Show $(\Omega, \mathscr{F}, P)$ where $\mathscr{F}=\{\text{all subsets of }\mathbb{R}\text{ such that either }A^c\text{ or }A\text{ is a countable set}\}$, and $P(A)=0$ if $A$ is countable, $P(A)=1$ if $A^c$ is countable, is a probability space.
I've shown that $\mathscr{F}$ is a $\sigma$-algebra and that $P(\Omega )=1$, but I still need to show countable additivity, i.e.
$$\displaystyle P\left(\bigcup_{n\in\mathbb{N}} A_n\right)=\sum_{n\in\mathbb{N}} P(A_n)$$ for any disjoint sequence $(A_n)_{n\in\mathbb{N}}$ of sets in $\mathscr{F}$.
Pf:
If $(A_n)_{n\in\mathbb{N}}$ are all countable and disjoint, then since $\mathscr{F}$ is a $\sigma$-algebra, the union $\bigcup A_n$ are countable as well and so $\bigcup A_n\in\mathscr{F}$.
But if, for some $j\in\mathbb{N}$, $A_j$ is uncountable, then $A_j^c$ is countable and $P(A_j)=1$.
There has to be only one such $j$ such that $A_j$ is uncountable, otherwise $\sum P(A_j)>1$. How can I relate this to the fact that $(A_n)_{n\in\mathbb{N}}$ are disjoint? A hint would be very appreciated!
Best Answer
If $A,B$ are disjoint and and co-countable, then $A\subset B^c$ is necessarily countable. But then $\Omega = A\cup A^c$ is also countable, in contradiction. Hence there could be at most a single co-countable $A_j$ (the rest follows as you say).