Finding examples of functions which are infinite or undefined with current extensions of the expected value?

Question

Preliminaries

Consider the expectations desribed in this paper, which is an extension of the Lebesgue density theorem; this paper which is an extension of the Hausdorff measure, using Hyperbolic Cantor sets; and this paper which applies a Henstock-Kurzweil type integral on a measure Metric Space. We also use conditional expectation; however, the result of this expectation depends on the choice of the "condition". Moreover, there is no "known" choice function choosing "conditions" which "naturally" extend the expected values of the previous sentences to be unique and finite.

Motivation

According to an article in Quanta Magazine Wood writes, "No known mathematical procedure can meaningfully average an infinite number of objects covering an infinite expanse of space in general. The path integral is more of a physics philosophy than an exact mathematical recipe."

I want to represent Wood's example with functions—I'm looking for two examples:

1. A function that matches Wood's description, such the expected values w.r.t any measure in the preliminaries is infinite or undefined.

2. A "non-fractal" function where none of the expected values in the preliminaries gives a unique, finite, expected value.

One example of 2. is $f:\mathbb{Q}\to\mathbb{R}$, where:

\begin{equation} f(x)=\begin{cases} 1 & x\in \left\{r/q:r\in\text{odd }\mathbb{Z},q\in\text{even }\mathbb{Z},q\neq 0,\gcd(r,q)=1\right\}\\ 0 & x\in \left\{r_1/(q_1):r_1\in\mathbb{Z},q_1\in\text{odd }\mathbb{Z},\gcd(r_1,q_1)=1\right\} \end{cases} \end{equation}

where we could find a unique average using conditional expectation of $f$ given a sequence of sets with a set-theoretic limit of $\mathbb{Q}$; however, the expectation depends on the sequence chosen. (Hence, the expected value can be any value and is undefined.)

Question: Does there exist an explicit function which answers 1. and 2.?

(This post might be able to help.)

Answer 1

Consider the function $f:\mathbb{R}\to\mathbb{R}$, where for a base-3 expansion of $x\in\mathbb{R}$, we take the "pseudo-random" iterations of function $\mathscr{F}:\{0,1,2\}\to\{0,1,2\}$ where for the first iteration, replace base-3 digit zero with one; one with two, and two with zero.

In mathematical terms, if $\mathscr{F}:=\left\{(0,1),(1,2),(2,0)\right\}$, such that equation: $$\mathscr{F}^{k}=\underset{k\text{ times}}{\underbrace{\mathscr{F}(\mathscr{F}(\cdots\mathscr{F}(x)))}}$$ is the $k$-th iteration of $\mathscr{F}$, we define a function with "pseudo-random" outputs in $\mathbb{N}$. If we define $g_{\varepsilon}:\mathbb{R}\to\mathbb{R}$ where:

$$g_{\varepsilon}(x)=\begin{cases} \quad\!\!(2\varepsilon)^{x/(2\varepsilon)} & \quad\!\! x<-\varepsilon\\ -1/x & \!-2\varepsilon\le x <0\\ \quad \! 0 & \!\!\quad x=0 \\ \quad\!{1}/{x} & \quad\!\! 0<x<\varepsilon \\ \quad\!\!\,\varepsilon^{-x/\varepsilon} & \quad\!\! x\ge \varepsilon\\ \end{cases}$$

such that $\left[\cdot\right]$ rounds to the nearest integer, then function $z_{\varepsilon}:\mathbb{R}^2\to\mathbb{R}$ or

$$z_{\varepsilon}(x,k)=\big[g_{\varepsilon}(x) k\sin(g_{\varepsilon}(x)k)\big]$$

contains "pseudo-random" outputs of $\mathbb{N}$, where furthermore:

$$\small{f_{\varepsilon}(x)=\left\{\sum\limits_{k=-\infty}^{\infty}{\text{sign}\left(\mathscr{F}^{z_{\varepsilon}(x,k)}(a_k)\right)\mathscr{F}^{z_{\varepsilon}(x,k)}(a_{k})}/{3^k}:a_k\in\left\{0,1,2\right\},x=\sum\limits_{k=-\infty}^{\infty}a_{k}/{3^k}\right\}}$$

which is the same as:

$$\scriptsize{f_{\varepsilon}(x)=\left\{\sum\limits_{k=-\infty}^{\infty}\Big(\text{sign}(\text{mod}(z_{\varepsilon}(x,k)+a,3))\cdot\text{mod}(z_{\varepsilon}(x,k)+a,3)\Big)/{3^k}:a_k\in\left\{0,1,2\right\},x=\sum\limits_{k=-\infty}^{\infty}a_k/{3^k}\right\}}$$

(i.e., $\mathscr{F}^{k}(a_k)=\text{mod}(a_k+k,3)$), such that for set $A\subseteq \mathbb{R}$ we want an $f:\mathbb{R}\to\mathbb{R}$ where:

\begin{equation} \small{\forall(\varepsilon_1>0)\exists(A\subseteq\mathbb{R})\forall(\varepsilon\in A)\exists(M>0)\forall(x\in\mathbb{R})\left(0<\varepsilon\le M\Rightarrow \left|f_{\varepsilon}(x)-f(x)\right|<\varepsilon_{1}\right)} \label{bjj}\tag{1} \end{equation}

Note: For $g_{\varepsilon}(x)$, we have $2\varepsilon$ (rather than $\varepsilon$) to avoid symmetry in the graph of $f$, and the $\text{sign}$ function in $f_{\varepsilon}(x)$ to allow negative outputs in $f$'s graph.

Using evidence, consider $f_{\varepsilon}(x)\mapsto f_{\varepsilon}(x,t)$ where:

$$\scriptsize{f_{\varepsilon}(x,t)=\left\{\sum\limits_{k=-\infty}^{t}\Big(\text{sign}(\text{mod}(z_{\varepsilon}(x,k)+a_{k},3))\cdot \text{mod}(z_{\varepsilon}(x,k)+a_{k},3)\Big)/{3^k}:a_{k}\in\left\{0,1,2\right\},x=\sum\limits_{k=-\infty}^{\infty}a_{k}/{3^k}\right\}}$$

and $x\in G_{r}:=\left\{c/d:c,d\in\mathbb{Z}, d\le r, -dr\le c\le dr\right\}$, such that when $x\in G_{10}$, we get the set of points $\left\{(x,f_{.1}(x,10)):x\in G_{10}\right\}$ is:

Question: If $f$ does not exist or satisfy the motivation, how do we change equation $\eqref{bjj}$ so it does?

I assume none of the expected values in the preliminaries give this function a unique and finite expected value. (Hopefully, someone can check.)