Does $L^2$ convergence and convergence on a countable dense subset, together imply almost everywhere convergence?

Question

Let $\Omega\subset\mathbb{R}^m$ be a domain with smooth boundary. Let $\psi:\Omega\to\mathbb{R}$ be a function of bounded variation. Let $D\subset\Omega$ be a countable dense subset such that $\psi$ is continuous at all $x\in D$. There is a sequence of smooth functions $\{f_n\}$ such that as $n\to\infty$,$\|f_n-\psi\|_{L^2(\Omega)} = 0$. It is also given that $\forall x\in D$ as $n\to\infty$,$f_n(x)\to\psi(x)$.

I want to prove that $f_n\to \psi$ almost everywhere. Appreciate any hints/help.

Edit: (after answer by Christian Remling, and after a close vote)

What if Fourier transform of every $f_n$ has a compact support? to make this scenario possible, Assume domain $\Omega$ is $\mathbb{T}^m$ and $\{f_n\}$ are all trigonometric polynomials of various degrees. In this case, will the $L^2$ convergence imply almost everywhere convergence, for a bounded variation function $\psi:\mathbb{T}^m\to\mathbb{R}$ ?

Answer 1

This is not true. Take $\Omega = (0,1)$ and consider moving bumps $f_1=\chi_{(0,1/2)}$, $f_2=\chi_{(1/2, 1)}$, then intervals of length $1/4$ for the next four functions etc. Then $f_n(x)=0$ eventually at all $x=k2^{-m}$, but $\limsup f_n(x)=1$ at all other $x$.

We now make these functions smooth in such a way that for each group of functions (the first two, then the next four, etc.) the set where $f_n=1$ for some function still has almost full measure in $(0,1)$. If the measures of the complements are summable, then the Borel-Cantelli lemma shows that still $\limsup f_n=1$ almost everywhere.