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 $\lambda (S)=0$?
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 $\lambda(S)=\infty$.
$$\dots\dots\dots$$
2. Lebesgue differentiation theorem applies to $d\nu$, so
$$ \frac {\lambda(B(x,r))}{\nu(B(x,r))} =
\frac{1}{\nu(B(x,r))} \int_{B(x,r)} L(y)\space d\nu(y)
\:\xrightarrow[r \rightarrow 0^+]{}\: L(x) \text{ a.e. $\nu$} $$
and
$$ \frac {\mu(B(x,r))}{\nu(B(x,r))} =
\frac{1}{\nu(B(x,r))} \int_{B(x,r)} M(y)\space d\nu(y)
\:\xrightarrow[r \rightarrow 0^+]{}\: M(x) \text{ a.e. $\nu$} .$$
Note that
$$0 \le \mu(M \le 0) = \int_{(M \le 0)} M(y)\space d\nu(y) \le 0 $$
therefore $ M > 0 \text{ a.e. $\nu$},$ so outside a Borel $\nu$-null set,
$$ \frac {\lambda(B(x,r))}{\mu(B(x,r))} =
\frac {\lambda(B(x,r))/\nu(B(x,r))}{\mu(B(x,r))/\nu(B(x,r))}
\:\xrightarrow[r \rightarrow 0^+]{}\: \frac{L(x)}{M(x)} \text{ a.e. $\nu$} $$
$$\dots\dots\dots$$ 3. Since $\nu(S^c) = 0$, then $\lambda(S^c)=0$, and even $|\lambda|(S^c)=0$ (i.e. any Borel subset of $S^c$ has null $\lambda$ measure).
$$\dots\dots\dots$$ 4. I can't show that $S$ is Borel. If the measure $\lambda$ is such that the functions: $$ x \mapsto \lambda(B(x,r)) $$ ($r>0$) are continuous, then $S$ is Borel. Since $\lambda$ might have singular support, these functions need not be continuous, but there might be some other way of showing $S$ is Borel.
