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



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


