I am reading the following theorem: If $f$ is almost everywhere finite and measurable function, then for $\delta>0$, there exists a closed subset $F\subseteq E$ such that $\mu(E\setminus F)<\delta$ and $f(x)$ is continuous on F.
Does this theorem hold when $\delta=0$ too?
The answer is negative. Here is a counter-example:
Choose $E = [-7, 7]$ and let $C \subseteq [0, 1]$ be a fat Cantor set. This $C$ is a closed and nowhere-dense set which has positive Lebesgue measure. Then we choose $f = \mathbf{1}_C$, which is the indicator function of $C$.
Now let $F \subseteq E$ be any measurable subset such that $Z := E\setminus F$ has zero Lebesgue measure. Then $C\cap F$ has positive Lebesgue measure, and in particular, it is non-empty. Moreover, for each point $x \in C\cap F$ and for each open neighborhood $U \subset E$ of $x$, the intersection
$$U \cap (F \setminus C) = (U \setminus C) \setminus Z$$
is non-empty. This is because $U \setminus C$ is a non-empty open set of $E$, which itself follows from the fact that $C$ is closed and nowhere-dense. This implies that $f$ takes both values $0$ and $1$ on $U$, and since $U$ was arbitrary, we conclude that $f$ is not continuous at $x$. Therefore $f$ cannot be continuous.
