Main Question:
Does there exist a function $f:[0,1]\to[0,1]$ such that:
- the function $f$ is measurable in the sense of Caratheodory
- the graph of $f$ is dense in $[0,1]\times[0,1]$
- the collection of all subsets of the range of $f$ with pre-images under $f$ that are measurable in the sense of Caratheodory is a non-perfect dense set in the collection of all subsets of the range of $f$, such that we define a topology where:
Let ${\mathbb P}[0,1]$ be the collection of all subsets of $[0,1]$ modulo the equivalence relation $\sim$ defined by $E \sim F \Leftrightarrow {\lambda^{*}(E \Delta F)} = 0,$ where $\lambda^{*}$ is Lebesgue outer measure and $\Delta$ is the symmetric difference operation on sets. The set ${\mathbb P}[0,1]$ can be made into a complete metric space by defining the distance function $d,$ where $d(E,F) = {\lambda^{*} (E \Delta F)}.$
- the graph of $f$ is non-uniform (i.e. without complete spacial randomness) in $[0,1]\times[0,1]$
- using the uniform probability measure, the expected value of $f$ is undefined?
Motivation: We want to define a 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, the expected value of $f$ is undefined so we can find an unique extension of the expected value which gives a finite value instead.
Question on motivation: If the function from the main question exists, does it give the same function as the one from the motivation?
Simplified Version of the Main Question:
In the main question, a non-perfect dense set in the collection of all subsets of the range of $f$ is “topologically large” in the collection of all subsets of the range of $f$, where the main question translates to the following:
Does there exist measurable function $f:[0,1]\to[0,1]$ such the graph of $f$ is
dense in $[0,1]\times[0,1]$, where the pre-image of a
“random subset” of the range of $f$ under $f$ is “likely” to be measurable in the sense of
Caratheodory, where the graph of $f$ is non-uniform (i.e. without complete spacial randomness) in $[0,1]\times[0,1]$, and w.r.t the uniform probability measure the expected value of $f$ is undefined?
Attempt to Solve Both Questions:
I can't prove an explicit example exists but here is my attempt from this question (note in the link, the previous question didn't give a function that satisified the motivation in this post):
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, such that $k(0)$ is a positive
number, and
the intermediate function (before $f$) or $f_{1}:[0,1]\to[0,10]$
satisfies the main question if the range is
$[0,10]$ instead of $[0,1]$.\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 the problem, i.e. if the
range is $[0,10]$, is $k(r)=10r+20$)What we are doing with $f_1$ is we converted 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 $20/3$. Further, we attempt to
make the function dense in $[0,1]\times[0,10]$ by adding the
$3^{r}$-th decimal fraction with the next $k$ decimal fractions;
however, we want to control the end-points of $[0,10]$ such that $f_1$
is dense in $\left[0,1\right]\times\left[0,1\right]$ (instead of
$\left[0,1\right]\times[0,10]$) 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. $k(r)=10r+20$) 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.*)
Unfortunately, I only studied up to intro to advanced mathematics. This could be non-sense. (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?
Best Answer
Preliminary observation: property 5 can never be satisfied, since every bounded and measurable functions on $[0,1]$ has a well-defined integral against the uniform (i.e. Lebesgue) measure.
The following example might work. Fix a sequence $(x_n,y_n)_{n} \in \mathbb Q^2$ dense in $[0,1]^2$ such that all the $x_n$ are distinct and no $y_n$ is $0$ or $1$. Let for every $n$ $$ 1_n(x) = \begin{cases} 1 &\text{ if }x=x_n \\0 &\text{ otherwise}\end{cases}$$ and define $$ f = \sum y_n 1_n . $$
$f$ is a sum of countably many Borel-measurable functions, thus is measurable.
The graph $G$ of $f$ is $$ [0,1]\times\{0\} \ \cup \ \{(x_n, y_n) : n\}\ \setminus\ \{(x_n, 0) : n\} . $$ It is dense because of our choice of $(x_n,y_n)_n$.
The preimage of any subset of $\mathbb R$ by $f$ is the union of a subset of $\mathbb Q$, together with (possibly) $[0,1]\setminus\{x_n:n\}$. They are thus Borel-measurable. This means that the set of image of Borel-measurable sets by $f$ is, in fact, the collection of every subset of the range of $f$.
The Lebesgue measure of G, given by $\int_G dxdy$, will always be zero by Fubini’s theorem: since the indicator function $[G]$ of $G$ is nonnegative, $$ \int_G dxdy = \int_{[0,1]^2} [G] dxdy = \int_{[0,1]} \left(\int_{[0,1]} [G] dy \right) dx = \int_{[0,1]} \left(\int_{\{f(x)\}} dy \right) dx = \int_{[0,1]} 0\, dx = 0 . $$ If instead we look at the distribution of the random variable $(U,f(U))$ where $U$ is uniform over $[0,1]$, i.e. the image by $\tilde f : x \mapsto (x,f(x))$ of the Lebesgue measure, i.e. the measure $\mathrm{Lebesgue}\circ \tilde f^{-1} : A \mapsto \mathrm{Lebesgue}(\tilde f^{-1}(A))$, then $\nu$ is a probability measure on $[0,1]^2$. Because $\mathbb Q$ has Lebesgue measure zero, we have in fact $$ \mathrm{Lebesgue}\circ \tilde f^{-1} = \mathrm{Lebesgue}\circ \tilde g^{-1} \qquad \text{with} \qquad g = 0. $$ This means that, from the point of view of the measure, it behaves in the same manner as the graph of the constant null function. In that sense, the graph is not homogeneous.
impossible, see the preliminary observation.
Edit: OP suggested considering a measurable set $X$ equipped with a non-trivial measure (finite or not), and $f:X\to [0,1]$. If $X$ is separable (i.e. there exists a dense sequence in $X$) and at least countable, the above construction works.