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 $$ H^d(Y) \in [c,\infty[\,\,\,? $$


2 Answers 2


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.


  • 2
    $\begingroup$ And no in general if you believe in AC and CH. Indeed, the $G_\delta$ sets of finite Hausdorff measure are continuum. Fix an enumeration $E_\alpha$ of them by a well-ordered set (AC) so that for every $\alpha$ the set of indices $\beta\le\alpha$ is countable (CH). For every $\alpha$ choose a point outside the union $\cup_{\beta\le\alpha} E_\beta$. The resulting set $E$ has a point outside every $E_\alpha$ and intersect every $E_\alpha$ by a countable set. I surmise that is written somewhere in the reference Bjorn gave but I got scared by the abstract already, so I decided to spell it out :-) $\endgroup$
    – fedja
    Commented Sep 20, 2016 at 11:14
  • $\begingroup$ @fedja that sounds like an answer! No it's not in the arxiv.org link I think $\endgroup$ Commented Sep 20, 2016 at 15:56

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.