Extensions of probability measures on fields I am trying to solve exercise 3.3 in Billingsley's "Probability and measure" and I am not sure I correctly understood the text. I am not going to copy all the text, but I will ask directly some questions.
Some background:
Let P be a probability measure on the field $F_0$ and $P^*$ be the outer measure defined for each subset A of $\Omega$ as $P^*(A)=inf \sum_n P(A_n)$, where $A \subset \cup_n A_n$.
A theorem (3.1 in Billingsley) guarantees that $P^*$ is the unique probability measure on the $\sigma$-field $\sigma(F_0)$ such that $P^*(B)=P(B)$ for each $B \in F_0$.
Now the questions:
1) If $\Omega$ is uncountable and $F_0$ is the field of finite and cofinite sets, and P(A)=0 if A is finite and P(A)=1 if A is cofinite, do $P^*$ (defined as above) and $P$ agree on $F_0$?
My answer: P is a probability measure on $F_0$ (it is possible to check that it is countably additive) and then theorem 3.1 holds and P and $P^*$ agree on $F_0$.
My doubt comes from the fact that the text of the exercise seems to suggest that this is not the case and to explain why...
2) If \Omega is uncountable and F_0 is the field of countable and cocountable  sets, and P(A)=0 if A is countable and P(A)=1 if A is cocountable, do $P^*$ (defined as above) and P agree on $F_0$?
My answer would be the same than before:
P is a probability measure on $F_0$ (it is possible to check that it is countably additive) and then theorem 3.1 holds and P^* agree on $F_0$.
3) Let $P(A)=I_A(\omega_0)$ for $A \in F_0$, and assume that ${\omega_0} \in \sigma(F_0)$, do $P^*$ (defined as above) and P agree on $F_0$?
My answer would be the same than before:
P is a probability measure on $F_0$ (it is possible to check that it is countably additive) and then theorem 3.1 holds and $P^*$ agree on $F_0$.
In this case I do not understand also why the text specifies that ${\omega_0} \in \sigma(F_0)$.
 A: The answer to 1 and 2 will be very similar since an uncountable set is never the union of a countable number of finite or countable sets. In this case, if $A$ is finite/countable, then $A \subset \cup A_n$ where for every $n$, $A_n = A$. Therefore $P^*(A) \leq \sum 0 = 0$. Since $P^*(A) \geq 0$ then you see that $P^*(A) = P(A) = 0$. On the other hand, if $B$ is a cofinite/cocountable set, then because $\Omega$ is uncountable, it must be that $B$ is uncountable. So if $A_n \in F_0$ such that $B \subset \cup A_n$ then for at least one of the $n$, $A_n$ must be uncountable (as remarked in the first sentence) which means that for this $n$, $P(A_n) = 1$ (since $A_n$ must in fact then be cofinite/cocountalbe). Therefore you will find that $P^*(B) \geq 1$. Choosing $A_1 = B$ and $A_n = \varnothing$ for $n >1$, you get $\sum P(A_n) = 1 \geq P^*(B)$, so $P^*(B) = 1 = P(B)$. 
If the assumption on the first problem was that $\Omega$ is countable, then $P^*$ and $P$ won't agree on $F_0$. This happens since if $A \in F_0$ is cofinite, then $A$ is countable and we can denumerate the elements of $A=\{a_1, a_2, ...\}$. Now, for each $i$, $\{a_i\} \in F_0$ and  $\cup\{a_i\} = A$. But then $0 \leq P^*(A) \leq \sum P(\{a_i\}) = 0$, so $P^*(A)=0$ while $P(A)=1$. What you should notice here is that since $A$ is the countable disjoint union of the $\{a_i\}$, you find that $1 = P(A) = P(\cup\{a_i\}) \neq \sum P(\{a_i\}) = 0$. So, $P$ is not a very nice candidate for a premeasure.
For question 3, we have $P(A)=1$ if $\omega_0 \in A$, otherwise $P(A) = 0$. If $A \subset \cup A_n$ and $\omega_0 \notin A$ then we can choose $A_n = A$ for every $n$ to find $P^*(A) = P(A) = 0$ using similar arguments as before. On the other hand, if $\omega_0 \in A$ and $A \subset \cup A_n$, then there is some $n$ such that $\omega_0 \in A_n$. Therefore you can infer that $P^*(A) \geq 1$. Since you can choose $A_1 = A$ and $A_n = \varnothing$ for every $n > 1$, you find that $P^*(A) \leq 1$ implying that $P^*(A) = P(A) =1$.
