Is the domain of the symmetric derivative of a borel measure a borel set? Let $\mu$ be the $n$-dimensional Lebesgue measure and $\lambda$ be a complex Borel measure on $\mathbb{R}^n$.
Let $S$ be the set of points $x\in \mathbb{R}^n$ where $\lim_{r\to 0} \frac{\lambda (B(x,r))}{\mu (B(x,r))}$ exists in $\mathbb{C}$.
Then, is $S$ a Borel set?
 A: Here is the positive measure case.  Taken from [1], (1.1).
Let $X$ be a metric space.  Let $\alpha$ be a positive finite Borel measure on $X$.  Write $B_r(x)$ for the open ball in $X$ with center $x$ and radius $r$.  Note in the OP, $\mu(B_r(x))$ is a constant times $r^n$.  Instead, let's more generally use a continuous strictly increasing gauge function $\phi : [0,\infty) \to [0,\infty)$ with $\phi(0)=0$.  Write
$$
\overline{D}^\phi_\alpha(x) = \limsup_{r\to 0^+}\frac{\alpha(B_r(x))}{\phi(r)},\qquad
\underline{D}^\phi_\alpha(x) = \liminf_{r\to 0^+}\frac{\alpha(B_r(x))}{\phi(r)} 
$$
for the upper and lower densities.
We will show that these two functions are Borel functions.  So the set where they agree is a Borel set.
(A) Fix $r>0$.  The function $x \mapsto \alpha(B_r(x))$ is Borel measurable.  (In fact, it is lower semicontinuous.)
Proof.
In fact, we claim: for any $t \in \mathbb R$,
$$
V = \big\{x \in X \;:\; \alpha\big(B_r(x)\big) > t\big\}
$$
is an open set.  Let $x_0 \in V$, so that $\alpha(B_r(x_0)) > t$.  Now
$\alpha(B_{r-1/n}(x_0)) \nearrow \alpha(B_r(x_0))$, so
there is $n$ with $\alpha(B_{r-1/n}(x_0)) > t$.  Then for any 
$x \in B_{1/n}(x_0)$, we have $B_r(x) \supseteq B_{r-1/n}(x_0)$, so
$\alpha(B_r(x)) \geq \alpha(B_{r-1/n}(x_0)) > t$.  So $x \in V$. 
This shows that $V$ is an open set.  Thus $x \mapsto \alpha(B_r(x))$ is
Borel measurable.
(B) For all $x \in X$ and all $\eta > 0$,
$$
\sup\left\{\frac{\alpha(B_r(x))}{\phi(r)} \;:\;0 < r < \eta\right\} =
\sup\left\{\frac{\alpha(B_r(x))}{\phi(r)} \;:\;0 < r < \eta,
r \in \mathbb Q\right\}
$$
Proof.
Inequality $\ge$ is clear.  Let 
$s < \sup\{\alpha(B_r(x))/\phi(r) : 0 < r < \eta\}$. 
So there exists $r_0 \in (0,\eta)$ with $\alpha(B_{r_0}(x))/\phi(r_0) > s$. 
Now for all $r < r_0$ sufficiently close to $r_0$ we
have $\alpha(B_r(x)) > s \phi(r_0) \ge s \phi(r)$; 
in particular there is a rational $r$ that satisfies 
this.  So $s < \sup\{\alpha(B_r(x))/\phi(r) : 
0 < r < \eta, r \in \mathbb Q\}$.  This proves inequality $\le$.
(C) For each fixed $r \in \mathbb Q$, the function $x \mapsto \alpha(B_r(x))/\phi(r)$ is Borel measurable, so the sup over all
$r \in (0,\eta)\cap\mathbb Q$ is Borel measurable.
The value $\sup\{\alpha(B_r(x))/\phi(r) : 0 < r < \eta\}$ decreases as
$\eta > 0$ decreases, so the limit may be taken over rational $\eta$.  So we conclude that $\overline{D}^\phi_\alpha(x)$ is Borel measurable.
The other case $\underline{D}^\phi_\alpha(x)$ is left for the reader... or see the references.  
See also [2] (2.4), where we consider possibly discontinuous gauge functions $\phi$ so that it matters whether we used open balls or closed balls in defining the densities.
References
[1] G. Edgar, "Fine variation and fractal measures." Real Analysis Exchange 20 (1994) 256-280  
[2] G. Edgar, "Centered densities and fractal measures." New York J. of Math. 13 (2007) 33-87
A: 
Theorem 1. Fix $r>0$ and let $\lambda$ be a positive semifinite borel measure on $\mathbb{R}^n$. Then, the map $\mathbb{R}^n\rightarrow \mathbb{R}:x\mapsto \lambda (B(x,r))$ is Borel measurable.

Proof> Define $g:\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n:(x,y)\mapsto x-y$.
Since $g$ is continuous it is borel measurable. Since $\mathbb{R}^n$ is second countable, $\mathscr{B}_{\mathbb{R}^n \times \mathbb{R}^n}=\mathscr{B}_{\mathbb{R}^n} \otimes\mathscr{B}_{\mathbb{R}^n}$. Hence, we have that $g:( \mathbb{R}^n \times \mathbb{R}^n, \mathscr{B}_{\mathbb{R}^n} \otimes\mathscr{B}_{\mathbb{R}^n} )\rightarrow (\mathbb{R}^n,\mathscr{B}_{\mathbb{R}^n})$ is measurable.
Define $h:=\chi_{B(0,r)}\circ g$.
By Tonelli's theorem, $(\mathbb{R}^n,\mathscr{B}_{\mathbb{R}^n})\rightarrow [0,\infty] : y\mapsto \int_{\mathbb{R}^n} h^y d\lambda$ is measurable. Since $\int_{\mathbb{R}^n} h^y d\lambda = \lambda (B(y,r))$, it proves that $\mathbb{R}^n \rightarrow [0,\infty]:x\mapsto \lambda(B(x,r))$ is Borel measurable. Since the range of this function is $[0,\infty)$, and this is contained in $\mathbb{R}$, we have that the map $\mathbb{R}^n \rightarrow \mathbb{R}:x\mapsto \lambda(B(x,r))$ is Borel measurable. Q.E.D.
Hence, now we know that the map $f_r:\mathbb{R}^n \rightarrow \mathbb{C}: x\mapsto \frac{\lambda(B(x,r))}{\mu(B(x,r))}$ is Borel measurable for each $r>0$. ( It is because $\lambda$ can be represented as a linear combination of bounded positive measures. Hence, what I have proven above also holds for $\lambda$.)
So the problem reduces to prove the set $\{x\in \mathbb{R}^n: \lim_{r\to 0} f_r(x) \text{ is convergent in } \mathbb{C}\}$ is Borel where $\{f_r\}_{r>0}$ is a family of borel measurable functions from $\mathbb{R}^n$ to $\mathbb{C}$.
Note that $\lim_{r\to 0} f_r(x)$ is convergent if and only if $\lim_{r\to 0} Re(f_r(x))$ and $\lim_{r\to 0} Im(f_r(x))$ are convergent. So wlog, assume that $\lambda$ is a real measure.

Theorem 2. The set of points where $\lim_{r\to 0} f_r(x)$ exist is a Borel set.

Proof > Define $F(x)=\limsup_{r\to 0} f_r(x)$ and $H(x)=\liminf_{r\to 0} f_r(x)$ for all $x\in \mathbb{R}^n$. It suffices to prove that $F,H$ are Borel measurable functions from $\mathbb{R}^n$ to $[-\infty,\infty]$.
Define $h_n(x)=\sup\{f_r(x)\in [-\infty,\infty]: r\in (0,1/n)\}$ for each $n\in \mathbb{Z}^+$. Since $F=\lim_{n\to \infty} h_n$, it suffices to prove that $h_n$'s are Borel measurable. (Here I am stuck and trying to figure out how to prove $h_n$'s are  Borel measurable)
..
