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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 the repeated elements $x$ of $B$ weights $w(x)$ 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.)

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.)

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 the repeated elements $x$ of $B$ weights $w(x)$ proportional to the multiplicity of $x$ in $B$.

3. So, 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.)

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 the repeated elements $x$ of $B$ weights $w(x)$ 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.)

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 the repeated elements $x$ of $B$ weights $w(x)$ proportional to the multiplicity of $x$ in $B$.

3. So, 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.

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 the repeated elements $x$ of $B$ weights $w(x)$ proportional to the multiplicity of $x$ in $B$.

3. So, 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.)