Past exam of the martingales and financial mathematics course of the University of Oxford / 2008 / A / 4 / b

Boundedness of $f$ makes $\phi$ finite and nonnegative. Suppose $\phi (x_0)<a$. By its infimum definition, some $n$ satisfies ${\phi }_n(x_0)<a$. If $|x-x_0|<1/(2n)$, thenThe supremum over the smaller interval cannot exceed that over the larger one. HenceThus $\{\phi <a\}$ is open for every $a$, proving $\phi$ is upper semicontinuous. Measurability of the original $f$ was never needed: these are purely pointwise oscillation bounds.

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.