Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 26J / e

Yes. In fact, measurability of $f$ is unnecessary. Define the oscillationFor every $\epsilon >0$, the set $\{x:{\omega }_f(x)<\epsilon \}$ is open, and $f$ is continuous at $x$ exactly when ${\omega }_f(x)=0$. Hence the continuity set of a function isa $G_{\delta }$ set and therefore a Borel set.