= Solution
{c}
If $f$ is continuous at $x_0$, choose a small interval on which $|f(s)-f(x_0)|<\epsilon/2$. The triangle inequality makes its oscillation less than $\epsilon$, so $\varphi(x_0)=0$.
Conversely, if $\varphi(x_0)=0$, then for every $\epsilon>0$ some $n$ has $\varphi_n(x_0)<\epsilon$. As $x_0$ itself belongs to the interval used in that supremum, $|f(x)-f(x_0)|<\epsilon$ for $|x-x_0|<1/n$. Thus $f$ is continuous at $x_0$.
The <continuity set of a function> is therefore
$$
\boxed{\{x:f\text{ is continuous at }x\}
=\bigcap_{m\geq1}\{x:\varphi(x)<1/m\}.}
$$
Each 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.
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