If $f_n \to f$ uniformly a.e. and $u_n \to u$ a.e., does $f_n(u_n) \to f(u)$ a.e.?

Question

Here $f_n$ and $f$ are real-valued functions of 1 variable.

Suppose that $f_n \to f$ uniformly on $\mathbb{R}\backslash \{0\}$. Let $u_n \to u$ pointwise a.e. on a manifold $X$.

Does it follow that $f_n(u_n) \to f(u)$ pointwise a.e.?

If the $f_n$ were uniformly convergent everywhere the answer is yes but here I'm not sure.

Answer 1

You will need some kind of regularity on $f$, something like $f$ is continuous a.e. Then the result follows from the inequality $$ |f_n(u_n(x))-f(u(x))|\le|f_n(u_n(x))-f(u_n(x))|+|f(u_n(x))-f(u(x))|. $$

Consider the following example: $$ f_n(x)=f(x)=\begin{cases}0 & \text{if }x\in\mathbb{Q},\\1 & \text{if }x\not\in\mathbb{Q},\end{cases}\qquad X=\mathbb{R},\quad u_n(x)=\frac{\pi}{n}. $$ Then $u_n$ converges uniformly to $0$ on $X$, $f_n(u_n(x))=1$ for all $x\in\mathbb{R}$ and $f(u(x))=0$ for all $x\in\mathbb{R}$.