Let $(X,d)$ be a metric space. If set $A\subseteq X$, let $H^{\alpha}$ be the $\alpha$-dimensional Hausdorff measure on $A$, where $\alpha\in[0,2]$ and $\text{dim}_{\text{H}}(A)$ is the Hausdorff dimension of set $A$.
Problem: If the set $A\subseteq[0,1]\times[0,1]$, I want to define a uniform $A$ w.r.t $\text{dim}_{\text{H}}(A)$. For instance,
- When $\text{dim}_{\text{H}}(A)=0$ and $A$ is finite, then for $n\in\mathbb{N}$, suppose we partition set $[0,1]\times[0,1]$ into $n^2$ squares with side-length $1/n$. If there exists $n\in\mathbb{N}$, such that there is only one point from $A$ at the center of each square, we then say $A$ is uniform in $[0,1]\times[0,1]$.
- When $\text{dim}_{\text{H}}(A)=2$, for all real $x_1,x_2,y_1,y_2$ and constant $\lambda$, if $0\le x_1<x_2\le 1$ and $0\le y_1<y_2\le 1$ with the Lebesgue measure (for sets in the Lebesgue sigma-algebra) of $([x_1,x_2]\times[y_1,y_2])\cap A$ having measure $\lambda(x_2-x_1)(y_2-y_1)$, then set $A$ is uniform in $[0,1]\times[0,1]$.
Here is the question I asked:
What about for $\text{dim}_{\text{H}}(A)\in(0,2)$?
According to @RyanBundey,
If the set has Hausdorff dimension less than $2$, presumably you would want to say the Hausdorff measure of the part of the set $A$ in your "box" is some constant multiple of the content of your box.
What does "content" mean? How do we define @RyanBundey's answer mathematically?