I'm reading Measure theory and fine properties of functions by Evans, 1st edition. In the page 39, Proof of Claim #3 defines a function $f_r(x)$
$$f_r(x)=\begin{cases} \frac{\nu(B(x,r))}{\mu(B(x,r))} \quad\text{ if } \mu(B(x,r))>0\\ \\ +\infty \quad \text{ if } \mu(B(x,r))=0 \end{cases}$$ where $\mu$ and $\nu$ are two Radon measures, $r$ is a positive constant and $B(x,r)$ denotes a closed ball centered at $x$ with radius $r$. Then the text states that
the functions $x\mapsto \mu(B(x,r))$ and $x\mapsto \nu(B(x,r))$ are upper semicontinuous and thus Borel measurable. Consequently, for every $r>0$, $f_r(x)$ is $\mu$-measurable.
I understand the first sentence. But I can't figure out why the text could claim easily that $f_r(x)$ is $\mu$-measurable ? Any hint ?