Subsets $X$ such that their Hausdorff outer measure is not finite

Question

Let $H^d:\mathcal{P}(\mathbf{R}^n) \to \mathbf{R}\cup \{\infty\}$ be the $d$-dimensional Hausdorff outer measure on $\mathbf{R}^n$, for some $0<d<n$ with $n$ integer, which is constructed in the following way:

For each $\delta>0$ and $X \subseteq \mathbf{R}^n$, define $$ H_\delta^d(X)=\inf\left\{\sum_{n=1}^\infty \left(\mathrm{diam}\,S_n\right)^d: \bigcup_{n=1}^\infty S_n \supseteq X, \mathrm{diam}\,S_n< \delta \right\}, $$ where $\mathrm{diam}\,S_n$ stands for the diameter of $S_n$. Then, $H^d$ is defined by $$ X\mapsto \lim_{\delta\to 0^+} H_\delta^d(X). $$

(The limit is meaningful because the function $\delta\mapsto H_\delta^d(X)$ is nonincreasing; moreover, it is well known that $H^d(X) \le H^d(Y)$ whenever $X\subseteq Y$.)

Question: Let $X\subseteq \mathbf{R}^n$ such that $H^d(X)=\infty$ and fix a constant $c$. Does there exist a subset $Y\subseteq X$ such that $$ H^d(Y) \in [c,\infty[\,\,\,? $$

Answer 1

Yes, if $X$ is an analytic set. See:

• Besicovitch, On existence of subsets of finite measure of sets of infinite measure, 1952
• R. O. Davies, Subsets of finite measure in analytic sets, 1951.

For discussion you may consult e.g.

http://arxiv.org/abs/1408.1999

Answer 2

The following is Corollary 7 of [1].

Theorem: For $X$ (an analytic subset of) a complete separable metric space, and $ s \in [0,\infty)$, the following is true about the Hausdorff measure $\mathcal{H}^s$: For every $ l < \mathcal{H}^s(X) $ there exists a compact subset $A \subset X$ such that $$ l < \mathcal{H}^s (A) < \infty \, . $$

[1] J. D. Howroyd, On dimension and on the existence of sets of finite positive Hausdorff measure, 1993