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For every Lebesgue measurable set $A\subseteq {\mathbb{R}}^n$, its Lebesgue density is $1$ at almost every point of $A$ and $0$ at almost every point of its complement.

The Lebesgue differentiation theorem states that if $f\in L_{\mathrm{loc}}^1({\mathbb{R}}^n)$, thenfor 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 $\nu (E)={\int }_{E}hd\mu$.

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 proportionso 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 givesalmost 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 $d\nu =hd\lambda$. 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 $h{1}_A$ givesThus the $\nu$-density also exists and belongs to $\{0,1\}$ at $\lambda$-almost every point.