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I just read that for $f: I \rightarrow [0, \infty),$ we can define $\sum_{i \in I} f(i)$ as $\sup\{\sum_{i \in F} f(i) \mid F \subset I, F$ finite $\}.$ It makes sense that we can define it in this way. It coincides with my notion for countable sums. However, how can we define such a sum where $f$ may include negative values. It suddenly does not make sense to take the supremum.

That is, if $I$ is an arbitrary set and $x_i \in \mathbb{R},$ how would we define $\sum_{i \in I} x_i?$

green frog
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  • Still the same way as above since the arbitraryness (if I may use the word) is in the index an not the values having the index. – Yadati Kiran Nov 21 '18 at 04:31
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    Short answer: in general, you can't. For instance, conditionally convergent sums can be rearranged to sum to any given number, so if $a_n$ converges conditionally and $I = \mathbb{N}$, then $\sum_{i \in \mathbb{N}} a_i$ isn't uniquely defined, because it's order-dependent. An interesting question to ask is, roughly speaking, how "many" negative terms can be included in the sequence before things begin to go awry. – MathematicsStudent1122 Nov 21 '18 at 04:31
  • @YadatiKiran but we are taking the supremum of the sum itself - so if $f$ is purely a negative valued function then taking the supremum of the finite sums would not work I think. – green frog Nov 21 '18 at 04:32
  • @伽罗瓦: As MathematicsStudent1122 suggests we see" how "many" negative terms can be included in the sequence before things begin to go awry." – Yadati Kiran Nov 21 '18 at 04:36
  • Just think about the countable case. Supremum of finite sums if of no use and trying to assign some value to the sum of an arbitrary sequence of real numbers is an exercise in futility. – Kavi Rama Murthy Nov 21 '18 at 04:53
  • @KaviRamaMurthy Right - that is what I thought but I was just wondering if there might have been some sort of convention for this. – green frog Nov 21 '18 at 04:57

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Compute the sum over the non-negative values and compute the sum over the non-positive values. If one of them is finite take the sum of the two results. As in the Lebesgue's integral.

Dante Grevino
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