Let $(X,\mathscr{F})$ be a measurable space.

Let $\mathscr{F}_\alpha \subseteq \mathscr{F}$, $\alpha \in \mathfrak{A}$ be a class of $\sigma$-subalgebras.

Let we have a measure $\mu_\alpha$ on every $(X,\mathscr{F}_\alpha)$.

Let the family $(\mu_\alpha, \alpha \in \mathfrak{A})$ be consistent, i.e. $\mu_{\alpha}(A)\le\mu_{\beta}(B)$ when $A\subseteq B$.

Now the question is if it is possible to construct such a measure $\mu$ on $(X,\mathscr{F})$ that $\mu(A)=\mu_{\alpha}(A)$ for all $A \in \mathscr{F}_\alpha$, $\alpha \in \mathfrak{A}$ ?

  • $\begingroup$ 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 $\mathcal F_\beta$. $\endgroup$ – Gerrit Begher Nov 26 '13 at 17:01
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    $\begingroup$ 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. $\endgroup$ – Monroe Eskew Nov 26 '13 at 17:15
  • $\begingroup$ @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. $\endgroup$ – m.a. Nov 26 '13 at 17:24
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    $\begingroup$ $X=\{0,1,2\}$, $\mathcal F=\mathcal P(X)$, $\mathcal F_0=\{\varnothing,X,\{0\},\{1,2\}\}$, $\mathcal F_1=\{\varnothing,X,\{0,1\},\{2\}\}$, $\mu_0(\varnothing)=\mu_0(\{1,2\})=0$, $\mu_0(\{0\})=\mu_0(X)=1$, $\mu_1(\varnothing)=\mu_1(\{0,1\})=0$, $\mu_1(\{2\})=\mu_1(X)=1$. $\endgroup$ – Emil Jeřábek Nov 26 '13 at 17:29
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    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Nov 26 '13 at 17:29

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