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changed the condition of consistency, thanks to Emil Jeřábek
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m.a.
m.a.
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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. if $A\in\mathscr{F}_{\alpha_1}\cap\mathscr{F}_{\alpha_2}$ then $\mu_{\alpha_1}(A)=\mu_{\alpha_2}(A)$$\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 $A \in \mathscr{F}_\alpha$, $\alpha \in \mathfrak{A}$ ?

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. if $A\in\mathscr{F}_{\alpha_1}\cap\mathscr{F}_{\alpha_2}$ then $\mu_{\alpha_1}(A)=\mu_{\alpha_2}(A)$.

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 $A \in \mathscr{F}_\alpha$, $\alpha \in \mathfrak{A}$ ?

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 $A \in \mathscr{F}_\alpha$, $\alpha \in \mathfrak{A}$ ?

Source Link
m.a.
m.a.
  • 1
  • 2

Recovering measure from the family of sigma-subalgebras

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. if $A\in\mathscr{F}_{\alpha_1}\cap\mathscr{F}_{\alpha_2}$ then $\mu_{\alpha_1}(A)=\mu_{\alpha_2}(A)$.

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 $A \in \mathscr{F}_\alpha$, $\alpha \in \mathfrak{A}$ ?