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Can someone give me an example of an algebra but not a sigma algebra, with a $\sigma$-finite measure $\mu$ on it?

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  • $\begingroup$ As far as I'm aware, $\sigma$-finiteness is only defined on sigma algebras - how are you defining it here? $\endgroup$ – B. Mehta Apr 14 '18 at 22:43
  • $\begingroup$ @B.Mehta I understood this to mean a measure on the sigma-algebra generated by the algebra. $\endgroup$ – Severin Schraven Apr 14 '18 at 22:44
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Let $X=[0, \infty)$ and consider $\mathcal{A}=\{ \text{finite union of }[a, b)\}$.

It's clear that $[a, b)\cap [c, d) = [\max\{a, c\}, \min\{b, d\})$ and $[a, b)^C = [0, a)\cup [b, \infty)$.

So $\mathcal{A}$ is an algebra but clearly not a $\sigma$-algebra since it doesn't contain any open interval.

Take the obvious measure $m([a, b)) = b-a$.

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  • $\begingroup$ +1 Neat example. $\endgroup$ – Severin Schraven Apr 14 '18 at 22:58
  • $\begingroup$ Thanks Jacky that’s very understandable! :D $\endgroup$ – JINGYA HAN Apr 14 '18 at 23:35
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Take all subsets of $\mathbb{N}$ which are either finite or have finite complement. It is an algebra, but not a sigma-algebra. For the sigma-finite measure you can pick the zero measure.

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  • $\begingroup$ Or for the measure pick counting measure. $\endgroup$ – GEdgar Apr 14 '18 at 23:01
  • $\begingroup$ Thanks for your example. In this context I don’t think the measure could be replaced by a counting measure since that would not always be finite. But say if we replace the measure by a counting measure, does it mean this measure does not have a unique extension on the whole natural number set, since it would meet the requirement of using Carathéodory Extension Theorem? $\endgroup$ – JINGYA HAN Apr 14 '18 at 23:34
  • $\begingroup$ @JINGYAHAN We could replace the measure with the counting measure, it would still be sigma-finite as we can write $$\mathbb{N}=\bigcup_{n\in \mathbb{N}} \{n\}$$ and the singeltons have counting measure equal to $1$. $\endgroup$ – Severin Schraven Apr 15 '18 at 0:26
  • $\begingroup$ @SeverinSchraven oh yes you’re right. I confused the definition. The point is you cannot assign a singleton with an infinity measure. Thanks for your help! $\endgroup$ – JINGYA HAN Apr 15 '18 at 0:31
  • $\begingroup$ @JINGYAHAN Exactly, we don't want concentration of mass on singeltons. By the way you should accept Jacky's answer (on the left you of the question you can find the respective button), unless you are not yet satisfied with the answers given and want to wait for another one. $\endgroup$ – Severin Schraven Apr 15 '18 at 0:36

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