= Solution
The <Lebesgue differentiation theorem> states that if $f\in L^1_{\mathrm{loc}}(\mathbb R^n)$, then
$$
\lim_{r\downarrow0}\frac1{\lambda(B_r(x))}\int_{B_r(x)}f(y)\,d\lambda(y)=f(x)
$$
for <almost everywhere>[Lebesgue almost every] $x$. The <Radon-Nikodym theorem> states that if two sigma-finite measures satisfy $\nu\ll\mu$, then there is a nonnegative measurable function $h=d\nu/d\mu$, unique $\mu$-almost everywhere, such that $\nu(E)=\int_Eh\,d\mu$.
For any $t\in[0,1]$, take $z=0$ and let $B$ be a planar sector of angle $2\pi t$. Every disc centred at the origin meets this sector in the proportion
$$
\frac{\lambda(B\cap B_r(0))}{\lambda(B_r(0))}=t,
$$
so its <measure density> at $0$ is $t$.
Apply the differentiation theorem to the <indicator function> $1_A$. Its averages over balls are exactly $\rho_{\lambda,A}(x)$, so the <Lebesgue density theorem> gives
$$
\rho_{\lambda,A}(x)=1_A(x)\in\{0,1\}
$$
almost everywhere. If $\lambda(A)=0$, every numerator vanishes, so the density is zero wherever its denominator is nonzero. Conversely, if the density vanishes almost everywhere, the displayed identity gives $1_A=0$ almost everywhere, hence $\lambda(A)=0$.
Finally suppose $\nu$ and $\lambda$ are <mutually absolutely continuous measures>. Write $d\nu=h\,d\lambda$. Then $0<h(x)<\infty$ almost everywhere by the <positive Radon-Nikodym derivative> result. At almost every $x$, the differentiation theorem applied to both $h$ and $h1_A$ gives
$$
\rho_{\nu,A}(x)
=\lim_{r\downarrow0}
\frac{\int_{B_r(x)}h1_A\,d\lambda}{\int_{B_r(x)}h\,d\lambda}
=\frac{h(x)1_A(x)}{h(x)}=1_A(x).
$$
\b[Thus the $\nu$-density also exists and belongs to $\{0,1\}$ at $\lambda$-almost every point.]
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