Is it true that for any Borel measurable function from $\mathbb{R}$ to $\mathbb{R}$, we can find a set $B \subset \mathbb{R}$, s.t. $m^*(\mathbb{R} - B) = 0$ (Lebesgue outer measure), and $f_{|B} : B \rightarrow \mathbb{R}$ is a continuous function? If not, what is a counterexample?
For example, for $f = \mathbf{I}(x \in {\mathbb{Q}})$ (indicator function), we can choose $B = \mathbb{R} - \mathbb{Q}$, so that $f_{|B}$ is a constant function (taking value 0), and therefore continuous.
Counterexample: let $f$ be the indicator function of a "fat Cantor set" $A$, i.e. a nowhere-dense compact set of positive measure. If $\left. f\right|_B$ is continuous and $a \in A \cap B$, there is $\delta > 0$ such that $B \cap (a-\delta, a+\delta) \subset A$. But $(a-\delta, a+\delta)$ contains an interval of $A^c$, and this interval has positive measure and is disjoint from $B$, implying $m^*(B^c) > 0$. On the other hand, if $B \subseteq A^c$, again $m^*(B^c) > 0$.
No, it's not true. As shown in this answer, the characteristic function of the Fat Cantor set is not continuous on any set whose complement has measure zero.
But we do have the next best thing: Lusin's Theorem, which states that if $f$ is measurable, then for any $\epsilon > 0$, there exists a set whose complement has measure less than $\epsilon$ on which $f$ is continuous.
