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I am having a bit of trouble understanding this definition. "A collection $\Sigma$ of subsets of S is called a $\sigma$-algebra on S if $\Sigma$ is an algebra on S such that whenever $F_n \in\Sigma (n \in N)$, then $\bigcup\limits_{n} F_{n} \in \Sigma$.

The part im not sure with is $\bigcup\limits_{n} F_{n} \in \Sigma$. Lets say n=3, then $F_3 \in\Sigma$, so then $\bigcup\limits_{3} F_{3} \in \Sigma$. Does $\bigcup\limits_{3} F_{3} \in \Sigma$ = $F_1+F_2+F_3$? With $F_1,F_2,F_3 \in \Sigma$?

Im just a bit confused with the union notation. What would $\bigcup\limits_{3} F_{3} \in \Sigma$ equals to? Thanks.

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The statement that ‘whenever $F_n\in\Sigma$ ($n\in N$), then $\bigcup\limits_n F_n\in\Sigma$’ means that if $F_0,F_1,F_2,\dots$ are elements of $\Sigma$, then $F_0\cup F_1\cup F_2\cup\dots$ is also an element of $\Sigma$. In other words, it’s talking about an infinite collection of $F_n$’s, one for each $n\in N$.

This is often expressed by saying that $\Sigma$ is closed under countable unions, meaning simply that the union of any finite or countably infinite collection of elements of $\Sigma$ is again an element of $\Sigma$.

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Thanks, just one question -- what do you mean by "one for each $n\in N$"? Do you mean for say $F_{10}$, its $F_0\cup\dots\cup F_{10}$? –  Thomas Nov 13 '11 at 1:27
    
@Thomas: No, I mean that there are infinitely many $F_n$’s. $\bigcup\limits_n F_n$ is short for $\bigcup\limits_{n=0}^\infty F_n$: $F_0\cup F_1\cup F_2\cup\dots\cup F_{100}\cup\dots\cup F_{1000}\cup\dots$ forever. –  Brian M. Scott Nov 13 '11 at 9:27
    
Ahh okay, thanks for clearing it up! –  Thomas Nov 13 '11 at 15:07
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