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This is connected to this question. The main difference is criteria 4.

Main Question:

Using the Lebesgue outer measure, does there exist an explicit and bijective function $f:[0,1]\to[0,1]$ such that:

  1. the function $f$ is measurable in the sense of Caratheodory
  2. the graph of $f$ is dense in $[0,1]\times[0,1]$
  3. the range of $f$ is $[0,1]$
  4. the pre-image of each sub-interval of $[0,1]$ under $f$ (where each sub-interval has some length $\ell\in[0,1]$) has a Lebesgue measure of $\ell$
  5. the graph of $f$ is non-uniform (i.e. without complete spacial randomness) in $[0,1]\times[0,1]$
  6. using the Lebesgue measure, the expected value of $f$ is computable?

Motivation: I wanted to define an explicit and bijective function $f:[0,1]\to[0,1]$ where the graph of $f$ is somewhat but not too evenly distributed (i.e. with complete spacial randomness) in $[0,1]\times[0,1]$, such that using the uniform probability measure, we want a subset $X\subseteq[0,1]$, where (when function $f:[0,1]\to[0,1]$ is restricted to $f:X\to[0,1]$) the expected value of $f$ is undefined so we can find an unique extension of the expected value of $f$ which gives a finite value instead.

Question on motivation: If the function from the main question exists, does it satisfy the motivation?

Attempt to Solve Both Questions:

I can't prove an explicit example exists. For example, the Conway base-13 function satisfies some criteria of the main question but is defined as $f:[0,1]\to\mathbb{R}$, not computable, and surjective.

Furthermore, here is an attempt from this, this, and this post (note in the links, neither of the main questions/answers satisfy the motivation, e.g. in this answer, the function wasn't very explicit; in this answer, the function was extremely “non-uniform”; and with the main question of this post, the answer may not satisfy the motivation):

In case one wants to read here, here's the attempt:

Suppose the base-$3$ expansion of real numbers, in interval $[0,1]$, have infinite decimals that approach $x\in[0,1]$ from the right side so when $0\le x_1,x_2\le 1$ (and $x_1=x_2$) we get $f(x_1)=f(x_2)$.

Furthermore, for $\mathbb{N}\cup\left\{0\right\}=\mathbb{N}_{0}$, if $r\in\mathbb{N}_{0}$ and $\text{digit}_{3}:\mathbb{R}\times \mathbb{Z}\to\left\{0,1,2\right\}$ is a function where $\text{digit}_{3}(x,r)$ takes the digit in the $3^{r}$-th decimal fraction of the base-$3$ expansion of $x$ (e.g. $\text{digit}_{3}(1.789,2)=\text{digit}_{3}({1.210022{\cdot\cdot\cdot}}_{3},2)=1$), then $\left\{{g_r}^{\prime}\right\}_{r\in\mathbb{N}_{0}}$ is a sequence of functions (and $\left[\cdot\right]$ is the nearest integer function) such that ${g_r}^{\prime}:\mathbb{N}_0\to\mathbb{N}_0$ is defined to be:

\begin{equation} g_r^{\prime}(x)=\left[\frac{10}{3}\sin(rx)+\frac{10}{3}\right] \end{equation}

then for some function $k:\mathbb{N}_{0}\to\mathbb{N}_{0}$, where $k$ is strictly increasing and $k(0)$ is a positive number, we want the the intermediate function (before $f$) or $f_{1}:[0,1]\to[0,10]$ to satisfy the main question (such that, in criteria 3. the range of $f_1$ is $[0,10]$).

\begin{alignat}{2} & f_{1}(x) = &&\left|\left(\sum\limits_{r=0}^{\infty} g_{r+1}^{\prime}\!\left(\sum\limits_{p=r}^{r+k(r)}\text{digit}_{3}(x,p)\right)\!\!\bigg/3^{r}\right)-10\right|= \label{eq:025} \\ & && \left|\left(\left(\sum\limits_{r=0}^{\infty}\left[\frac{10}{3}\sin\left(\left(r+1\right)\left(\sum\limits_{p=r}^{r+k(r)}\text{digit}_{3}(x,p)\right)\right)+\frac{10}{3}\right]\right)\!\!\bigg /3^{r}\right)-10\right| \nonumber \end{alignat}

(One example of $k(r)$ that may satisfy main question, when with criteria 3. the range of $f_1$ is $[0,10]$ instead of $[0,1]$, is $k(r)=10r+20$)

What we’re doing with $f_1$ is we are converting every digit of the base-$3$ expansion of $x$ into a pseudo-random number that is non-equally likely to be an integer, including and in-between, $0$ and $20/3$. Furthermore, we attempt to make the function dense in $[0,1]\times[0,10]$ by adding the $3^{r}$-th decimal fraction with the next $3^k$-th decimal fractions; however, we also want to control the end-points of $[0,10]$, such that $f_1$ is dense in $\left[0,1\right]\times\left[0,1\right]$ by manipulating $f_1$ to get:

\begin{alignat}{2} & f(x) = && 1-\frac{1}{10}f_1(x)\label{eq:109}\\ & && 1-\left(\frac{1}{10}\right)\left|\left(\left(\sum\limits_{r=0}^{\infty}\left[\frac{10}{3}\sin\left(\left(r+1\right)\left(\sum\limits_{p=r}^{r+k(r)}\text{digit}_{3}(x,p)\right)\right)+\frac{10}{3}\right]\right)\!\!\bigg/3^{r}\right)-10\right| \nonumber \end{alignat}

(e.g. where $k(r)=10r+20$). You can use programming to visualize $f$ though I don't know if you can graph the entire function. (The programming I used is mathematica.)

Clear["Global`*"]
k[r_] := k[r] = 
  20 (* You can adjust k[r]; however, mathematica is unable to graph \
f when k[r] is steepy increasing e.g. for this function, k[r] must be \
less than 25 for the code to show a graph. Instead, it should be k[r]=10r+20 *)

g1[xr_, r_] := g1[xr, r] = Round[(10/3) Sin[r xr] + (10/ 3)] (Converts the (3^r)th decimal fraction,in the base 3
expansion of the x-values in[x1,x2] (defined as xr or x_r not x
r)
into a psuedo-random number that's non-equally likely to spit a
number between,and including, 0 and 20/3 *)

f[x_] := f[x] = N[1 - ((1)/(10)) RealAbs[ Sum[g1[Sum[ RealDigits[x, 3, k[r], -r][[1]][[z]], {z, r + 1, k[r]}], r + 1]/(3^r), {r, 0, 8}] - 10]] (Defines function f,I assume the larger k[r]'s values, the more
the function appears dense in [0,1]x[0,1]
)

p = .00005 (Incremement between the x-values in the points of the
graph below
)

ListPlot[Table[{x, f[x]}, {x, p, 1, p}]] (Graphs countable points of the functions but is not a
complete accurate graph. There are uncountably many points that need
to be included.
)

Image of Code

Unfortunately, I only studied up to intro to advanced mathematics. (Without a deep undestanding of math I'm unable to prove if the function gives what I'm looking for.)

Is there a simpler example?

Arbuja
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    What exactly do you mean by "computable" in this context? The usual notion of computability of real functions entails continuity. Also, "bijective" (specifically, the injectivity part) is contradicted by "the preimage of every sub-interval of $[0,1]$ is $[0,1]$" (maybe you just want the preimage to be dense in $[0,1]$?). – Noah Schweber May 30 '23 at 05:27
  • @NoahSchweber I made edits. Is this clearer? – Arbuja May 30 '23 at 14:58
  • Given its current form, why is Conway's base-13 function not an answer? – Noah Schweber May 30 '23 at 15:13
  • @NoahSchweber Conway’s base-13 function is explicit but it’s not computable. At least, that’s what this answer says. How do we define “explicit” and “computable” according to the answer? – Arbuja May 30 '23 at 15:21
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    The default notion is that of computable analysis: basically, $f$ is computable iff there is a Turing machine which produces arbitrarily-accurate approximations to $f(x)$ given arbitrarily-accurate approximations to $x$ (see Chapter 3 of the linked notes for a precise definition). Importantly, and per my original comment, every computable function in this sense is continuous. So if you want a hope of a positive answer to your question you need a weaker notion of computability. At that point the onus is on you to articulate what you want. – Noah Schweber May 30 '23 at 16:23
  • @NoahSchweber Turns out from this comment, the Conway base-13 function doesn’t satisfy the main question. – Arbuja May 30 '23 at 20:29
  • Turns out the main question of this post doesn't satisfy my motivation either (i.e. in this answer the points in the graph of $f$ are extremely "non-uniform"). Hence, I made a new question. – Arbuja May 31 '23 at 23:14

1 Answers1

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Edit: From this reddit comment, the answer to the main comment doesn't satisfy my motivation.

Let D denote the dyadic rationals (whose denominators are powers of 2); note that D has Lebesgue measure 0, since D is countable. Hence it suffices that f is a bijection from D onto itself such that every interval in D has dense image in D, and f(x)=x when x is not in D.

Here's a possible construction for f on D. Every x in D has a finite binary expansion

$x = 0.a_1 a_2 ... a_{n-1} 1$,

and in this case we will define

$f(x) = 0.a_{n-1} ... a_2 a_1 1$

ie. f reverses the binary expansion of x, excluding the last digit (which is always a 1).

The function is almost completely non-uniform. Therefore, it does not satisfy my motivation.

Therefore, here is a link to the new question (note the main difference is criteria 4. where the sub-interval of $[0,1]$, intersected with the pre-image of another sub-interval of $[0,1]$ (under $f$), must equal the length of the original sub-interval times the length between the other sun-interval).

Arbuja
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