Suppose that $f_{m,n}$ is continuous alomst everywhere, and $f_{m,n}$ converges to $g_{m}$ alomst everywhere as $n \to \infty$, $g_{m}$ converges uniformly to $g$ almost everywhere as $m \to \infty$. Can we find $\{h_{k}\}_{k=1}^{+\infty}$ such that $h_{k}$ is continuous almost everywhere and $h_{k}$ converges to $g$ almost everywhere as $k \to \infty$?
I want to prove that $g_{m}$ is continuous almost everywhere, but I failed. I also tried to choose $h_{k}=f_{k,k}$, but it didn't work, either.
Could you please give me some ideas? Thank you very much!
I find a proof and it doesn't use the uniform convergence. We have $$ \lim_{n \to \infty}f_{m,n}(x)=g_{m}(x), ~a.e. $$
$$ \lim_{m \to \infty}g_{m}(x)=g(x), ~a.e. $$
$\forall k \geqslant 1$, $g_{m}$ converges to $g$ almost everywhere on $[-k,k]$, by Egoroff's Theorem we know there exists $E_{k}$ such that $\mu(E_{k}) < 2^{-k}$ and $g_{m}$ converges uniformly to $g$ on $[-k,k] \setminus E_{k}$, so there exists $m_{k}$ such that
$$ |g_{m_{k}}(x)-g(x)| < 2^{-k}, ~\forall x \in [-k,k] \setminus E_{k} $$
In the same way, $f_{m_{k},n}$ converge to $g_{m_{k}}$ almost everywhere on $[-k,k] \setminus E_{k}$, by Egoroff's Theorem there exists $n_{k}$ and $F_{k}$ such that
$$ \mu(F_{k}) < 2^{-k},~ |f_{m_{k},n_{k}}(x)-g_{m_{k}}(x)| < 2^{-k}, ~\forall x \in [-k,k] \setminus (E_{k} \cup F_{k}) $$
Let $G_{k}=\cup_{l \geqslant k}(E_{l} \cup F_{l})$, then $\mu(G_{k}) < 2^{2-k}$, $G_{k}$ is decreasing,
$$ |f_{m_{k},n_{k}}(x)-g(x)| < 2^{1-k}, ~\forall x \in [-k,k] \setminus G_{k} $$
so $f_{m_{k},n_{k}}$ converges to $g$ almost everywhere.