# Show $(\Omega, \mathscr{F}, P)$ is a probability space.

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!

• I like this question! Maybe you could use that if there's an $A_j$ such that is not countable then $A_j^c$ is countable, then $(\bigcup_{i=1}^\infty A_i)^c= \bigcap_{i=1}^\infty A_i^c\subset A_j^c$, I'm still thinking but this might help Sep 16, 2013 at 17:48

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).
If $P$ is a probability measure, then you already now that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, but since $A,B\in\mathscr{F}$ and if $A\cap B=\varnothing$ then $P(A\cap B)=0$, then you can use that $P(A\cup B)=P(A)+P(B)\dots\mathbf {(a)}$
Now, if $A_i\in\mathscr{F}$, then $P(A_1\cup\dots\cup A_j\cup\dots)=P(A_1)+\dots+P(A_j)+\dots =\sum_{i=1}^{j-1}P(A_i)+P(A_j)+\sum_{i=j+1}^\infty P(A_i)$ and to show why that equality is true, you have to use $\mathbf {(a)}$, with $A_j$ not countable.