 Solution

If $f$ is continuous at $x_0$, choose a small interval on which $|f(s)-f(x_0)|<ϵ/2$. The triangle inequality makes its oscillation less than $ϵ$, so $\phi (x_0)=0$.

Conversely, if $\phi (x_0)=0$, then for every $ϵ>0$ some $n$ has ${\phi }_n(x_0)<ϵ$. As $x_0$ itself belongs to the interval used in that supremum, $|f(x)-f(x_0)|<ϵ$ for $|x-x_0|<1/n$. Thus $f$ is continuous at $x_0$.

The continuity set of a function is thereforeEach set in the intersection is open by part (i), so this is a G-delta set and in particular a Borel set, even if $f$ itself is not measurable.