Bounds on Hausdorff measure when extending a set

Question

Let $(X, \rho)$ be a metric space that has Hausdorff dimension $d$, and let $S$ be an arbitrary Borel subset such that $H^d(S) = \alpha > 0$, where $H^d$ denotes the $d$-dimensional Hausdorff measure (assume this is nonzero). Is it possible to show that the quantity $$H^d(\left\{x \in X \mid \rho(x, S) < \delta\right\})$$ must be nonzero and finite? In other words, can we show that some nonzero and finite amount of mass must be picked up in order to extend the set $S$ by a distance of $\delta$? Feel free to make any further assumptions on the space $(X, \rho)$ as needed.

Answer 1

Denote $S_\delta = \{x \in X : \rho(x,S) < \delta \}$. The claim that $H^d(S_\delta) < \infty$ is false -- a counterexample is given by $S = \mathbb{Z} \subseteq \mathbb{R}$. In this case $d=1$, $H^d(S) = 0$, and $S_\delta$ is a countable union of intervals of length $2\delta$, so $H^d(S_\delta) = \infty$.

The claim that $H^d(S_\delta) > 0$ is also false. Let $X = \{0\} \cup [1,2]$ and $S = \{0\}$. Then $X$ is $1$-dimensional but $S_{1/2} = \{0\}$ is still measure zero according to $1$-dimensional Hausdorff measure.