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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.

ndurvasula
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  • Thanks for the counterexample -- if we enforce that $\alpha > 0$, then can such a statement be proved? – ndurvasula Jan 24 '21 at 07:18
  • No (I think I accidentally deleted my previous comment and I can't restore it, oops). The same basic idea works, take $S = [-1,0] \cup \mathbb{Z}$. Now $S$ has positive measure but the $\delta$ neighborhood still has infinite measure. I'm going to try to write up something a bit more complete and make it an answer – Adam Jan 24 '21 at 07:22
  • Great, thanks! A follow-up: if $S$ is an arbitrary open set rather than a Borel set, does this end up working out? – ndurvasula Jan 24 '21 at 07:29
  • Still no -- take $S = (0,1) \cup (1, 1+1/2) \cup (2, 2+1/4) \cup (3,3+1/8) \cup \dots$ – Adam Jan 24 '21 at 07:32
  • Makes sense -- if you happen to find some condition on the metric space/subset that makes this work, I'd really appreciate it :) – ndurvasula Jan 24 '21 at 07:41
  • Well one obvious condition is if the whole space $X$ has finite $H^d$ measure – Adam Jan 24 '21 at 07:46

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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.

Adam
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  • Sorry for the barrage of follow-ups: if $X$ is connected/complete, is $H^d(S_\delta) > 0$? – ndurvasula Jan 24 '21 at 08:06
  • Still no, you can modify this example a bit. Let $X$ be a subset of $\mathbb{R}^2$ that looks like a filled in square with a line segment sticking out of the side. Now it's connected and complete but still a neighborhood in the line segment part will have zero 2-dimensional measure – Adam Jan 24 '21 at 08:19
  • Got it, thanks! Does any condition on X come to mind here that would make this work? – ndurvasula Jan 24 '21 at 08:28
  • Well I guess the notion you want to capture is some kind of "uniformity" of $X$. You could try to formulate it in terms of measures supported on $X$. For example something like "there exists a probability measure $\mu$ on $X$ such that the $\mu$-measure of any ball of radius $r$ is $\lesssim r^d$" would probably work, and this should capture most non-weird spaces – Adam Jan 24 '21 at 08:33
  • In the $R^2$ with a stick example, wouldn't the uniform measure on that set (which attributes no measure to stick) satisfy the property, since any interval of radius r on the stick has measure zero? I might be misunderstanding the condition. – ndurvasula Jan 24 '21 at 08:42
  • Good point, maybe add the condition that $\mu$ is globally supported (gives positive measure to every open set in $X$). Basically you are just trying to disallow $X$ from having any "lower-dimensional pieces" – Adam Jan 24 '21 at 08:52
  • Let us continue this discussion in chat. – ndurvasula Jan 24 '21 at 20:57
  • If $H^d$ is globally supported, do we get a lower bound on $H^d(S_\delta)$ in terms of $\delta$ (assuming $H^d(S)$ is fixed), or can it be arbitrarily close to zero? I think I'm missing the picture for why $S_\delta$ has an open ball. – ndurvasula Jan 25 '21 at 05:32
  • If $x \in S$, then by definition the ball of radius $\delta$ centered at $x$ is contained in $S_\delta$ – Adam Jan 25 '21 at 05:37
  • Ah I'm so sorry -- I messed up my quantifier -- I meant $H^d({x \in X \setminus S \mid \rho(x,S) < \delta })$ – ndurvasula Jan 25 '21 at 05:39