Let $\mu$ and $\lambda$ be locally finite Borel measures on $\mathbb{R}^n$. We may define an upper resp. lower derivative of $\mu$ w.r.t $\lambda$ at $x\in\mathbb{R}^n$ by $$ \underline{D}(\mu,\lambda,x)=\liminf_{r\to0}\frac{\mu(B(x,r))}{\lambda(B(x,r))},\quad\text{resp.}\quad\overline{D}(\mu,\lambda,x)=\limsup_{r\to0}\frac{\mu(B(x,r))}{\lambda(B(x,r))},\quad(x\in\mathbb{R}^n). $$ Here $B(x,r)$ is the closed ball centered at $x$ with radius $r$, although I think it may as well be the corresponding open ball, and we set $0/0=0$.
The goal is to show that $\underline{D}(\mu,\lambda,\,\cdot\,)$ and $\overline{D}(\mu,\lambda,\,\cdot\,)$ are Borel functions.
It is clear to me that the function $x\mapsto\mu(B(x,r))$ for fixed $r>0$ is not necessarily continuous, as we can take $\mu$ concentrated on the point $0$, and then if e.g. $r=1$, then the function will not be continuous at any point from the unit circle, looking e.g. at any point $y$ arbitrarily close to a point on the unit circle, but with $\lVert y\rVert>1$.
This eliminates the easiest way to see Borel-measurability, since now I need to find some other way of showing measurability for each $r>0$ and then deal with the $\liminf$, since without continuity, it may not be possible to take $\inf$ over say rational points $q>0$.
Anyone with an idea?