Is it true that $\mu\left(\left\{x\in\partial K : \limsup_{r\to 0^+}\ \frac{\mu(B_r(x)\cap K^c)}{\mu(B_r(x))}>0\right\}\right)=0?$

Question

Let $(X,d)$ be a complete separable metric space. If $x\in X$ and $r>0$, denote by $B_r(x)$ the ball centered in $x$ of radius $r$ of $(X,d)$. Suppose that $\mu$ is a finite Borel measure of $(X,d)$. Define $\operatorname{supp}(\mu):=\{x\in X : \forall r>0, \mu(B_r(x))>0\}$. Let $K\subset X$ be a compact of $(X,d)$ such that $\mu(\partial K)>0$.

Is it true that $\left\{x\in\partial K \cap \operatorname{supp}(\mu) : \limsup_{r\to 0^+}\ \frac{\mu(B_r(x)\cap K^c)}{\mu(B_r(x))}>0\right\}$ is contained in a Borel set of $\mu$-measure zero?

I know that the result holds true if the Lebesgue's differentiation theorem holds true in $(X,d)$ (a characterization of metric spaces that have this property can be found here https://arxiv.org/pdf/1802.02069.pdf). However, does it hold true in general (or maybe under less stringent conditions)?

Answer 1

This is a partial answer: The usual proof of the Lebesgue differentiation theorem uses the Vital covering argument and therefore depends on the properties of the metric. In fact, one can prove Lebesgue differentiation theorem for doubling spaces or more general if the condition $$\tag{1}\limsup_{r \downarrow 0} \frac{\mu(B_{2r}(x))}{\mu(B_r(x))} <\infty$$ is satisfied for $\mu$-almost all $x \in X$. See Sectin 3.4 (in particular Theorem 3.4.3) in "Sobolev Spaces on Metric Measure Spaces" by Heinonen, Koskela, Shanmugalingam, Tyson.

Thus, a counterexample (if one exists) should violate (1).

Answer 2

The paper "Invalid Vitali theorems" by Preiss contain a (pretty messy) counterexample, i.e. he showed that in any separable Hilbert space we can find a finite Borel measure $\mu$ and a compact $K$ such that $\mu(K)>0$ and for all $x\in K$ we have that: $$\lim_{r\to 0^+} \frac{\mu(K\cap B_r(x))}{\mu(B_r(x))} = 0.$$ Then each $x\in K$ is a boundary point of $K$ (if $x$ is an interior point of $K$, then $\lim_{r\to 0^+} \frac{\mu(K\cap B_r(x))}{\mu(B_r(x))} = 1$), so $\mu(\partial K)=\mu(K)>0$ and for all $x\in\partial K$ we have that: $$\lim_{r\to 0^+} \frac{\mu(K^c\cap B_r(x))}{\mu(B_r(x))} = 1-\lim_{r\to 0^+} \frac{\mu(K\cap B_r(x))}{\mu(B_r(x))} = 1.$$

It seems that what prevents the existence of such pathological measures is some kind of $\sigma$-finite dimensionality property that can be used to prove a generalization of the Besicovitch covering lemma.