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Is the "average" of the function $\mathcal{G}:\mathbb{R}\to\mathbb{R}$ non-zero and finite?

This is a restatement of this post, since it is difficult to read and this post since the latter might later be closed. I don't know where else to share.

Motivation: Consider the function $\mathcal{G}:\mathbb{R}\to\mathbb{R}$,

(Example of $\mathcal{G}$) Let $(q_t)_{t\in\mathbb{N}}$ be a numbering of the rational numbers and $k_t=2^{2^t}$, and define functions $s_t$ as follows:

If $s_0=0$ everywhere and: $$\small{ s_{t+1}(x)=\begin{cases} q_{t/2} & x\in\bigcup_{j\in\{1,\cdots,t\}}(q_j-1/k_t,q_j+1/k_t), \, t \text{ is even}\\ k_t^2 & x \in\bigcup_{j\in\{1,\cdots,t\}}(q_j-1/k_t,q_j+1/k_t), \, t \text{ is odd}\\ s_t(x) & \text{otherwise} \end{cases}}$$ each $s_{t+1}$ agrees with $s_t$ at all real numbers except a set of measure $<1/2^t$ for big $t$, so one could consider $\mathcal{G}$ the pointwise limit of the functions $s_t$, which is defined everywhere except measure $0$ (in those bad points just define $\mathcal{G}=0$).

Here is the partial code of the example using Mathematica:

Clear["Global`*"]

enumerateRationals[n_Integer?Positive] := 
  Module[{posRationals, 
    fullList},(*Generate enough positive rationals using the Calkin-
   Wilf step*)

   posRationals = NestList[1/(2 Floor[#] - # + 1) &, 1, Ceiling[n/2]];
   (*Interleave:0,q1,-q1,q2,-q2...*)

   fullList = Riffle[posRationals, -posRationals];

   Prepend[fullList, 0][[1 ;; n]]];

enumerateRationals[0] = {};

q[t_] := q[t] = enumerateRationals[t][[t]]
(*Takes the t-value in an enumeration*)

k[t_] := k[t] = 2^(2^t)

s[0, _] = 0;
s[t_Integer, x_] := 
 s[t, x] = 
  Piecewise[{{q[(t - 1)/2], 
     EvenQ[t - 1] && 
      0 <= Min[Abs[x - enumerateRationals[t - 1]]] < 1/k[t - 1]}, {k[
       t - 1]^2, 
     OddQ[t - 1] && 
      0 <= Min[Abs[x - enumerateRationals[t - 1]]] < 1/k[t - 1]}}, 
   s[t - 1, x]]

FullSimplify[PiecewiseExpand[s[5, x]]]
(*Example of s[t,x]*)

(I want to compute s[t,x] as t->Infinity to get $\mathcal{G}(x)$ or bigG[x].)

In the usual sense: the mean of $\mathcal{G}$, w.r.t. the Lebesgue measure on a family of bounded sets converging to $\mathrm{dom}(\mathcal{G})=\mathbb{R}$, is always undefined since $\mathcal{G}$ satisfies two properties,

  1. The restriction of $\mathcal{G}$ to any interval has infinite area both above and below $y=0$.
  2. For all $a<b$ and $c<d$ real numbers, $\{x\in (a,b):\mathcal{G}(x) \in (c,d)\}$ is a set with positive Lebesgue measure.

Furthermore, the mean w.r.t. the Hausdorff measure in its dimension of a family of bounded functions with different bounded domains converging to $\mathcal{G}$ is defined but has different values depending on the family of bounded functions chosen. Hence, the former mean is non-unique.

Question: In my concept paper (see the below summary), assuming $\mathbf{R}$ is the origin and $E=1/2$, what is the average of the example of $\mathcal{G}$ using Mathematica? Is the "average value" non-zero and finite? If not, explain why?

Every reference in the next section refers to the Sections, Definitions, and pages of the concept paper.

Summary of The Concept Paper

My attempt to average of $\mathcal{G}(x)$ on $\small{(-\infty,+\infty)}$ is defined w.r.t. to four definitions:

For every chosen reference point, there should be a unique, "satisfying" (i.e., see the section "Modeling Question Summary" at the bottom), and either:

  1. a finite mean
  2. if not finite, then infinite mean
  3. if not a finite or infinite, then undefined mean.
  • The "measure" (Definitions 20-22 (pg. 20-25), where $\{G_r^{\star}:r\in\mathbb{R}^{+}\}=\{\mathrm{graph}(f_r^{\star}):r\in\mathbb{R}^{+}\}$ and $\{f_r^{\star}:r\in\mathbb{R}^{+}\}$ is a family of functions with different bounded domains converging to $\mathcal{G}$):

The "measure" partitions each graph into equal measure sets which take a sample point from each partition, pathways of line segments between sample points, lengths of line segments in each pathway, removed lengths which are outliers, remaining lengths which are converted into a probability distribution, and the entropy of the distribution. (There is more to read: i.e., Definitions 20-22 (pg. 20-25).)

  • The expected rate of expansion

The expected rate of expansion is an arbitrary fixed constant and is written as the function $E:\mathcal{A}(A)\to\mathbb{R}$ (i.e., $\mathcal{A}(A):=\mathbb{R}^{+}$ is the chosen index set of a family of functions or sets).

  • The actual rate of expansion (Definition 23, pg. 25-26)

The actual rate of expansion of a family of each bounded function's graph is a function of the "average" $2$-dimensional Euclidean distance between every point in each bounded function's graph and the reference point $\mathbf{R}\in\mathbb{R}^{2}$


Modeling Question Summary:

Combining "the measure" and the actual rate of expansion, we get a general notion of the choice function in the modeling question (Section 3.1, pg. 27-30). The choice function should pick "equivelant" families (Definition 15-17, pg. 17-18) of bounded functions converging to $\mathcal{G}$ which satisfy all the criteria in the modeling question, such that:

  • the "measure" (Definition 20-22, pg. 20-25) of each of the chosen families of each bounded function's graph increases at a rate linear or superlinear (Definition 22, pg. 23-25) to that of "non-equivelant" families (Definition 18-19, pg. 19-20) of each bounded function's graph (Section 3.1 criterion 2)
  • the absolute difference between the $y$-coordinate of the reference point $\mathbf{R}\in\mathbb{R}^{2}$ and the mean of each chosen families of bounded functions converging to $\mathcal{G}$ (Definition 13 [pg. 13-14], Section 3.1 criterion 4) is minimized
  • the absolute difference between the expected rate of expansion and the actual rate of expansion of each chosen families of each bounded function's graph (Definition 23 [pg. 25-26], Section 3.1 criterion 4) is minimized,

so that it's more likely the choice function, which answers the modeling question, chooses the desired families. Hence, we take the mean of the chosen "equivelant" (Definition 15-17, pg. 17-18) families as the new mean (i.e., the "equivelant" families should have the same mean).

POSTED BY: Bharath Krishnan
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