Measurability of upper and lower derivatives of Radon measures Let $\mu$ and $\nu$ be Radon measures in $\mathbb R^N$. Define their upper and lower derivatives by
$$
\overline{D}_\nu\mu(x):=\limsup_{r\to0}\frac{\mu(B_r(x))}{\nu(B_r(x))},\qquad
\underline{D}_\nu\mu(x):=\liminf_{r\to0}\frac{\mu(B_r(x))}{\nu(B_r(x))},\qquad
$$
if $\nu(B_r(x))>0$ for all $r>0$, and $\overline{D}_\nu\mu(x):=\underline{D}_\nu\mu(x):=\infty$ otherwise. Here, $B_r(x):=\{y\in\mathbb R^N\ |\ |x-y|\leq r\}$ is the closed ball of radius $r$ with center $x$.
I have found several books and lecture notes which state/prove that the above functions are Borel measurable, but I don't see why. One of the proofs which is almost clear to me can be found here (see Proposition 6.5). At the end of this proof, they form the limsup of an uncountable family of Borel measurable functions and conclude that this limsup is Borel measurable as well. But why? Can anyone explain this step to me please? Or give me another proof for my claim?
Any help would be great!
EDIT: I know proofs where outer measures are used. But I am looking for some proof without outer measures.
Thx in advance.
 A: I would do it as follows; say for $\overline D_\nu\mu$.
First, observe that for each fixed $r>0$, the functions $x\mapsto \mu(B_r(x))$ and $x\mapsto \nu(B_r(x))$ are Borel. There is a reference for this (not completely trivial) fact in the link you give.
It follows that the set $\Omega=\{ x\in\mathbb R^N;\; \nu(B_r(x))>0\;\hbox{for all}\;r>0\}$ is Borel, because $$x\in\Omega\iff\forall n\in\mathbb N\;:\; \nu(B_{1/n}(x))>0\, .$$
It follows also that the functions $f_r:\Omega\to \mathbb R^+$ defined by $f_r(x)=\frac{\mu(B(x,r))}{\nu(B(x,r)}$ are Borel. 
We have $\overline D_\nu\mu=f$ on $\Omega$ and $\infty$ outside $\Omega$. Since a function obtained by "gluing" 2 Borel functions is again Borel, the proof will be complete if we can show that the function $f:=\limsup_{r\to 0} f_r$ is Borel on $\Omega$.
Note that something has to be done because, as you told it, an "uncountable limsup" of Borel functions has no reason for being Borel. 
It is enough to show that for any $\alpha\in\mathbb R^+$, the set $\{ x\in\Omega;\; \limsup_{r\to 0} f_r(x)>\alpha\}$ is Borel. So we fix $\alpha\in\mathbb R^+$. 
Let us fix a decreasing sequence $(r_n)_{n\in\mathbb N}$ tending to $0$. Let us also set $\alpha_k:=\alpha+\frac1k$ for each $k\in\mathbb N$. Then, for any $x\in\Omega$, the following equivalence holds true:
$$\limsup_{r\to 0} f_r(x)>\alpha\iff \exists k\in\mathbb N\;\forall n\in\mathbb N \;\exists r<r_n\;:\;\Bigl(\mu(B_r(x))>\alpha_k\nu(B_r(x)) \Bigr)$$
Moreover, denoting by $\mathcal Q_{k}$ the set of all pairs of positive rational numbers $(u,v)$ such that $\frac{u}v>\alpha_k$, the condition "$\mu(B_r(x))<\alpha_k\nu(B_r(x))$" can be written as follows:
$$\exists (u,v)\in\mathcal Q_{k}\;:\;\Bigl( \mu(B_r(x))>u\;{\rm and}\; \nu(B_r(x))<v\Bigr) \, .$$
Finally, observe that for a given pair $(u,v)\in\mathcal Q_k$ and a given $n\in\mathbb N$, one can find $r<r_n$ such that $\Bigl( \mu(B_r(x))>u\;{\rm and}\; \nu(B_r(x)<v)\Bigr)$ if and only if one can find a rational number $r'<r_n$ such that this property holds true. Indeed, assume that $r<r_n$ satisfies the property. Then any rational number $r'$ such that $r<r'<r_n$ satisfies $\mu(B_{r'}(x))>u$ by monotonicity of $\mu(B_t(x))$ with respect to $t$. Moreover, since $\nu(B_r(x))<v$, we have $\nu(B_{r'}(x))<v$ for all $r'>r$ close enough to $r$ because $\nu(B_{r'}(x))\to \nu(B_{r}(x))$ as $r'\to r^+$. So any rational number $r'$ close to $r$ an such that $r<r'<r_n$ has the required property.
Putting everything together, we obtain the following equivalence (for any $x\in\Omega$):
\begin{eqnarray}
\limsup_{r\to 0} f_r(x)>\alpha&\iff& \exists k\in\mathbb N\;\forall n\in\mathbb N \;\exists r'\in\mathbb Q^+\;{\rm with}\; r'<r_n\;\exists (u,v)\in\mathcal Q_k\;:\;\\
& &\Bigl(\mu(B_{r'}(x))>u\;{\rm and}\; \nu(B_{r'}(x))<u\Bigr)
\end{eqnarray}
Since the functions $x\mapsto \mu(B_{r'}(x))$ and $\nu(B_{r'}(x)$ are Borel and since all the quantifiers run along countable sets, this shows that the set $\{ x\in\Omega;\; \limsup_{r\to 0} f_r(x)>\alpha\}$ is indeed Borel.
Edit. There were lots of typos in the first version of this answer. I hope it's better now.
