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