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In my measure theory class, I believe the professor made the claim that if $X$ was a countable or finite set and $\mathcal M$ was an algebra on $X$, then $ \mathcal M$ was a $\sigma$-algebra.

I am unclear how to prove this. I know that given any countable collection of sets, $\{E_j\}_{j \in \mathbb N}$, I can say that $G_n$ := $\bigcup_{j=1}^{n} E_j$ is in $\mathcal M$. But I don't know if it is OK to simply say that since by induction, all of the $G_n$ are in $\mathcal M$ implies the entire countable union is. My problem in doing this is that there are set properties that can hold for any finite number of sets that will not hold for a countably infinite collection (like for example, the finite union of closed sets are closed, but a countably infinite union may not be).

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    $\begingroup$ The family of all subsets of $\mathbb{N}$ that are either finite or have a finite complement is an algebra, but not a $\sigma$-algebra. $\endgroup$ – Michael Greinecker Aug 24 '14 at 2:35
  • $\begingroup$ @MichaelGreinecker Yes, I think I can see that. Let X be the natural numbers and take the countable collection of sets in M (as you defined it) to be $E_p$ = {p} where p is prime. Their countable union is neither finite, nor is their complement, hence it is not in M. Does that sound right? $\endgroup$ – Gremlin Brenneman Aug 24 '14 at 2:45
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    $\begingroup$ Yes, that's right. $\endgroup$ – Michael Greinecker Aug 24 '14 at 2:52
  • $\begingroup$ @MichaelGreinecker I apologize, but I cannot find the icon to check that you have answered my question. The FAQ says there should be a check mark icon below the "up-vote" icon, but on my page the icon below the up-vote icon is one that flags the comment. $\endgroup$ – Matt Brenneman Aug 25 '14 at 12:26
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    $\begingroup$ @MattBrenneman 1. You cannot accept comments as answers. I reposted the comment as an answer. 2. You can only accept the answer from the account from which you asked the question "Gremlin Brenneman". $\endgroup$ – Michael Greinecker Aug 25 '14 at 16:59
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The family of all subsets of $\mathbb{N}$ that are either finite or have a finite complement is an algebra, but not a $\sigma$-algebra.

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