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