Let $S = \{a,b,c,d\}, \mathcal{F}= 2^{\{a,b,c,d\}}, \mathcal{F} = σ(\mathcal{C})$.
- What is the example of two distinct probability measures $P_1$ and $P_2$ agree on $\mathcal{C}$, a collection of sets? Here $\mathcal{C} = \{\{a,b\},\{c,d\},\{a,c\},\{b,d\}\}$.
-
$(S, \mathcal{F})$ a measurable space, and $\mathcal{C}$ is closed under finite union. How to prove that two measures $P_1$ and $P_2$ agree on $\mathcal{C}$ also agree on $\mathcal{F}$?
-
Can I still prove that they also agree on $\mathcal{F}$, if $\mathcal{C}$ is closed under complements? Or is there other counterexample?
Best Answer
1) example:
$P_1(\{x\})=\frac14$ for $x\in\{a,b,c,d\}$.
$P_2(\{x\})=\frac18$ for $x\in\{a,d\}$ and $P_2(\{x\})=\frac38$ for $x\in\{b,c\}$.
2) Define $\mathcal E:=\{A^c\mid A\in\mathcal C\}$. Then $\mathcal E$ is a $\pi$-system with $\sigma(\mathcal E)=\sigma(\mathcal C)=\mathcal F$.
Also $\mathcal E$ is contained in $\lambda$-system $\mathcal D:=\{A\in\wp(S)\mid P_1(A)=P_2(A)\}$ so that $\mathcal F\subseteq\mathcal D$.
3) The $\mathcal C$ under 1) is closed under complements and satisfies $\sigma(\mathcal C)=\mathcal F$, so serves as counterexample.