Why is Derivative of a Radon measure measurable

Question

We have $\mu$, $\nu$ Radon measures on $\mathbb{R}^d$. We remember that upper derivative is defined $$ \overline{D}_{\mu}\nu(x) = \begin{cases} \limsup\limits_{r\to 0} \frac{ \nu(B(x,r))}{\mu(B(x,r))} & \text{if } \mu(B(x,r))>0 \;(\forall r>0), \\ +\infty & \text{otherwise} \end{cases}. $$ I already know that mapping $x \mapsto B(x,r)$ is upper semicontinous for all $r>0$. But how to prove, that upper derivative is actually measurable? Please help.

Answer 1

First, any upper-semicontinuous function is measurable with respect to the Borel sigma algebra.

Second, if $f$ and $g$ are both measurable functions, then the function defined by

$$\begin{cases} \frac{f(x)}{g(x)}\text{ if } g(x)\neq 0 \\ a \text{ otherwise} \end{cases}$$

is measurable. The easiest way to see this is that the set $\{g(x)=0\}$ is measurable, and if $A$ is a measurable set, a function is measurable if and only if its restrictions to $A$ and $\Omega\setminus A$ are measurable (with respect to the subset measurable space structures).

Finally, the $\limsup$ of a sequence of measurable functions is measurable, and $\limsup_{r\to 0} f(r)=\limsup_{n\to \infty} f(\frac1n)$.