Let $\nu$ be a Radon measure and $A$ a Borel set.

We know the expression for the approximate density of a set given by:

$\lim _{r\rightarrow 0} =\frac{\nu (A \cap B_r(x))}{\nu ( B_r(x))}$

if $\nu (B_r(x)) >0$ and $0$ else. By Lebesgue differentiation theorem we can easily show that this limit exists. But I am asking whether one can prove directly that there exists a null set such that the limit above exists and is only $0$ and $1$ outside this null set. So somewhat a reverse direction.

Let $\nu$ be a Radon measure on $\mathbb{R}^d$ and $A$ a Borel set with $A \subset \mathbb{R}^d$.

We know the expression for the approximate density of a set given by:

$\lim _{r\rightarrow 0} =\frac{\nu (A \cap B_r(x))}{\nu ( B_r(x))}$

if $\nu (B_r(x)) >0$ and $0$ else. By Lebesgue differentiation theorem we can easily show that this limit exists. But I am asking whether one can prove directly that there exists a null set such that the limit above exists and is only $0$ and $1$ outside this null set. So somewhat a reverse direction.