I will copy the following definitions of Exercise 1.1.14 of the book "An introduction to measure theory" by Terence Tao.
Define a dyadic cube to be a half-open box of the form \begin{align} \left[\frac{i_1}{2^n}, \frac{i_1+1}{2^n}\right)\times\cdots\times \left[\frac{i_d}{2^n}, \frac{i_d+1}{2^n}\right) \end{align} for some integers $n,i_1,\ldots,i_d$. Let $E\subset\mathbb{R}^d$ be a bounded set. For each integer $n$, let $\mathcal{E}_{\star}(E, 2^{-n})$ denote the number of dyadic cubes of sidelength $2^{-n}$ that are contained in $E$, and let $\mathcal{E}^\star(E, 2^{-n})$ denote the number of dyadic cubes of sidelength $2^{-n}$ that intersect $E$.
The exercise then wants the reader show that $E$ is Jordan measurable if and only if \begin{align} \lim_{n\rightarrow\infty} 2^{-dn}(\mathcal{E}^\star(E, 2^{-n}) - \mathcal{E}_\star(E, 2^{-n})) = 0, \end{align} which is fine.
Now, let us forget about Jordan measurability, etc. Given $E$, I am interested in how $f(n) = \mathcal{E}^\star(E, 2^{-n}) - \mathcal{E}_\star(E, 2^{-n})$ behaves. For example, let $E = [0,\pi]$, then $f(n) = 1,\,\forall n$. Or, pick $E$ to be a sufficiently nice subset of $\mathbb{R}^2$, for example the unit Euclidean disk, then $f(n) \sim 2^n$. It seems to me that for a nice $E$, we have $f(n) \sim 2^{n(d-1)}$, so that the "bad" dyadic cubes (that intersect $E$ but are not contained in $E$) have total (say Lebesgue) measure $\sim 2^{n(d-1)}2^{-dn} \sim 2^{-n}$. I like this exponential decay, and would like to find out which class of sets has this property (I am aware of counterexamples such as $E = $ The Smith-Volterra-Cantor set.)
Thanks in advance for any comments/references.
