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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.

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  • $\begingroup$ $\nu$ is a Radon measure where? $\endgroup$ Commented Dec 8, 2016 at 21:31
  • $\begingroup$ yes, as mentioned above ;) $\endgroup$ Commented Dec 8, 2016 at 21:34
  • $\begingroup$ You have misunderstood Fedor's question. On what set does this Radon measure live? An arbitrary metric space? The real line? $\endgroup$
    – Yemon Choi
    Commented Dec 8, 2016 at 21:35
  • $\begingroup$ oh sorry, you are right... on $\mathbb{R}^d$ and $A$ is a subset of that, sorry again! $\endgroup$ Commented Dec 8, 2016 at 21:36
  • $\begingroup$ Please edit that into the body of the question – people shouldn't have to go through the comments to understand the question. $\endgroup$ Commented Dec 8, 2016 at 22:04

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