# Summation with non integer induces

All of the sums I've encountered so far have been functions $f:Z\rightarrow$ (any other set). Or in other words the sums are in the form of $\sum_{j\in Z\lor j\in N}^n a_j=a_0+a_1+...+a_n$. This process is pretty straight forward but what if let the lower index be a member of a non-discrete set. Say $j\in R$. We now have set which does not obey the well ordered principle and is non discrete. So how would we compute the sum $\sum_{j\in R \lor j\in C}^n a_n$? Is this even possible?

An algebraic definition of $$\sum_{i\in I} a_i$$ is only possible in cases that amount to finite sums, i.e. if $a_i=0$ for all but finitely many $i\in I$. Even in the countable case (i.e. when $a_i\ne 0$ for countably many $i\in I$), the order is essential - just think of conditionally convergent series (and did you notice we call these things series not sum?), which makes it an analytic, not an algebraic definition. Like, assume we want to compute $$\sum_{i\in\mathbb Q}a_i$$ and that the summands are $\frac{(-1)^n}n$ in some order; in which order should we run over the rationals? "From left to right" is not an option!

However, if $a_i\ge0$ for all $i\in I$ and there exists a number $S$ such that $\sum_{i\in J}a_i\le S$ holds for all finite subsets of $I$, then we can safely define $$\sum_{i\in I}a_i:=\sup\left\{\,\sum_{j\in J} a_i\Biggm| J\subseteq I, J\text{ finite}\,\right\}$$ In the case of countable $I$ this corresponds computing the sum like a series with $\mathbb N$ as index set (in arbitrary order, i.e. using an arbitrary bijection $\mathbb N\to I$).

What happens if $a_i>0$ for uncountably many $i\in I$ (and even assuming $a_i\ge 0$ for all $i\in I$, just for safety)? Since $\{\,i\in I\mid a_i>0\,\}=\bigcup_{n\in\mathbb N}\{\,i\in I\mid a_i>\frac1n\,\}$, there exists $n\in\mathbb N$ such that there are uncountably many $a_i>\frac 1n$. Whatever we expect of a sum (more precisely: of an extension of the concept of sum to uncountably many summands, so we might take a bit of freedom here and cast some permanence principles overboaurd), this should surely amount to the sum being infinite. Thus an uncountable sum makes sens only if it essentially is not uncountable.

You may ask, what about the integral? But there's a reason why it is written with differnt symbols than $\Sigma$. Wha twe sincerely expect of sums is that $I\cap J=\emptyset$ implies $\sum_{i\in I}a_i+\sum_{j\in J}a_j=\sum_{i\in I\cup J}a_i$. You can't keep this up when interpreting uncountable sums as integral and still keeping the well-known notion of finite sums.

I can provide some intuitive arguments for impossibility of such a summation in an uncountable sets: let associate the numbers with line segments of certain lengthes. Then the sum is associated with the measure of set defined by placing them all in the real line without intersection. But it could be shown that a set of disjoint intervals in the real line must be countable. Nevertheless, in some areas of mathematics this question arise naturally, for example, take generalizing the notion of dot product from finite-dimensional vector spaces to function space. We take this sum to be equal to the integral over $\mathbb{R}$ or $[0;1]$. Note this is not like usual summation, because adding one or two numbers does not increase the sum at all ($\int \chi (\mathbb{Z}) = 0$, i mean characteristic function of integers).