I would like to construct a set $\Sigma\subset\mathbb{R}^n$ such that $$ mr^\alpha\leq\mathscr{H}^\alpha(B(0,r)\cap\Sigma)\leq M r^\alpha, $$ for all $0\leq r\leq1$ and some $0<m\leq M<\infty$. Here $0\leq\alpha\leq n$ and $B(0,r)$ denotes the ball in $\mathbb{R}^n$ of radius $r$ centred at $0$.
The question is very easy for integer dimension (simply take a hiperplane of dimension $\alpha$), so one can also assume that $n-1<\alpha< n$, otherwise the question reduces to a lower dimension, by constructing $\Sigma$ as a subset of a hyperplane of dimension $\lceil \alpha\rceil$.
I should add that it would be very nice if $\Sigma$ is trapped in a given open cone with apex at $0$ (in which case one should impose $\alpha>0$).
Does anyone have any suggestions? Many thanks!
Let $\Sigma=\mathbb{R}^{n-1}\times F$ where ${\rm dim}\ F=\gamma$ and $n-1<\alpha =n-1+\gamma <n$. Let $F_1=[0,1]$ and $F_2$ be a union of $n_1$ intervals of length $l_1$ s.t. $n_1l_1^\gamma=1$.
Similarly, $F_3$ has $n_1$ components in a component of $F_2$ s.t. $F_3$ is union of $n_1^2$ intervals of length $ l_2$ and $n_1^2l_2^\gamma =1$.
Hence $F_n$ is $n_1^{n-1}$ intervals of length $l_n$ with $n_1^{n-1}l_{n-1}^\gamma=1 $. So let $F=\bigcap_{i=1}^\infty\ F_i$.
[Add] In $\mathbb{R}^n$, note that there is a subset $A$ in $ \mathbb{S}^{n-1} $ s.t. $[0,1]^n$ and $[r,R]\cdot A$ are bi-Lipschitz.
Hence your construction holds.