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I wanted to add a quick follow up question to one I asked earlier here.

To summarize, there I let $\mathcal{R}$ be the set ring of subsets of $\mathbb{Q}$ consisting of finite unions of left-open, right closed intervals on the rational line.

With George Lowther's help, I know there exists a finitely additive $\mu\colon\mathcal{R}\to[0,\infty]$ where $\mu=\nu\circ f^{-1}$ is a function such that $\mu((a,b])=b-a$, where $\nu$ is the restriction of the Lebesgue measure on $\mathbb{R}$ to finite unions of half-open real intervals $(a,b]$, and $f$ is the function mapping finite unions of half-open real intervals to their intersection with $\mathbb{Q}$. That is, $f(A)=A\cap\mathbb{Q}$ where $A$ is a finite union of left-open, right-closed intervals on the real line.

One additional thing I'm curious about, with the $\mu$ constructed previously, is it possible to extend $\mu$ to a measure on the $\sigma$-algebra $\mathcal{T}$ generated by $\mathcal{R}$, that is, the intersection of all $\sigma$-algebras containing $\mathcal{R}$?

My suspicion is that countable additivity does not come along with $\mu$, so no such extension is possible, but I'm not 100% sure.

Thanks.

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@Rasmus: $\mu$ is not $\sigma$-additive on $\mathcal{R}$. To see this, enumerate the rationals as $\{q_n\}$, and let $A_n = (q_n-2^{-n}, q_n+2^{-n}]$. Then $\bigcup A_n = \mathbb{Q}$ but $\sum \mu(A_n) = 2$. –  Nate Eldredge Nov 2 '11 at 12:14
    
@NateEldredge: You are right. Thank you for the explanation! –  Rasmus Nov 2 '11 at 12:26

1 Answer 1

up vote 2 down vote accepted

The sigma-algebra generated by $\mathcal{R}$ is just $\mathcal{P}(\mathbb{Q})$, because $\{q\} = \cap_n [q, q+1/n)$, so it contains all singletons etc. The measure of $\{q\}$ in such an extension would be $0$, using the same equation, and so such an extension would be $0$ on all subsets of $\mathbb{Q}$. So such an extension cannot exist.

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