I was trying to solve the following exercise. Let $K$ a closed subset of $\mathbb{R}$. $$X:=\left\{f\in L^2[0,1]:f(x)\in K \mbox{ a.e. }\:x\in [0,1] \right\}$$ Then:
1) $X$ is closed under strong convergence in $L^2$.
2) $X$ is in general not closed under weak convergence (find a counterexample).
I proved 1) and also that X is closed in $\ell^2$ if $K$ is a closed interval. So the counterexample for 2) should use a closed subset $K$ which is not an interval. But i've not found it.
Consider a sequence of Borel subsets of the unit interval $\left(A_n\right)_{n\geqslant 1}$ such that for each Borel subset $B$ of the unit interval, $\lim_{n\to +\infty}\lambda\left(A_n\cap B\right)=\lambda(B)/2$. For example, we can choose $A_n=\bigcup_{j=0}^{2^n-1}\left[j2^{-n},(j+1)2^{-n}\right)$. Then define $f_n(x):=2\mathbf 1_{A_n}(x)-1$. We have the weak converge to $0$, but we can take $K:=\{-1,1\}$.