The Classical Egoroff's Theorem asserts that given a sequence of functions $(f_{n})$ defined on a subset $E$ of $R$ of finite Lebesgue measure and if $(f_{n})$ converges to $f$ everywhere on $E$, then for all $\varepsilon>0$ there is an $F_{\varepsilon}$ such that $m(F_{\varepsilon})<\varepsilon$ and that $(f_{n})$ converges uniformy to $f$ on $E-F_{\varepsilon}$.

Question: Under what condition the set $\bigcap_{\varepsilon>0}F_{\varepsilon}$ is a union of isolated points?

Here $F_{\varepsilon}=\bigcup_{k=1}^{\infty}E_{n(k),k}$, $E_{n,k}=\bigcup_{i=1}^{\infty}\{x\in E:|f_{i}(x)-f(x)|\geq 1/k\}$. And $n(k)$ is chosen so that $m(E_{n(k),k})<\varepsilon/2^k$.