Timeline for Recovering measure from the family of sigma-subalgebras
Current License: CC BY-SA 3.0
15 events
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| Nov 26, 2013 at 19:21 | comment | added | Emil Jeřábek | I’m sorry about that. On the positive side, you could try to take the outer measure $\mu^*$ generated by the union of all the $\mu_\alpha$’s, and the induced measure $\mu$, and see whether there are reasonable conditions under which $\mu^*$ extends every $\mu_\alpha$, and every $A\in\mathcal F_\alpha$ is $\mu$-measurable. The first part is in fact easy: it amounts to the condition that if $A\subseteq\bigcup_nA_n$, where $A\in\mathcal F_\alpha$ and $A_n\in\mathcal F_{\alpha_n}$, then $\mu_\alpha(A)\le\sum_n\mu_{\alpha_n}(A_n)$. The second part is more of a problem. | |
| Nov 26, 2013 at 18:05 | comment | added | m.a. | @EmilJeřábek thanks, you've finally ruined my conception of the subject. | |
| Nov 26, 2013 at 17:57 | comment | added | Emil Jeřábek | And for an example of another thing that can go wrong, let $X=\mathbb N$, $\mathcal F=\mathcal P(\mathbb N)$, $\mathcal F_n$ be the algebra of subsets of $\{0,\dots,n\}$ and their complements (in $\mathbb N$), and $\mu_n$ be the unique probabilistic measure on $\mathcal F_n$ such that $\mu_n(\{i\})=3^{-i-1}$ for $i\le n$. Note that $\{(\mathcal F_n,\mu_n):n\in\mathbb N\}$ form an increasing chain, hence they are pairwise (or finitewise) compatible in any sense you can think of, but they have no common $\sigma$-additive extension. | |
| Nov 26, 2013 at 17:51 | comment | added | Emil Jeřábek | That’s not going to help. With the same $X$ and $\mathcal F$ as above, let $\mathcal F_i=\{\varnothing,X,\{i\},X\smallsetminus\{i\}\}$ for $i=0,1,2$, and let $\mu_i(\{i\})=0$, $\mu_i(\{X\smallsetminus\{i\})=1$. | |
| Nov 26, 2013 at 17:50 | history | edited | m.a. | CC BY-SA 3.0 |
changed the condition of consistency, thanks to Emil Jeřábek
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| Nov 26, 2013 at 17:42 | comment | added | m.a. | @EmilJeřábek thanks, it seems I have to demand of $\mu_\alpha(A) \le \mu_\beta(B)$ when $A\subseteq B$. | |
| Nov 26, 2013 at 17:34 | comment | added | Emil Jeřábek | In my counterexample, $\mathcal F=\sigma(\mathcal F_0\cup\mathcal F_1)$. | |
| Nov 26, 2013 at 17:31 | comment | added | m.a. | @MonroeEskew Thanks for a good counterexample. Well, as far as I can see, I have to ask another question: is it possible to continue the measure to $\sigma (\mathscr{F}_{\alpha}, \alpha \in \mathfrak{A})$? | |
| Nov 26, 2013 at 17:29 | comment | added | Nate Eldredge | See mathoverflow.net/questions/118636/… where it is shown that the Kolmogorov extension theorem can fail for measurable spaces which are not standard Borel. It seems like this may provide a counterexample for your claim. | |
| Nov 26, 2013 at 17:29 | comment | added | Emil Jeřábek | $X=\{0,1,2\}$, $\mathcal F=\mathcal P(X)$, $\mathcal F_0=\{\varnothing,X,\{0\},\{1,2\}\}$, $\mathcal F_1=\{\varnothing,X,\{0,1\},\{2\}\}$, $\mu_0(\varnothing)=\mu_0(\{1,2\})=0$, $\mu_0(\{0\})=\mu_0(X)=1$, $\mu_1(\varnothing)=\mu_1(\{0,1\})=0$, $\mu_1(\{2\})=\mu_1(X)=1$. | |
| Nov 26, 2013 at 17:24 | comment | added | m.a. | @GarlefWegart Well, it might be not as simple as it seems. To begin with, in some sence it's a generalization of Kolmogorov theorem. Also, as I can see, your property of compatibility is too strong and I'm not sure I want to use it. | |
| Nov 26, 2013 at 17:15 | comment | added | Monroe Eskew | Not necessarily. $\mathscr{F}$ could be $\mathscr{P}(\mathbb{R})$, and $\mathfrak{A}$ could have one element $0$, where $\mathscr{F}_0$ is the Borel sets, and $\mu_0$ is the Lebesgue measure. | |
| Nov 26, 2013 at 17:01 | comment | added | Gerrit Begher | This seems to be a homework question: However, i guess this is possible and straightforward if $\mathcal F$ is generated (as a $\sigma$-agebra) by the union $\bigcup\mathcal F_\alpha$ and furthermore the family $\mathcal F_\alpha$ is compatible in the following sense: If $A\in\mathcal F_\alpha$ and $B\in\mathcal F_\beta$ then is $A\cap B$ is in both $\mathcal F_\alpha$ and $\mathcal F_\beta$. | |
| Nov 26, 2013 at 16:54 | review | First posts | |||
| Nov 26, 2013 at 17:19 | |||||
| Nov 26, 2013 at 16:39 | history | asked | m.a. | CC BY-SA 3.0 |