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Reading a probability theory book, and it says that if we have a sample space $\Omega$, then some class $F$ of subsets of $\Omega$ makes a $\sigma$-algebra and has the following properties:

1) $\Omega\in F$

2) if $A\in F$, then $\complement A\in F$

3) if $A_1,A_2,...\in F$, then ${\stackrel{\infty}{\cup}}_{i=1}A_i\in F$

I don't understand how the last property works, how exactly do we do that infinite number of unions with finite number of elements?

If we have $A_1,A_2,A_3,A_4\in F$, does it mean that $A_1\cup A_3\cup...\cup A_3\in F$ ("..." in that case is $A_3$ unioned with itself an infinite number of times)?

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  • $\begingroup$ In ${\stackrel{\infty}{\cup}}_{i=1}$ how do I write $i=1$ under $\cup$? $\endgroup$
    – Pavel
    Jun 12, 2016 at 2:42
  • $\begingroup$ You can use \displaystyle to force the entire scope in which it occurs to be set in display style, which puts limits above and below. $\endgroup$
    – user21820
    Jun 23, 2016 at 14:30

2 Answers 2

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Yes, there is nothing saying that the countably many $A_i$ need to be distinct. So while we could posit a separate axiom for finite unions, there's no need to do so.

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The first and second properties imply that the empty set is always contained in a $\sigma$-algebra: $\varnothing=\Omega^{\mathsf c}\in F$. Consequently, any union of finitely many sets from $F$ can be represented as an infinite union by taking infinitely many copies of the empty set: if, say, $A_1,A_2,A_3,A_4\in F$, then $$A_1\cup A_2\cup A_3\cup A_4=A_1\cup A_2\cup A_3\cup A_4\cup\varnothing\cup\varnothing\cup\varnothing\cup\cdots\in F.$$

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  • $\begingroup$ what about my second question? do I have to union consecutive elements or can i skip some? Like I asked if $F$ contains $A_1\cup A_3$ when $A_1,A_2,A_3,A_4\in F$ $\endgroup$
    – Pavel
    Jun 12, 2016 at 2:47
  • $\begingroup$ @paulpaul1076 Yes, you can always “skip” elements. Property 3 says that if you take countably many elements from $F$ (which countable family need not be the whole of $F$ and there can be repetitions), then their union is always in $F$. You can represent $A_1\cup A_3 $ as $A_1\cup A_3\cup A_3\cup A_3\cup\cdots$ as you wrote, or as $A_1\cup A_3\cup\varnothing\cup \varnothing\cup\cdots$. $\endgroup$
    – triple_sec
    Jun 12, 2016 at 2:50
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    $\begingroup$ Oh, so if you have uncountable $F$, then the union of any countable subsets also belongs to $F$, right? $\endgroup$
    – Pavel
    Jun 12, 2016 at 2:54
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    $\begingroup$ @paulpaul1076 Exactly. $\endgroup$
    – triple_sec
    Jun 12, 2016 at 2:54
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    $\begingroup$ @paulpaul1076 Yeah, I agree: the terminology can be confusing if we’re talking about sets of sets. Just to make sure we’re on the same page, let me clarify: If $F$ is a $\sigma$-algebra over a non-empty set $\Omega$, then the typical element of $F$ is some subset of $\Omega$. Now, that a $\sigma$-algebra is closed under countable unions means the following: if $A_1,A_2,A_3,\ldots$ is a countable collection of elements of $F$ (there can be repetitions), then the union $A_1\cup A_2\cup A_3\cup\cdots$ must be an element of $F$. Nothing here is assumed about the cardinality of $F$. $\endgroup$
    – triple_sec
    Jun 12, 2016 at 2:59

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