Solution
= Solution
\b[Yes.] In fact, measurability of $f$ is unnecessary. Define the oscillation
$$
\omega_f(x)=\inf_{r>0}\sup\{|f(y)-f(z)|:y,z\in(x-r,x+r)\}.
$$
For every $\varepsilon>0$, the set $\{x:\omega_f(x)<\varepsilon\}$ is open, and $f$ is continuous at $x$ exactly when $\omega_f(x)=0$. Hence the <continuity set of a function> is
$$
\boxed{\bigcap_{n=1}^{\infty}\{x:\omega_f(x)<1/n\},}
$$
a $G_\delta$ set and therefore a <Borel set>.