# Borel measurable function is continuous when restricted to a large subset?

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 $$m^*(B^c) > 0$$.
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.