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? Moreover, is $\lambda (S)=0$?



Is the domain of symmetric derivative borel set?
Yes, $S$ is Borel. Assume, to be specific, that $B(x,r)$ denotes the open ball of radius $r$ centered at $x$. 


Lemma 1. The function $(0,\infty)\ni r\mapsto\ell(r):=\lambda (B(x,r))$ is left-continuous, for each $x\in \mathbb{R}^n$. 


Proof. By the Hahn decomposition theorem, $\lambda$ is a linear combination (possibly with complex coefficients) of nonnegative measures. So, in the rest of the proof of Lemma 1, without loss of generality (wlog) $\lambda$ itself may be assumed to be nonnegative. Take any $x\in \mathbb{R}^n$ and then any $r\in(0,\infty)$. The sequence of balls $B(x,r-1/n)$ is increasing, with $\bigcup\_n B(x,r-1/n)=B(x,r)$. So, $\lambda(B(x,r-1/n))\to\lambda(B(x,r))$. Since $\lambda$ is nonnegative, the function $(0,\infty)\ni r\mapsto\lambda (B(x,r))$ is nondecreasing. Now Lemma 1 follows. 


Lemma 2. The function $\mathbb{R}^n\ni x\mapsto\lambda (B(x,a))$ is Borel, for each real $a>0$. 


Proof. Wlog $\lambda$ is nonnegative in the proof of Lemma 2 as well. Take any real $c$. It is enough to show that the set $L\_c:=\{x\in\mathbb{R}^n\colon\lambda(B(x,a))>c\}$ is open (and hence Borel). To do this, take any $x\in L\_c$, so that $\lambda(B(x,a))>c$. By Lemma 1, we can find some real $\delta>0$ such that $\lambda(B(x,a-\delta))>c$. But for all $y\in B(x,\delta)$ one has $B(y,a)\supseteq B(x,a-\delta)$ and hence $\lambda(B(y,a))\ge \lambda(B(x,a-\delta))>c$. So, for any $x\in L\_c$ one has $B(x,\delta)\subseteq L\_c$. Thus, indeed the set $L\_c$ is open, and Lemma 2 follows. 


We are now ready to show that $S$ is Borel.  

Indeed, for any real $r>0$, let $Q\_r:=\mathbb{Q}\cap(0,r)$, a countable set. 
Then, again by Lemma 1,  

\begin{equation}
 S=\bigcap\_{k\in\mathbb N}\bigcup\_{m\in\mathbb N}\bigcap\_{a,b\in Q\_{1/m}}
 \Big\{x\in\mathbb{R}^n\colon\Big|\frac{\lambda (B(x,a))}{\mu (B(x,a))}-\frac{\lambda (B(x,b))}{\mu (B(x,b))}\Big|<\frac1k\Big\} \tag{1}
\end{equation}
and, by Lemma 2, the sets $\big\{x\in\mathbb{R}^n\colon\Big|\frac{\lambda (B(x,a))}{\mu (B(x,a))}-\frac{\lambda (B(x,b))}{\mu (B(x,b))}\Big|<\frac1k\big\}$ are Borel. 
Thus,  

$S$ is indeed Borel. 


Of course, in general $\lambda(S)\ne0$. E.g., take $\lambda=\mu$. Then $\lambda(S)=\infty$.


