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



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> For a given $0<\beta <\alpha$ can we find a subset $B\subset A$ with Hausdorff dimension $\beta$?
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In case this is true, could you provide a reference for this statement?


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



Existence of subset with given Hausdorff dimension
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 [1] there is a compact $K\subset A$ such that $0<H^\alpha(K)<\infty$.


More generally, if $A$ is *Souslin* then by Theorem 5.6 from [1] again there is a compact $K\subset A$ such that $0<H^\alpha(K)<\infty$ (as @SeverinSchraven noted).




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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](https://mathoverflow.net/questions/250275/subsets-x-such-that-their-hausdorff-outer-measure-is-not-finite#comment615483_250280) to [a closely related post](https://mathoverflow.net/questions/250275/subsets-x-such-that-their-hausdorff-outer-measure-is-not-finite), where an interesting paper [2] 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 [1]). 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 $\gamma < \omega\_1$.




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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>


