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Suppose $\mathcal{I}$ is an arbitrary indexing set, and suppose we look at a family $A_i$ of subsets of some set $E$. Let's say we fix an element $x \in E$ and we define the function $f: \mathcal{I} \to \{0,1\}$ by \begin{equation} f_x(i) = \begin{cases} 1 &\text{if } x \in A_i\\ 0 &\text{otherwise} \end{cases} \end{equation}

Now, does the function $F: E \to \mathbb{R} \cup \{\infty\}$ given by \begin{equation} F(x) = \sum_{i \in I} f_x(i) \end{equation} make sense?

I am asking this because I came upon a comment that says the summation symbol $\sum$ cannot be used in cases where the indexing set $\mathcal{I}$ is uncountable, instead, on has to use integration theory for this (and therefore, the symbol $\int$).

Yet to me it doesn't look as if the above function $F$ is ill - defined. What am I missing?

Thanks for your feedback!

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    The way it is defined, it does make sense as a map from $E$ into $\mathbb{R}\cup \left\{\infty\right\}$, when $\infty$ is used as a formal symbol to mean the value of $f$ whenever for infinitely (countably or uncountably) many $i\in I$, $f_x(i) = 1$. I don't see a problem.2012-02-05

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If $f: \mathcal{I} \to [0,\infty)$ is a function, with $\mathcal{I}$ a set, you can define $ \sum_{ x \in \mathcal{I}} f(x) = \sup \{ \sum_{x \in F } f(x) \,|\, F \subseteq \mathcal{I}, F \text{ finite } \}.$ In other words, we take the supremum over all finite partial sums. Note that this is equivalent to taking the integral of $f$ with respect to the counting measure, but no integration theory is required to understand this; it is simply the supremum over a set of real numbers, and that always exists (provided we allow $+ \infty$).

Edit: As others have pointed out, though, it isn't all that interesting (but it is probably useful in certain contexts to have a definition for indexing sets other than $\mathbb{N}$). Consider the set $E = \{ x \in I \,|\, f(x) > 0 \}$. If this is uncountable, then the sum (as defined above) is infinite. If $E$ is countably infinite, then it can be written as a series (using a bijection between $E$ and $\mathbb{N}$ and ignoring $\mathcal{I} \setminus E$).

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    Fair enough :).2012-02-05