$u_n\to u$ a.e. in $\mathbb{R}^N \implies u_n^2\to u^2$ a.e. in $\mathbb{R}^N$?

Question

Let $(u_n)_n$ be a sequence in $H^1(\mathbb{R}^N)$ such that $$u_n\to u\quad\mbox{ a.e. in } \mathbb{R}^N.$$

My question is: we can say that $u_n^2\to u^2$ a.e. in $\mathbb{R}^N$, too? I would say that it is true because I am considering the composition between $u_n$ and $u$ through a continuous function, but I am not sure about my answer.

Could someone please tell me if I am wrong?

Thank you in advance!

Answer 1

You are right. Define $M=\{ x \in \mathbb{R}^N : u_n(x) \mbox{ does not converge to } u(x) \}$. Thus $\mathcal{L}^N(M)=0$. If $f : \mathbb{R} \to \mathbb{R}$ is continuos then $f(u_n(x))$ converges to $f(u(x))$ at least for $x \not \in M$ and this proves that $f(u_n(x))$ converges to $f(u(x))$ a.e.