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I'm having trouble desribing the following:

Suppose, using the Lebesgue Outer measure, we define a function $f:[0,1]\to[0,1]$ that is measurable in the sense of Caratheodory.

I want to make the function, that answers the main question of this post, as non-uniform as possible in $[0,1]\times[0,1]$ yet satisfies criteria 4. Here is my attempt to define a measure of non-uniformity:

For $n,j\in\mathbb{N}$ (and $j<n$), suppose we partition set $[0,1]\times[0,1]$ into $n^2$ squares, such that we combine these squares to form rectangles/squares with area $1/(j^2)$. Furthermore, suppose we define $i,r,m,t\in\mathbb{N}$, where the side of each of the rectangles/squares (parallel to the y-axis) is the interval $[y_i,y_{i+r}]$ with length $1/(n+r)$ such that $0\le i+r \le n-1$. Also, suppose the side of each rectangle/square (parallel to the x-axis) is interval $[x_m,x_{m+t}]$ with length $1/(n+t)$ such that $0\le m+t \le n-1$.

I would like to divide the Lebesgue measure of each $f^{-1}([y_{i},y_{i+r}])\cap[x_m,x_{m+t}]$ such that $(1/(n+r))\cdot(1/(n+t))={1}/{(j^2)}$ and divide each one by the total area of all the squares/rectangles with area $1/(j^2)$. This results in a discrete probability distribution.

We then want to take the entropy of the discrete distribution from the previous paragraph.

Note the smaller $\lim_{j\to\infty}\lim_{n\to\infty}$ of the entropy of the previous equation divided by $\log_2$ of the total number of squares/rectangles with area $1/(j^2)$, the more non-uniform the points in $[0,1]\times[0,1]$.

Question:

How do we rigorously describe my definition in the block-quote of non-uniformity?

Arbuja
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1 Answers1

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Here is my second attempt (someone correct me if I'm wrong):

For $n,j\in\mathbb{N}$ (and $j<n$), suppose we partition set $[0,1]\times[0,1]$ into $n^2$ squares with length $n$, such that we combine these squares to form larger rectangles/squares with area $1/(j^2)$. Furthermore, we also define $i,r,m,t\in\mathbb{N}\cup\left\{0\right\}$, where $y_{i+1}-y_i=1/n$ and $0\le i+r \le n-1$ such that the length of each of the rectangles/squares (parallel to the y-axis) is interval $[y_i,y_{i+r}]$ with length $1/(n+r)$. Also, suppose $x_{m+1}-x_m=1/n$, where $0\le m+t \le n-1$ and the side of each rectangle/square (parallel to the x-axis) is interval $[x_m,x_{m+t}]$ with length $1/(n+t)$.

If the Lebesgue measure on the sigma algebra of caratheodory-measurable sets is $\lambda$, then I would like to divide the Lebesgue measure of each $f^{-1}([y_{i},y_{i+r}])\cap[x_{m},x_{m+t}]$ such that the area of $[y_i,y_{i+r}]\times[x_{m},x_{m+t}]$ is ${1}/{(j^2)}$ and we divide each of those measures by the total area of all the "squares/rectangles with area $1/(j^2)$". This results in a discrete probability distribution $\mathbb{P}$:

\begin{alignat}{3} & S(j,n)=\{&&(i,r,m,t): i,r,m,t\in\mathbb{N}\cup\left\{0\right\}, y_{i+1}-y_i=x_{m+1}-x_m=1/n, \\ & && \text{Area}([y_i,y_{i+r}]\times[x_{m},x_{m+t}])=1/j^2\} \end{alignat}

$$\mathbb{P}(j,n)=\left\{\frac{\lambda(f^{-1}([y_{\text{i}},y_{\text{i}+\text{r}}])\cap[x_{\text{m}},x_{\text{m}+\text{t}}])}{1/(j^2)\cdot|S|}:\text{i},\text{r},\text{m},\text{t}\in S(j,n)\right\}$$

We want to then take the entropy of the discrete distribution.

$$\text{E}(\mathbb{P}(j,n))=\sum\limits_{x\in\mathbb{P}(j,n)}-x\log_{2}x$$

Next, we wish to define the non-uniformity i.e. $\mathcal{U}(f,[0,1]\times[0,1])$ of a distribution using the limit in terms of $j$ and $n$ of the relative similarity of two distributions i.e., the subtraction of one and the relative change between $\text{E}(\mathbb{P}(j,n))$ and the entropy of a discrete uniform distribution with cardinality $|S|$ i.e. $\log_{2}{|S|}$:

$$\mathcal{U}(f,[0,1]\times[0,1])=1-\lim\limits_{j\to\infty}\left(\lim\limits_{n\to\infty}\frac{\log_{2}(|S|)-\text{E}(\mathbb{P}(j,n))}{\log_{2}(|S|)}\right)$$

Summary: The value $\mathcal{U}(f,[0,1]\times[0,1])$ should be as small as possible.

Arbuja
  • 53