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Halmos developed measure theory based on $\sigma$-rings. Nowadays measure theory is based on $\sigma$-algebras. I would like to know how to bridge the two theories if possible.

Namely, let $(X, \mathcal R, \mu)$ be a measure space, where $\mathcal R$ is a $\sigma$-ring. Let $\mathcal A$ be the smallest $\sigma$-algebra containing $\mathcal R$. Can $\mu$ be extended to a measure on $\mathcal A$?. If yes, is the extension unique?

My guess is that if $\mu$ is $\sigma$-finite, i.e. every member of $\mathcal R$ is covered by a countable union of members of finite measure, then the extension is unique.

JHW
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  • It just means assigning a measure to $X$ and to complements of sets in $\cal R$. – Asaf Karagila Jul 23 '15 at 08:37
  • @AsafKaragila How to do that? If $\mathcal R\neq\mathcal A$ (or equivalently $X\notin\mathcal R$) then the collection of complements of sets in $\mathcal R$ ($X$ is one of them) is not a $\sigma$-ring. It misses the empty set. You are probably thinking of exploiting $\mu(R)+\mu(R^c)=\mu(X)$ for $R\in\mathcal R$. But what value should $\mu(R^c)$ get if $\mu(R)=\infty$? – drhab Jul 23 '15 at 09:19
  • Surely $\mu$ can be extended. If you define $\mu^(B)$ as the infinum of (countable) sums $\sum\mu(R_i)$ where the $R_i\in\mathcal R$ cover $B$ then $\mu^$ is an outer measure that coincides with $\mu$ on $\mathcal R$. Then there is a $\sigma$-algebra that contains $\mathcal R$ hence also contains $\mathcal A$ such that $\mu^*$ restricted to it is a measure. – drhab Jul 23 '15 at 09:45
  • @drhab: Since $\cal R$ is closed under countable unions and relative complements, this is the same as defining it as $\inf{\mu(R)\mid R\in\mathcal R\land B\subseteq R}$. – Asaf Karagila Jul 23 '15 at 09:47
  • @AsafKaragila Yes, you are right. That is in fact more elegant and it is nice for the OP that you remark this. But also in my view it is good to keep the link with the general theory (outer measures etc.) in sight. – drhab Jul 23 '15 at 09:53
  • I realized afterwards that the method I reached unto you will give you $\mu(R^c)=\infty$ for each $R\in\mathcal R$ if $X\notin\mathcal R$. This because in that case the collection of the sums will be empty. I restrict to comments because uptil now nothing has been said about the uniqueness of the extension. And maybe that is even your main question. – drhab Jul 23 '15 at 10:10
  • @drhab Someone has just written to me in his comment to my question(http://math.stackexchange.com/questions/1370814/langs-treatment-of-product-of-radon-measures) that Dynkin's $\pi- \lambda$ Theorem(http://math.stackexchange.com/questions/812715/if-two-measures-agree-on-generating-sets-do-they-agree-on-all-measurable-sets/813414#813414) is useful to prove the uniqueness of an extension of a measure. I don't want to answer my own question. Maybe you can. – JHW Jul 24 '15 at 23:17
  • @JHW I will have a look at it, but not today (no time). If I have done that and succeed then I will answer. If I do not succeed then I will let you know in a comment. – drhab Jul 25 '15 at 07:17
  • @drhab Maybe I was mistaken. I found a counter-example. Let $(X, \mathcal A, \nu)$ be a finite measure space where $\mathcal A$ is a $\sigma$-algebra. Let $Y$ be a member of $\mathcal A$ such that $X \neq Y$. Let $\mathcal R = {E: E\subset Y, E\in \mathcal A}$. Let $\mu$ be the restriction of $\nu$ on $\mathcal R$. Then $\mu$ is a measure on $\mathcal R$ and $\nu$ is an extension of $\mu$, but $\nu(X) \lt \infty$. – JHW Jul 25 '15 at 09:26

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