{n \geq 1}$, $f$, and $g$ be Borel measurable functions from $\mathbb{R} \to \mathbb{R}$. Suppose $g$ is continuous almost everywhere and $f_n \to f$ almost everywhere, does $g \circ f_n \to g \circ f$ almost everywhere?

I saw a similar question in Cohn's measure theory book, but Cohn's question assumes $g$ is continuous at $f(x)$ for almost all $x$, but the version I found here only assumes $g$ is continuous almost everywhere. May I ask how to solve it?

Thanks in advance!

Answer 1

This is not true. Take for example $f_n\equiv\frac 1n$ and $$g(x):=\begin{cases}1 &\text{if }x=\frac 1n\text{ for some }n\in\mathbb N\\0 &\text{otherwise}.\end{cases}$$ Clearly, $g$ is continuous almost everywhere, but we have $f_n\to 0$ uniformly while $g\circ f_n\equiv 1\neq 0=g(0)$.