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 sigmaalgebra 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 Karagila Sep 24 '12 at 23:58
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The notion of sigmaalgebra 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. 

