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

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 $$\psi:\mathbb{T}^m\to\mathbb{R}$$ ?

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.
• What if Fourier transform of every $f_n$ has a compact support? – Rajesh D Jan 24 at 11:29
• If Fourier transforms of all of them are supported on the SAME compact set, then the answer is evidently positive. If supports of transforms of $f_n$ become larger and larger, then this is not really a restriction.. – Alexandre Eremenko Jan 24 at 15:05