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Edit: I believe the below (original) problem can be reduced to the following one

Let $(X, \rho)$ be a metric space, where $X$ has Hausdorff dimension $d$. Under what conditions on $X$ can we say that for any $x_0 \in X$ and $c > 0$

$$\lim_{\delta \to 0} H^d(\{x \in X \mid c \le \rho(x,x_0) \le c + \delta\}) = 0$$

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Suppose that $(X, \rho)$ is a metric space, and let $S$ be an open ball about a point $x_0$. Further, suppose that $X$ has Hausdorff dimension $d$, and that $\mu \ll H^d$ is a probability measure that is absolutely continuous w.r.t the Hausdorff measure, and $\mu(S) > 0$. I have two related questions.

(1) Can we show that $$\lim_{\delta \to 0} \mu(\{x \in X \setminus S \mid \rho(x,S) < \delta\}) = 0$$

If the answer to (1) is positive, can the result be extended to the case where we have an uncountable family of probability measures $\{\mu_t\}_{t \in T}$ each absolutely continuous with respect to $H^d$? In other words,

(2) Can we show that $$\lim_{\delta \to 0} \sup_{t \in T} \mu_t(\{x \in X \setminus S \mid \rho(x,S) < \delta\}) = 0$$ If necessary, we may assume that the Radon-Nikodym derivative $\frac{d\mu_t}{dH^d}$ is universally bounded for any $\mu_t$

These questions are follow-ups to an earlier question that I asked: Bounds on Hausdorff measure when extending a set

ndurvasula
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  • I think I'm misunderstanding something. What if $X = [0, 1]$ with the usual metric, $S = \mathbb{Q} \cap [0, 1]$, and $\mu$ is the Lebesgue measure on $[0, 1]$? Then $\mu({x \in [0, 1] \backslash \mathbb{Q} ; | ; d(x, \mathbb{Q}) < \delta}) = 1$ for all $\delta > 0$, since the rationals are dense and have measure 0. – Michael Jesurum Jan 26 '21 at 21:41
  • Ah I'm sorry, I forgot to specify that $S$ must have positive $\mu$ measure – ndurvasula Jan 27 '21 at 00:02
  • For your first question, the sets are nested and $\bigcap_{\delta >0} {x \in X \setminus S , \mid , \rho(x,S) < \delta} = \overline{S} \setminus S$ so $\mu({x \in X \setminus S , \mid , \rho(x,S) < \delta}) \to \mu(\overline{S} \setminus S)$. I don't see why $\mathcal{H}^{d}(\partial S) = 0$, though --- are you assuming this? –  Jan 27 '21 at 00:13
  • Edited $S$ to be an open ball -- in this case can $H^d(\partial S) = 0$ be shown? – ndurvasula Jan 27 '21 at 03:06

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