Defining a measure of uniformity for measurable subsets of $[0,1]^2$ w.r.t dimension $\alpha\in[0,2]$

Question

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 $A$.

Motivation: If the set $A\subseteq[0,1]\times[0,1]$, I would like to measure set $A$'s deviation from a uniform subset $A^{\prime}$ of $[0,1]\times[0,1]$ with Hausdorff dimension $\alpha$, where:

If set $A^{\prime}\subseteq[0,1]\times[0,1]$ has Hausdorff dimension $\alpha$ less than $2$; for all real $x_1,x_2\in J_n$ where sequence of sets $(J_n)_{n\in\mathbb{N}}=(\left\{1/n,2/n,3/n,\cdot\cdot\cdot,1\right\})_{n\in\mathbb{N}}$, if $0\le x_1<x_2\le 1$ and $\mu$ is the counting measure where:

$$\mu^{\star}(A)=\begin{cases} 1/\mu(A) & \mu(A)<+\infty\\ 0 & \mu(A)=+\infty \end{cases}$$

with $\mu^{\star}(A)< x_2-x_1\le 1$; we get $A^{\prime}$ is uniform in $[0,1]\times[0,1]$, when for $j\in\mathbb{N}$ and $j< n$, the number of squares $[x_1,x_2]\times[x_1,x_2]$ with area $j^2/n^2$ satisfying:

$$H^{\alpha}(([x_1,x_2]\times[x_1,x_2])\cap A^{\prime})=H^{\alpha}(\text{dom}(A^{\prime})){H}^{\alpha}(\text{range}(A^{\prime}))(x_2-x_1)^2$$ divided by the total # of squares with area $j^2/n^2$ approaches $1$ as $n,j\to\infty$ and $j/n\to 0$.

Moreover, I want to define a measure of uniformity to be between (and including) zero and one (or zero and infinity) such that the larger the measure, the smaller the non-uniformity.

Question: How do we define such a measure?

Answer 1

Here is another possible approach, perhaps closer to what the OP had in mind.

Let $S:=[0,1]^2$ be the unit square. "Partition" $S$ naturally into four congruent squares $S_{1,j}$ (with side length $1/2$ each), where $j=1,\dots,4$; the quotation marks are used here because the $S_{1,j}$'s will have some common boundary points. Next, "partition" each $S_{1,j}$ naturally into four congruent squares (with side length $1/2^2$ each), so that we get $4^2$ squares $S_{2,j}$ for $j=1,\dots,4^2$. Continue doing so, so that at the $k$th step we get $4^k$ squares $S_{k,j}$ for $j=1,\dots,4^k$, for each $k=1,2,\dots$.

Take any subset $A$ of $S$. For each $k=1,2,\dots$ and each $j=1,\dots,4^k$, let $$A_{k,j}:=(A\cap S_{k,j})-s_{k,j},$$ where $s_{k,j}$ is the southwest vertex of the square $S_{k,j}$, so that $A_{k,j}\subseteq S_k:=2^{-k}S$.

Suppose that for each $k$ we have a "measure" $D_k$ of dissimilarity for subsets of $S_k$, so that for any two subsets $B$ and $C$ of $S_k$ we have a nonnegative real number $D_k(B,C)$, which is the greater the more "dissimilar" $B$ and $C$ are (and is $0$ if $B=C$); here the term "measure" is used in the general sense, not necessarily in the sense of measure theory. For instance, $D_k(B,C)$ may depend on the Hausdorff distance between $B$ and $C$ or on some "measure" of the symmetric difference of the sets $B$ and $C$ or on some combination thereof.

Then the distance of the set $A$ from uniformity can be defined by the formula $$D(A):=\sum_{k=1}^\infty\frac1{L^k}\sum_{j=1}^{4^k}\sum_{m=1}^{4^k} \frac{D_k(A_{k,j},A_{k,m})}{1+D_k(A_{k,j},A_{k,m})},$$ where $L$ is a real number $>16$ (to ensure the convergence of the series). Then $D(A)$ will be small if, for "most" levels $k$ of "zooming", "most" of the intersections of the set $A$ with all the "$k$-level" small squares $S_{k,j}$ "look similar" to one another. (Of course, $D(A)$ will depend on the choices of $L$ and the dissimilarity "measures" $D_k$.)

For instance, for any $L$ and any $D_k$'s we have $D(S)=0$ -- of course, the unit square $S$ is at distance $0$ from uniformity (in itself).

As another example, for the uniform grid $G_n$ (defined in the previous answer) with $n=2^K$ for a natural $K$, any real $L>16$, and any $D_k$'s we have $$D(G_n)\le \sum_{k=K+1}^\infty\frac1{L^k}\,16^k =Cn^{-p}\to0$$ as $n=2^K\to\infty$, where $C:=\dfrac{16}{L-16}$ and $p:=\log_2\dfrac L{16}$. So, we see that $G_n$ is close to uniformity for large $n$, even though the Hausdorff dimension of $G_n$ is $0$ for all $n$ (in big contrast with the Hausdorff dimension of $S$, which is $2$). Thus, again we see that the Hausdorff dimension can hardly have anything to do with the idea of uniformity.

Answer 2

Responding to the latest comment by the OP: "How would you suggest measuring the uniformity of measurable subsets of the unit square?" :

1. I think the idea of uniformity has hardly anything to do with the Hausdorff dimension (HD). E.g., I think almost everyone would agree that the entire unit square $S:=[0,1]^2$ (of HD$=2$) is maximally uniform (in itself), whereas a fine uniform grid of $n^2$ points (of HD$=0$) in $S$ is approximately uniform in $S$ if $n$ is large. This, again, suggests that the HD has little (if anything) to do with the idea of uniformity.

2. Moreover, it makes more sense to talk, not about subsets of $S$, but about probability distributions over $S$ being close to uniform -- that is, to the uniform probability distribution over $S$.
   In particular, if $A$ is a finite subset of $S$, we can attach a weight/probability mass $w(x)>0$ to each $x\in A$ so that $\sum_{x\in A}w(x)=1$, and thus we get a probability distribution over $S$ supported on $A$. This way, we can identify any finite subset $A$ of $S$ with the uniform probability distribution supported on $A$. This also allows us to consider multisets $B$ in $S$, by giving each of the (possibly repeated) elements $x$ of $B$ the weight $w(x)$ proportional to the multiplicity of $x$ in $B$. Also, we can, somewhat similarly, identify a measurable subset $A$ of $S$ of positive Lebesgue measure with the uniform distribution over $A$. Similarly we can do for, say, rectifiable curve images $C\subset S$, to identify $C$ with the uniform distribution over $C$ in the length sense; e.g., we can identify a circle contained in $S$ with the uniform distribution over the circle. Yet more generally, for any subset $M$ of $S$, if we can somehow define the uniform distribution over $M$, we can identify $M$ with that uniform distribution.

3. Anyhow, if $U$ denotes the uniform probability distribution over $S$ and $P$ is any probability distribution over $S$, then we can measure the closeness of $P$ to $U$ by any probability metric (say, by one metrizing the weak convergence of probability distributions over $S$). For instance, we can use the Wasserstein distance $W_1(P,U)$ from $P$ to $U$ (based, say, on the Euclidean distance between points of $S$), which is the cost of the optimal transportation of the probability mass distribution $P$ to the uniform probability mass distribution $U$. So, e.g., if $P_n$ is the uniform distribution on the uniform grid $G_n:=\{(\frac in,\frac jn)\colon i=0,\dots,n,\;j=0,\dots,n\}$ in $S$, then $W_1(P_n,U)\asymp\frac1n$, which is small for large $n$ -- that is, the uniform distribution on the uniform grid $G_n$ is close to uniformity if $n$ is large; this seems to make sense. (Recalling that any finite subset $A$ of $S$ can be identified with the uniform probability distribution supported on $A$, we can now also say that the uniform grid $G_n$ itself is close to uniformity if $n$ is large.)