use of $\sum $ for uncountable indexing set I was wondering whether it makes sense to use the $\sum $ notation for uncountable indexing sets. For example it seems to me it would not make sense to say
$$
\sum_{a \in A} a \quad \text{where A is some uncountable indexing set e.g. some $A \subset \mathbb{R}$ }
$$
Would it be better to avoid the above notation in general for uncountable indexing sets ? Any help in making better sense of this would be very appreciated.
 A: For sums of real numbers, I have always thought of all sums as integrals with respect to a suitable counting measure (which is somewhat limiting, I know, since I am handling everything in the Lebesgue sense, which messes up conditional convergence). At the very least, that interpretation reduces to something most graduate students have seen a well-developed theory for.
As an aside, it is not uncommon to see sums over possibly uncoutnable index sets with interpretation in the above sense.  One of the classical examples (you can find it in Rudin's Real and Complex Analysis) arises in the study of Fourier series over a Hilbert Space and if you would like an example in both the statement and proof of the Parseval identity.  Consider $\mathcal{l}^2 (\mathbb{R})$ (that is, the Lebesgue space of $\mathbb{R}$ equipped with the counting measure).  Appealing to Zorn's Lemma we know this space has a maximal Hilbert Space basis (uncountable) $e_\alpha$ and that for any $x \in \mathcal{l}^2 (\mathbb{R})$ we have $||x||^2$ = $\sum_\alpha |\langle e_\alpha, x\rangle|^2$
A: Suppose $I$ is an  index set, and let $n_i$ be a real number for all $i$ in $I$. If $n_i$ is almost always zero, i.e., there exists a finite subset $S\subset I$ such that $n_i =0$ for all $I\backslash S$, then I see no harm in using the notation $\sum_{i\in I} n_i$. 
In general, it's not well-defined. 
A: There's no problem with summing over an uncountable indexing set, but over the reals you don't really get anything interesting by moving from the countable case to the uncountable case.
Suppose we have such an uncountable summation of nonnegative real numbers: $\sum_{r \in \Gamma} x_r$.  Then we have a few cases:
(i) One of the $x_r = \infty$. Then our sum, $\sum_{r \in \Gamma} x_r$, is also $\infty$. 
(ii) Suppose none of the $x_r =\infty$.  Now we utilize dyadic decomposition to write the positive reals as a countable union of sets: $(0, \infty] = \cup_{j \in \mathbb{Z}} (2^j, 2^{j+1}]$.
By the pigeonhole principle, either (a) There is a $j$ such that there are uncountably many nonzero $x_r$ in $(2^j, 2^{j+1}]$, in which case our sum is $\infty$ again or (b) there are only countably many nonzero $x_r$ in each interval  for all $j$.  In the latter case, we're back to our countable sum, as a countable union of countable sets is still countable.  
Now, if our indexing set $\Gamma$ is essentially countable in this way, we can motivate our sum as usual by defining $\sum_{r \in \Gamma} x_r = \sup_{E \subset \Gamma} \sum_{r \in E} x_r$, where $E$ is a finite set.  So an uncountable sum can be defined in the same way, without really picking up many problems (or interesting cases, sadly).
A: It makes sense when it is given a precise definition that makes sense.  If $I$ indexes a set of elements of a Hausdorff topological abelian group, then $\sum\limits_{i\in I}a_i$ can be used to denote the limit of the net of finite sums, with the finite subsets of $I$ directed by inclusion, when this limit exists.
In particular, if each $a_i$ is a nonnegative real number, then $\sum\limits_{i\in I}a_i$ exists as an element of $\mathbb R$ if and only if $\sup\left\{\sum\limits_{i\in F}a_i:F\text{ is a finite subset of }A\right\}<\infty$, and in that case $\sum\limits_{i\in I}a_i$ equals that supremum.  Finiteness of this sum implies that $\{i\in I:a_i\neq 0\}$ is countable.
A: You can define the sum of the elements of an infinite set $S \subseteq \mathbb{R}_{>0}$ by
$$ \sum_{s \in S} s = \sup \left \{ \sum_{s \in F} s\, :\, F \subseteq S \text{ finite} \right \} $$
However, the sum of uncountably many strictly positive reals is always infinite, so this often isn't too useful. However the definition still coincides with the countable case.
When negative numbers are included it becomes impossible to define what you mean by the sum. With sums of countably many reals the idea is that you add one element at a time in a particular order, so that sums like $\displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}$ are well-defined despite the fact that by taking the elements of the sequence $-1, \frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, \dots$ in a different order you can obtain any real number. For an uncountable set this is harder to make sense of.
