I am wondering if the concept of a summation over all $x \in [0, 1]$ would be useful. It would generalize the traditional concept of series. For instance:

$\sum_{x \in [0, 1] \cap \{1, 1/2, 1/3, \cdots\}} x^2 = \pi^2/6.$

The definition I have in mind is as follows. Define level-0 numbers as $L_0 =\{0\}$, level-1 numbers as $L_1 =\{1/2\}$, level-2 numbers as $L_2 =\{1/4, 3/4\}$, level-3 numbers as $L_3=\{1/8, 3/8, 5/8, 7/8\}$, level-4 numbers as $L_4=\{1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16\}$ and so on. Then

$\sum_{x\in [0, 1]} f(x) = \sum_{k=0}^\infty \sum_{x \in L_k} f(x).$

You can define absolute convergence easily. Let's denote $y_k = \sum_{x \in L_k} f(x)$. Then the series converges absolutely if $\sum_{k=0}^\infty y_k$ converges absolutely.

A possible applications is to define a summation over all rational numbers in $[0, 1]$. Is my concept new or old, interesting or not, or does not make any real sense?

  • $\begingroup$ Not sure I follow. You appear to only be summing over rational numbers with denominators of the form $2^k$. Was that your intent? $\endgroup$ – lulu Jul 29 at 14:47
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    $\begingroup$ And, in any ordinary sense, the sum of all the rationals between $0$ and $1$ would diverge (as, for instance) the sum would contain infinitely many terms $>\frac 12$). $\endgroup$ – lulu Jul 29 at 14:48
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    $\begingroup$ No...$\frac 13$ would never appear in your sum. $\endgroup$ – lulu Jul 29 at 14:54
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    $\begingroup$ @VincentGranville $\frac13\neq \frac p{2^n}$, hence $\frac13\not\in L_k$ for any $k$. It is true that closure of union of $Lk$'s $\textrm{cl}\bigcup_{k\in\mathbb{N}}L_k$ is equal to the whole interval $[0,1]$, but you don't take any closures here. $\endgroup$ – Maja Blumenstein Jul 29 at 15:05
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    $\begingroup$ @BenedictW.J.Irwin Well, the integral isn't the sum of the values of $f(x)$ but the limit of sum of the areas of rectangles of height $f(x)$ and base approaching infinitismal. Example: $\int_0^1 x dx=\frac 12 < 1$ but $\sum_{x\in[0,1]} x \le \sum_{x=\frac 1n|n\in \mathbb N} x =\infty$. ... I'm not sure there is any use to this question. But I haven't given it enough thought for such a blanket condemnation. The OP seems to be only doing countable sums and I'm not sure how to even define uncountable sums and unless $f(x)\ne 0$ only countably many times the sum will be infinite. $\endgroup$ – fleablood Jul 29 at 15:31

Here is some exposition on the framework of uncountable sums. Maybe this will scratch your itch.

Let $A$ be a set with some unknown cardinality, and let $x_\alpha$ be elements of a topological vector space (for simplicity assume $\{x_\alpha\}_{\alpha \in A} \subset \mathbb{R}$.) Our aim is define the unordered sum:

$$ \sum_{\alpha \in A} x_\alpha $$ First, let us nail down some set-theoretic terminology. A binary relation $\prec$ on a set $\mathcal{F}$ is called a a partial order on $\mathcal{F}$ if for evert $I,J,K \in \mathcal{F}$, we have that:

  • $I \prec I$ (reflexivity)
  • $I\prec J, \ J \prec K \implies I \prec K$ (Transitivity)
  • $I \prec J , \ J \prec I \implies I = J$ (Antisymmetry)

We call this a partial order because not all elements of $\mathcal{F}$ may be comparable with this relation. For the subset $A$ in question, let us consider :$$ \mathcal{F} = \{B \subset A : B \text{ is finite}\} $$ I.e., all finite subsets of $B$. You can check that this is a partially order set with $\prec = \subset$, i.e. $A\prec B \iff A \subset B$. Note that if $I,J \in \mathcal{F}$, we have some $K$ such that $ I \subset K$, $J\subset K$, and hence $I \prec K$ , $J\prec K$ (consider $K = I \cup J$). Any ordered set where any two elements have an element "greater than them", is known as a directed set. $\mathcal{F}$ is a directed set.

Given a finite subset $I \subset A$, we can define a partial sum: $$ S_I = \sum_{\alpha \in I} x_\alpha $$ Thus, we may say that $\sum_{\alpha \in A}x_\alpha$ converges to $x \in \mathbb{R}$ if for all $\epsilon >0$ , we have $I\subset A$ with $I$ finite, such that:$$ \left |x - \sum_{\alpha \in I} x_\alpha \right| < \epsilon $$ We have the following result: If $\sum_{\alpha \in A} x_\alpha$ converges, then $x_\alpha \neq 0 $ for at most countably many $\alpha$. The main idea of the proof is to consider sets $I_n \in \mathcal{F}$ with:

$$ \left |\sum_{\alpha \in I_n} x_\alpha \right| < \frac{1}{n} $$ The sets $I_n$ are finite, so their union is countable, and thus so is the set of elements with terms larger than $0$.


Firstly, $$\sum_{x\in [0, 1]} f(x) \neq \sum_{k=0}^\infty \sum_{x \in L_k} f(x)$$ because $$\bigcup_{k=0}^\infty L_k\neq[0,1].$$ Secondly, for any uncountable set $S$ sum $\sum_{x\in S} x$ diverges because the series contains infinitely many terms strictly greater than $\varepsilon\neq0$.

Summation over all rationals from $[0,1]$ (or even all rationals) is nothing new, it's just rephrasing theory of the ordinary series, as all $\mathbb{Q}$, $\mathbb{Q\cap[0,1]}$ and $\mathbb{N}$ are countable. Say you have a set of real numbers indexed by rationals $\{x_r|r\in\mathbb{Q}\}$. Because $\mathbb{Q}$ is countable, there is a sequence of all rationals, say $q_1,q_2,q_3,...$. Then, instead of the sum $$\sum_{r\in\mathbb{Q}}x_r$$ we can work with sum $$\sum_{n=1}^\infty x_{q_n}$$ which is "the usual" series over naturals. (Here I talk only about absolute convergence, so we can rearrange the terms, I can't think of a reasonable definition of conditional convergence of sums over rationals.)

  • $\begingroup$ I see your point. It needs to be defined differently: $\sum_{x \in [0, 1]} f(x) = \lim_{y\rightarrow x} \sum_{k=0}^\infty \sum_{y\in L_k} f(y)$. This assumes $f$ is continuous. $\endgroup$ – Vincent Granville Jul 29 at 15:18

One way of defining such a thing is to define a measure $\tau$ on $(\mathbb{R},\mathcal{P}(\mathbb{R}))$ by $$\tau(A)=\begin{cases} |A| & \text{if A has finitely many elements} \\ \infty & otherwise\end{cases}$$

and then define $\sum_{x\in I} f(x) := \int_I f d\tau$. Now a relevant question to ask would be: "What functions are in $\mathcal{L^1}(\tau)=\{f:\mathbb{R}\rightarrow \mathbb{R} \: | \: \int|f|d\tau < \infty\}$??".

If there exist an interval I such that $f(x)>\varepsilon$ for all $x\in I$, then $\sum_{x\in I} f(x) \geq \int_I \varepsilon d\tau = \varepsilon \tau(I) = \infty$, hence any continous function, which is not 0 everywhere is not in $\mathcal{L^1}(\tau)$.

Now suppose that $f\in \mathcal{L}^1(\tau)$ and set $K=\int_I |f| d\tau$, then the set $\{x \: | \: |f(x)| \geq K/n \}$ has atmost $n$ elements, and is therefore finite, this means that

$$ X = \{x \: | \: |f(x)|> 0\} = \bigcup_{n\in \mathbb{N}} \{x \: | \: |f(x)| \geq K/n \}$$ is a countable set, which means that $X$ can be written as a sequence $\{x_n \: | \: n \in \mathbb{N}\}$ and $f$ can be written as $f=\sum_{n=1}^\infty f(x_n) \cdot \chi_{x_n}$ and by the dominated convergence theorem $$ \sum_{x\in\mathbb{R}} f(x) = \int_\mathbb{R}fd\tau = \int_\mathbb{R}\sum_{n=1}^\infty f(x_n) \cdot \chi_{x_n} d\tau = \sum_{n=1}^\infty\int_{x_n}f(x_n)d\tau = \sum_{n=1}^\infty f(x_n)\tau(\{x_n\}) = \sum_{n=1}^\infty f(x_n) $$


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