Suppose $(\Omega,\mathcal F,P)$ is a probability space. Set $P^*(A)=\inf\{P(B):A\subseteq B,B\in\mathcal F\}$ for any $A\subseteq\Omega$. Take $A_0\notin \mathcal F$ such that $P^*(A_0)>0$ and define $P_0(A\cap A_0)=\dfrac{P^*(A\cap A_0)}{P^*(A_0)}$ for any $A\in\mathcal F$. Show that $P_0$ is a probability on $\mathcal F\cap A_0:=\{A\cap A_0:A\in\mathcal F\}$.
Countable additivity is what is turning out to be tricky to prove. Clearly I want to show that if $A_n,n\geq1$ are such that $A_n\in\mathcal F$ for all $n$ and for $k\neq l$, $A_k\cap A_l\cap A_0=\phi$ then $P^*(\cup_n(A_n\cap A_0))=\sum_n P^*(A_n\cap A_0)$. Note that $P^*$ is an outer measure and so combining with countable subadditivity of $P^*$, we only need to show that $P^*(\cup_n(A_n\cap A_0))\geq\sum_n P^*(A_n\cap A_0)$.
So what I tried was, I took any $B\in\mathcal F$ such that $B\supset \cup_n (A_n\cap A_0)$ and I wanted to show $P(B)\geq \sum_n P^*(A_n\cap A_0)$. I thought that if I define $B_n=B\cap A_n\in\mathcal F$ then $B_n$ would be disjoint and I could write $P(B)\geq\sum_n P(B_n)$, and $B_n\supset A_n\cap A_0$, but the issue is that $B_n$ are not necessarily disjoint.
Any help is appreciated.
Best Answer
I have a suggestetion which might help you. Define an outer measure
$$\nu(A):=P^*(A\cap A_0) \quad \text{for all} \quad A\subseteq \Omega.$$
Since an outer measure is a measure when restricted to its $\nu$-measurable sets, it will be enough to show that any $A\in \mathcal{F}$ satisfies
$$ \nu(D)= \nu(D\cap A)+\nu(D\cap A^c) $$
for all $D\subseteq \Omega$. By definition of $P^*$, for all $\epsilon>0$ there exists $D\cap A_0 \subseteq D_\epsilon \in \mathcal{F}$ such that $P(D_\epsilon)< P^*(D\cap A_0)+\epsilon$. Hence,
$$ P(D_\epsilon) \overset{A\in \mathcal{F}}{=} P(D_\epsilon \cap A)+P(D_\epsilon\cap A^c)\geq $$
$$ \geq P^*(D\cap A_0 \cap A)+P^*(D\cap A_0 \cap A^c). $$ Hence, $ P^*(D\cap A_0)\geq P^*(D\cap A_0 \cap A)+P^*(D\cap A_0 \cap A^c) $, which shows that $A$ is $\nu$-measurable and implies countable additivity.