# Existence of subset with given Hausdorff dimension

Let $$A\subseteq \mathbb{R}$$ be Lebesgue-measurable and let $$0<\alpha<1$$ be its Hausdorff dimension.

For a given $$0<\beta <\alpha$$ can we find a subset $$B\subset A$$ with Hausdorff dimension $$\beta$$?

In case this is true, could you provide a reference for this statement?

Added: Actually I am happy if $$A$$ is compact.

First of all, $$\dim_{H} (A) = \alpha$$ iff $$H^k(A)=\infty$$ for all $$k<\beta$$ and $$H^k(A) = 0$$ for all $$k>\beta$$. Then $$H^\alpha(A) = \infty$$ for all $$\alpha \in (0,\beta)$$.

If $$A$$ is closed then by Theorem 5.4 from  there is a compact $$K\subset A$$ such that $$0.

More generally, if $$A$$ is Souslin then by Theorem 5.6 from  again there is a compact $$K\subset A$$ such that $$0 (as @SeverinSchraven noted).

If $$A$$ is not Souslin then (at least assuming AC and CH) it may happen that $$H^\beta(B) = 0$$ for any $$B\subset A$$ and any $$\beta \in (0, \alpha)$$. In order to justify this claim let me add adopt the argument from a comment by fedja to a closely related post, where an interesting paper  is also mentioned.

Let $$C$$ denote the standard Cantor set in $$[0,1]$$ and take $$\alpha = \dim_H(C) = \frac{\ln 2}{\ln 3}$$. Take $$\beta \in (0,\alpha)$$.

Consider the family $$\mathcal E$$ of all $$G_\delta$$-subsets $$E \subset C$$ such that $$H^\alpha(E)=0$$. Then for any countable family $$(E_i)_i\in \mathcal E$$ the set $$C \setminus \bigcup_{i=1}^\infty E_i$$ has infinite $$\beta$$-dimensional Hausdorff measure and in particular is not empty.

Assuming AC+CH one can enumerate the family $$\mathcal E$$ using ordinals as $$\mathcal E = \{E_\gamma : \gamma < \omega_1\}$$, where $$\omega_1$$ is the first uncountable ordinal.

For each $$\gamma < \omega_1$$ take $$x_\gamma \in C \setminus \bigcup_{\lambda < \gamma} E_\lambda$$. Now we finally define the set $$A:= \{x_\gamma : \gamma < \omega_1\}$$.

If $$H^\alpha(A)$$ was zero then one could cover $$A$$ with $$E_\gamma$$ for some $$\gamma$$ (see e.g. Theorem 1.6 in ). This contradicts the definition of $$A$$, hence $$H^\alpha(A)>0$$.

Now if $$B \subset A$$ and $$H^\beta(B)\in (0,\infty)$$ then $$H^\alpha(B)=0$$ and then there exists $$\gamma<\omega_1$$ such that $$B \subset E_\gamma$$, hence $$H^\beta(B) = H^\beta(B\cap E_\gamma) \le H^\beta(A\cap E_\gamma) = 0$$ since the intersection $$A\cap E_\gamma$$ is at most countable for any $$\gamma < \omega_1$$.

References.

1. The Geometry of Fractal Sets by K.J. Falconer (1985)

2. Finding subsets of positive measure by Bjørn Kjos-Hanssen and Jan Reimann (2014) https://arxiv.org/abs/1408.1999

• That's exactly what I was looking for, thanks very much. Mar 15, 2019 at 22:01
• Actually it is worth mentioning that in the same reference in Theorem 5.6, this is generalized to Souslin spaces. Mar 16, 2019 at 0:06
• @SeverinSchraven Thank you, I have included this remark into my answer. Mar 16, 2019 at 17:56

The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see  and references therein.

Theorem. If a compact set $$A\subset\mathbb{R}^n$$ has non-$$\sigma$$-finite measure $$\mathcal{H}^\beta$$, then there exists a subset $$B\subset A$$ such that $$0<\mathcal{H}^\beta(B)<\infty$$.

 R.O. Davies, A theorem on the existence of non-σ-finite subsets. Mathematika 15 (1968), 60–62.

• I actually have a question (which is the reason why I wrote that my answer is partial): in the OP it is assumed that $A$ is Lebesgue measurable. Strictly saying this does not imply that $A$ is Borel, so it is not immediate that the result we both eventually refer to can be used (if $A$ were Borel or at least Souslin then yes). Mar 15, 2019 at 20:56
• @Skeeve Good point. I do not know what happens in general, but perhaps the compact case is sufficient for the needs of OP. Let's hear from him. I will changes my answer to emphasize that it only answers the compact case. By the way you know a lot of geometric measure theory so you should be more active :) There are not too many of us. Mar 15, 2019 at 21:23
• @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer. Mar 15, 2019 at 22:01
• @PiotrHajlasz It seems to me that in the general case under certain axioms bad things might happen. I have updated my answer with some thoughts on that. By the way thank you, I will try to be more active :) Even though you probably overestimate my knowledge of GMT, I hope to learn it better in nice discussions on this site (in particular with you, of course!) Mar 16, 2019 at 18:03

The following is Corollary 7 of .

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 $$l < \mathcal{H}^s (A) < \infty \, .$$

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