# Prove or disprove this integral of a function, defined on a countable set with infinite limit points, converges to zero [closed]

Edit: I got rid of my old definitions. Everything should be clear now

Since no one has answered my question on MSE, I’m hoping to get an answer here. I apologize if you dislike my writing. I am an undergraduate student and I don’t know whether this is a research question.

## Definition

Consider a continuous $$f:A\to[0,1]$$ where $$A\subseteq[0,1]$$.

Edit: I did not use @Mathworker21's answer because I assumed it would not give the result I am looking for. Now I realize this is not the case. Here is his definition.

We can assume $$A \subseteq [a,b]$$, since that's where everything is happening. For ease, I'll have $$[a,b] = [0,1]$$. Based on the above discussion, I'll just have $$P$$ be defined on all of $$[0,1]$$ and continuous. Let $$E_1 = [0,1], E_2 = [0,\frac{1}{2}], E_3 = [\frac{1}{2},1], E_4 = [0,\frac{1}{4}], E_5 = [\frac{1}{4},\frac{1}{2}], E_6 = [\frac{1}{2},\frac{3}{4}]$$, $$E_7 = [\frac{3}{4},1], E_8 = [0,\frac{1}{8}]$$, etc.

If $$A$$ is finite, it's obvious how to define the average of $$P$$ (just do $$\frac{1}{|A|}\sum_{x \in A} P(x)$$). So, assume $$A$$ is infinite. Consider the sets $$A\cap E_1, A\cap E_2, \dots$$. Let $$x_1$$ be a point in the first nonempty one of these sets. Let $$x_2$$ be a point in the second nonempty of these sets, etc. Look at the measures $$\delta_{x_1}, \frac{\delta_{x_1}+\delta_{x_2}}{2}, \dots, \frac{\delta_{x_1}+\dots+\delta_{x_N}}{N},\dots$$. Since $$[0,1]$$ is a compact metric space, there is some probability measure $$\mu$$ on $$[0,1]$$ that is a weak* limit of some subsequence of these measures, i.e. there is some $$(N_k)_k$$ with $$\frac{1}{N_k}\sum_{j=1}^{N_k} f(x_j) \to \int_0^1 f d\mu$$ for each $$f \in C([0,1])$$.

We then define the average of $$P$$ over $$A$$ to be $$\int_0^1 Pd\mu$$.

For concreteness, suppose that $$f=\operatorname{id}$$ and $$A=\left\{\frac{1}{2^x}+\frac{1}{2^y}+\frac{1}{2^z}:x,y,z\in\mathbb{Z}^{+}\right\}\cap[0,1]$$

My guess is, in this case, the average will converge to $$0$$. How do we prove whether I am right or wrong? Could we create a better definition that is easier to compute or gives an exact value?

## Possible evidence that the average is zero

Here is a number line plot of all elements of set $$A$$

generateA[n_Integer] :=
Select[Union@
Flatten[Table[
1/2^x + 1/2^y + 1/2^z, {x, 1, n}, {y, x, n}, {z, y, n}]],
0 <= # <= 1 &]
NumberLinePlot[generateA[50], PlotStyle -> PointSize[0.003]]


It appears the points become "denser" near $$0$$. If the 'densest point' carries the most weight in the average, it is possible that the average should be $$f(0)=0$$.

Also by analysis, if we take the function inside $$A$$ and set $$x$$ as constant $$a$$ and $$y$$ as constant $$b$$, taking $$z\to\infty$$ we get

$$\left\{\frac{1}{2^a}+\frac{1}{2^b}:a,b\in\mathbb{Z}^{+}\right\}$$

This represents all the limit points. Since the limit points are "infinitely denser" than finite points, they should have infinitely greater weight for the average.

If we set $$b\to\infty$$ we get

$$\left\{\frac{1}{2^a}:a\in\mathbb{Z}^{+}\right\}$$

These represent second-order limit points "closely approximated" by first-order limit and finite points. The second-order limit points should have infinitely more weight than the lower order limit points.

As $$a\to\infty$$, we get $$0$$, the third order limit point and the densest point in the set. Third-order limit points should have infinitely more weight than second order, first order, and finite limit point.

From our analysis, it's possible, the average is $$f(0)=0$$.

Finally, consider the following code (from Wolfram Mathematica). This replicates $$L(f,P)$$ and $$U(f,P)$$ which I mentioned earlier:

partition[a_List, s_] := Module[{f, r}, f[{}, x_] := {x};
f[l_List, x_] := If[x - l[[1]] < s, Append[l, x], Sow[l]; {x}];
r = Reap[Fold[f, {}, a]];
Append[r[[2, 1]], r[[1]]]]
partition[{0, 1, 2, 7, 10, 11, 12}, 5]
(*{{0,1,2},{7,10,11},{12}}*)
calculate[p_, a_, s_] :=
Module[{parts = partition[a, s], n, inf, sup}, n = Length[parts];
inf = Total[Min[p /@ #] & /@ parts];
sup = Total[Max[p /@ #] & /@ parts];
{inf/n, sup/n}]
calculate[Identity, generateA[500], N[10^(-170)]] // N


I get:

{0.00598798, 0.00598798}


But it took a long time to compute. In fact, the person who gave me this answer doesn't think the sum converges to $$0$$ when $$f=\operatorname{id}$$? How do we prove otherwise?

## Deleted Definition

Let $$\require{enclose} \enclose{horizontalstrike}{A \subseteq [0,1]}$$, and let $$\require{enclose} \enclose{horizontalstrike}{P}$$ be a partition of $$\require{enclose} \enclose{horizontalstrike}{[0,1]}$$ such that it is a finite set of sub-intervals $$\require{enclose} \enclose{horizontalstrike}{X}$$ with disjoint interiors and each subinterval has the same length. Define $$\require{enclose} \enclose{horizontalstrike}{P' = \{ X\in P: X\cap A \neq \emptyset\}}$$. Define $$\require{enclose} \enclose{horizontalstrike}{n' = |P'|}$$ (the cardinality of a finite set, or in this case, the number of sub-intervals whose intersection with $$\require{enclose} \enclose{horizontalstrike}{A}$$ is non-empty).

Calculate/define the following:

$$\require{enclose} \enclose{horizontalstrike}{L_{f,P} = \frac{1}{n^{\prime}} \sum_{X \in P^{\prime}} \bigg(\inf_{t \in X}f(t) \bigg)}$$

$$\require{enclose} \enclose{horizontalstrike}{U_{f,P} = \frac{1}{n^{\prime}} \sum_{X \in P^{\prime}} \bigg(\sup_{t \in X}f(t) \bigg)}$$

Define the limits under refinements of $$P$$ like so: $$\require{enclose} \enclose{horizontalstrike}{L_f = \lim_{\|P\| \to 0}(L_{f,P})}$$ $$\require{enclose} \enclose{horizontalstrike}{U_f = \lim_{\|P\| \to 0}(U_{f,P})}$$

Where $$\require{enclose} \enclose{horizontalstrike}{\|P\|=\sup_{X\in P}\|X\|}$$.

$$\require{enclose} \enclose{horizontalstrike}{L_f}$$ is the 'lower average' of $$\require{enclose} \enclose{horizontalstrike}{f}$$ on $$\require{enclose} \enclose{horizontalstrike}{[0,1]}$$ (with respect to partition $$\require{enclose} \enclose{horizontalstrike}{P}$$). Likewise, $$\require{enclose} \enclose{horizontalstrike}{U_f}$$ is the 'upper average' of $$\require{enclose} \enclose{horizontalstrike}{f}$$ on $$\require{enclose} \enclose{horizontalstrike}{[0,1]}$$ with respect to partition $$\require{enclose} \enclose{horizontalstrike}{P}$$.

If these lower and upper averages limits converge to the same value (id est: are equal), we are given "my definition of average" of $$f$$. If they do not converge, then the average is undefined. Notice I define "upper" and "lower" averages to show when an average can not exist.

Note I describe $$\require{enclose} \enclose{horizontalstrike}{L(f, P)}$$ and $$\require{enclose} \enclose{horizontalstrike}{U(f, P)}$$ as "Riemman-like" because original Riemman-sums have upper and lower sums. However, this doesn't mean they are the same. Here we discard empty sub-interval with no points. This means my average could be anywhere between $$\require{enclose} \enclose{horizontalstrike}{f(0)}$$ and $$\require{enclose} \enclose{horizontalstrike}{f(1)}$$ depending on $$\require{enclose} \enclose{horizontalstrike}{A}$$, and possibly initial $$\require{enclose} \enclose{horizontalstrike}{P}$$.

• You probably wanted to link to this question: How do prove the Darboux-like sum of a function, defined on a countable set with infinite limit points, converges to zero? You link goes to a picture instead: "Since no one has answered [my question][(i.stack.imgur.com/9oD9M.png) on MSE..." (BTW the question on Mathematics still has a bounty.) – Martin Sleziak May 1 at 17:09
• @MartinSleziak Fixed it. – Arbuja May 1 at 17:38
• The generation procedure of the partition P is unclear to me. Suppose initially $A = \{ 1/6, 1/4, 3/4, 5/6\}$. And you start with $n = 3$. You end up with the first and third intervals having 2 points each, and the middle interval with no point. Do you or do you not modify the middle interval? Is $n'$ here supposed to be $2$ or $3$? – Willie Wong May 1 at 17:39
• If your answer is $n' = 2$, then it contradicts what you have written as your procedure. If your answer is $n' =3$, then your definition of the lower sum will involve taking the $\inf$ over an empty set. – Willie Wong May 1 at 17:40
• @WillieWong You do not modify the middle interval so $n^{\prime}=3$. So we take the $\text{inf}$ over an empty set which is "undefined". According to one of the lines in my post, if $t_i$ is undefined, then $P(t_i)$ should be zero. How do I make this more clear in my post? – Arbuja May 1 at 17:46

For the concrete question you ask, the answer is yes.

Your set $$A$$ is in fact the set of all numbers between $$0$$ and $$1$$ with binary expansion that has no more than 3 bits equaling 1.

Let $$\ell_+ = \lceil \log_2 n \rceil$$ and $$\ell_- = \lfloor \log_2 n \rfloor$$. You can estimate

$$n' \geq (\ell_- - 1) + \binom{\ell_- - 1}{2} + \binom{\ell_- - 1}{3} \approx \ell_-^3$$

On the other hand, the sum inside either the upper/lower sum is between

$$\sum_{i = 1}^{\ell_+} 2^{-i} + \sum_{i = 1}^{\ell_+} \sum_{j = i+1}^{\ell_+} (2^{-i} + 2^{-j}) + \sum_{i = 1}^{\ell_+} \sum_{j = i+1}^{\ell_+}\sum_{k = j+1}^{\ell_+} (2^{-i} + 2^{-j} + 2^{-k})$$

(which is the sum of all numbers in $$A$$ with no more than 3 bits equaling 1 and least significant bit at least $$2^{-\ell_+}$$) and

$$\sum_{i = 1}^{\ell_+} (2^{-i} + 2^{-\ell_-}) + \sum_{i = 1}^{\ell_+} \sum_{j = i+1}^{\ell_+} (2^{-i} + 2^{-j}+ 2^{-\ell_-}) + \sum_{i = 1}^{\ell_+} \sum_{j = i+1}^{\ell_+}\sum_{k = j+1}^{\ell_+} (2^{-i} + 2^{-j} + 2^{-k}+ 2^{-\ell_-})$$

(which is the sum of the same set of numbers of the first sum, increased by the width of the interval).

The first sum is easily evaluated to be $$O(\ell_+^2)$$ (here we take advantage of the fact that at least one level of the summation converges using geometric series). The second sum is $$O(\ell_+^2) + O(\ell_+^3 2^{-\ell_-})$$.

So this means that both the upper sum and the lower sum behave like $$\frac{C}{\log_2(n)} \to 0$$ as $$n \to +\infty$$.

(This explains why your numerical computation goes very slow. At $$n = 10^{-170}$$, $$\log_2(n)$$ is something like 800. In fact, I don't even know if there is a reasonable numerical method for this, since to get convergence to the 10th decimal place, you will need a floating point arithmetic with a 2^30 bit mantissa to avoid numerical errors, or alternatively invent a new way to do exact arithmetic with rational numbers with very large numerators and denominators.)

• Thank you. Do you have a better way of defining my sums (in general, not the example)? Perhaps you can edit my post. If I present my version of the definition, no one will show interest. My guess is the "highest-order limit points" will determine the average. – Arbuja May 1 at 20:13
• Can't help you there, because I have no idea why you even care about this sum. The procedure of "merging" is very strange. It would make more sense (to me) to define $n'$ from a given partition $P$ as "number of intervals in $P$ with non-empty intersection with $A$"; this makes the number a lot easier to count. Your definition on the other hand can take values between this one and twice that, and is a lot harder to count. – Willie Wong May 1 at 20:19
• I can only answer this question because the expected sum is 0. If you look at $f(x) = 1 + x$, I don't know how even to prove either the upper or the lower sum has converging limits. – Willie Wong May 1 at 20:24
• There are fractal like constructions that allows very frequently half of the intervals to be empty, and not in a way you can merge. Let $B$ denote all infinite subsets of the square numbers {1, 4, 9, ...}. Let $A: \{ \exists b \in B: x = \sum_{i\in b} 2^{-i} \}$. Then for any $s = 2^{-k^2 - 1}$, the corresponding $P$ has exactly half of its intervals containing infinitely many points, and half of its intervals exactly empty. – Willie Wong May 1 at 21:20
• Incidentally, for this last example, every point in $A$ is an "infinite-order" limit point. If you define $n'$ to only range over all non-empty intervals, then the "average" process you define should be able to be expressed in terms of integrating $f$ against some singular continuous measure supported on the closure of $A$. – Willie Wong May 1 at 21:36