# Summation $\sum_{x\in [0, 1]} f(x)$ over all real numbers $x \in [0, 1]$.

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?

• Not sure I follow. You appear to only be summing over rational numbers with denominators of the form $2^k$. Was that your intent? – lulu Jul 29 at 14:47
• 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 $>\frac 12$). – lulu Jul 29 at 14:48
• No...$\frac 13$ would never appear in your sum. – lulu Jul 29 at 14:54
• @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. – Maja Blumenstein Jul 29 at 15:05
• @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. – fleablood Jul 29 at 15:31

## 3 Answers

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.)

• 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. – 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 $$\sum_{x\in\mathbb{R}} f(x) = \int_\mathbb{R}fd\tau = \int_\mathbb{R}\sum_{n=1}^\infty f(x_n) \cdot \chi_{x_n} d\tau = \sum_{n=1}^\infty\int_{x_n}f(x_n)d\tau = \sum_{n=1}^\infty f(x_n)\tau(\{x_n\}) = \sum_{n=1}^\infty f(x_n)$$