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

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

Yes, if $X$ is an analytic set. See:
• 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 :-) – fedja Sep 20 '16 at 11:14