# why do we need to set a countable union only in the definition of sigma algebra? [closed]

I'm sorry if the question is too basic but at least it would be good to discuss it. Well, I was reading the defition of a sigma algebra and I was trying to understand why the definition of it. If the definition is that (1) the whole space is in the sigma algebra, (2) the complement of any element is also in the sigma algebra, and (3) the countable union of elements of the sigma algebra is in the sigma algebra as well. Well, for me, (1) makes sense because I would like to be able to define a probability over elements on the sigma algebra and the whole space should be "measurable" so in other words, it makes sense to set a measure (probability) to the event that everything happens (which of course it should be one). Number (2) makes sense as well because if an event A occurs then the complement will be the event that doesn't happen so I can set a probability for the complement as 1-(probability A) and this will make sense since the complement is an element of the sigma algebra so can be "measured" (evaluated in my probability set function). My problem is with (3), why do we need a countable union? For instance, there are set that are not borel measurable and they can be obtained by taking the union of single sets containing the elements of it (and these single sets are measurable) but OF COURSE this argument is totally wrong since just saying the word "not measurable" refered to the definition of sigma-algebra which is using explicitly the word "countable union only".
Thank you in advance.

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## closed as off topic by Evan Jenkins, Steven Landsburg, Tony Huynh, Tom Leinster, Asaf KaragilaSep 24 '12 at 23:58

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One can give meaning to a sum of uncountable non-negative real numbers, but it turns out that unless all but countably many of them are zero, the sum ends up being $\infty$. So, for the applications to measure theory, being closed uncountable unions are not going to be of much help (one does complete $\sigma$-algebras w.r.t. measures, but this is, in a sense, related to the fact that one can add uncountably many zeroes...) –  Mariano Suárez-Alvarez Sep 24 '12 at 23:29
If you re-ask this question on another site, I suggest clarify whether or not you mean to ask whether "Why only countable unions and not more?" or whether you mean to ask "Why do we take countable unions to begin with and not just finite unions?", either questions is valid, but it takes some effort to get the former from your text. –  Asaf Karagila Sep 24 '12 at 23:58
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## 1 Answer

The notion of sigma-algebra was abstracted from previously known special cases. For example Lebesgue measurable sets in the real line. In that setting, one cannot prove that uncountable unions of measurable sets are measurable. But all that is needed for measure and integration theory is countable operations.

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your answer is very deficient and unsatisfactory –  Fer Sep 25 '12 at 1:22
+1: Fer's question is ambiguous, and this answer is good (even if it's not satisfactory to Fer). Maybe Fer would've wanted and explanation of how it can be proven that a countable union of measurable sets is measurable? If so, then that would be a good thing to be "deficient" of, as that can easily be found from a search engine. –  Ricky Demer Sep 25 '12 at 2:28
"one annot prove that uncountable unions of measurable sets are measurable" If you say that then you are using the definition of measurability and that is my question. Ricky probably needs to take some classes of comprehension because my question was clear: why do we need that a countable union of elements in the sigma algebra is still in the sigma algebra? what is the problem with an arbitrary union? –  Fer Sep 26 '12 at 16:57
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