I'll use the version of this question I posted on Stakexchange to replace the former version. To simplify the situaton, we assume the measure space we are dealing with right now is a closed interval $I$ of the real line.

When we are using the Egoroff's theorem, little attention has been paid to the exceptional set, i.e. the set $E\subset I$ on which $(f_n)$ fails to converge uniformly. The only thing we know about $E$ is that $m(E)$ can be arbitrary small. Taking the topology on the real line into account, we can also assume that $E$ is closed.

Recall that in the proof of this theorem, we constructed the exceptional set $E_{\varepsilon}$ with respect to a fixed $\varepsilon>0$ in the following way: \begin{equation}E_{\varepsilon}=\bigcup_{k=1}^{\infty}\bigcup_{i=n(k,\varepsilon)}^{\infty}\{x\in I:|f_{i}(x)-f(x)|\geq 1/k\},\end{equation} where $n(k,\varepsilon)$ is chosen so that $m(\bigcup_{i=n(k,\varepsilon)}^{\infty}\{x\in I:|f_{i}(x)-f(x)|\geq 1/k\})<\varepsilon/2^k$.

Now assume $(f_n)$ is a sequence of smooth functions, and $f\in L^1(I)$ is the limit function. Since $f$ is only determined up to a null set, the set $E_{\varepsilon}$ can only be determined up to a null set. Thus it is very natural to require $f$ to be the "best choice" to make $E_{\varepsilon}$ the as small as possible.

A satisfatory and notable case is that the family $\{E_{\varepsilon}\}$ be a sequence of nested closed intervals, thus we have $\bigcap_{\varepsilon}E_{\varepsilon}=\{x_{0}\}$, where $x_{0}\in I$. Or more generally, we can ask **in which case the set $\bigcap_{\varepsilon}E_{\varepsilon}$ is a union of isolated points?**

However, this is not true in general case, GTM 2 (p.38) contains an example in which case the set $I-E$ is nowhere dense, of course this coincides with the well-known result that every subset of the line can be represented as a union of a null set and a set of first category.