Can I please get help proving the problem below? I feel like I wrote a lot of superfluous stuff. Thank you.
Let $\Omega = \mathbb{R}, \mathcal{F} = $ all subsets so that $A$ or $A^c$ is countable, $P(A) = 0$ in the first case and $ = 1$ in the second. I wish to show that $(\Omega, \mathcal{F}, P)$ is a probability space.
$\textit{Proof.}$ Observe that $\mathcal{F}$ is a sigma field, and $\emptyset \in \mathcal{F}$ as $\emptyset$ is countable by definition, $\Omega \in \mathcal{F}$ as $\Omega^c = \emptyset$ is also countable by definition. If $A\in \mathcal{F}$ then $A$ is countable as $A^c$ is countable. Now, if $A$ is countable then $A^c \in \mathcal{F}$ as $(A^c)^c$ is countable and if $A^c$ is countable then clearly $A^c \in \mathcal{F}.$ If $A_n's \in \mathcal{F}$ then if all $A_n's$ are countable then $\cup A_n's$ are countable then $\cup A_n \in \mathcal{F}.$ If there exists $N$ such that $A_N^c$ is countable then $$\left(\bigcup_{n=1}^\infty A_n\right)^c = \bigcap_{n=1}^\infty A_n^c \subseteq A_N^c$$ is countable then $\bigcap_{n=1}^\infty A_n^c$ is countable then $\cup_{n=1}^\infty A_n \in \mathcal{F}.$ Hence $\mathcal{F}$ is a sigma field.
By definition, $P(\emptyset) = 0$ as $\emptyset$ is countable and $P(\Omega) = 1$ as $\Omega^c = \emptyset$ is countable. To show that if $A_n's$ are disjoint and $A_n's \in \mathcal{F}$ then $$P\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty P(A_n)$$ if all the $A_n's$ are countable then $P(A_n) = 0$ for all $n.$ As $\cup_{n=1}^\infty A_n$ is countable then $P(\cup A_n) = \sum P(A_n) = 0.$ Now suppose there exists $N_0$ such that $A_{N_0}$ is uncountable as $A_{N_0} \in \mathcal{F}$ then $(A_{N_0})^c$ is countable and $A_n \subseteq (A_{N_0})^c$ for all $n \ne N_0$ as $A_n's$ are pairwise disjoint then $P(A_n) = 0$ for all $n=\ne N_0$ and $P(A_{N_0}) = 1.$ Now $\cup_{n=1}^\infty A_n$ is uncountable implies $$P(\cup_{n=1}^\infty A_n) = 1 = P(A_{N_0}) = \sum_{n=1}^\infty P(A_n)$$ as $P(A_n) = 0$ for all $n\ne N_0$ and $P(A_{N_0}) = 1.$ This shows $(\Omega, \mathcal{F}, P)$ is a probability space as $P(\Omega) = 1.$
(Note that $\mathcal F$ is sometimes called the countable-cocountable $\sigma$-algebra.)
There are two things we must show - that $\mathcal F$ is a $\sigma$-algebra and that $\mathbb P$ is a probability measure. As for the first, clearly $\Omega^c=\varnothing$ is countable, so $\Omega\in\mathcal F$. If $E\in\mathcal F$ is countable, then $(E^c)^c=A$ is countable, so $E^c\in\mathcal F$. If $E\in\mathcal F$ such that $E^c$ is countable, then by similar reasoning $(E^c)^c$ is co-countable, and hence $E^c\in\mathcal F$. Now let $(E_n)_n$ be a sequence of sets in $\mathcal F$. If each $E_n$ is countable, then $E:=\bigcup_{n=1}^\infty E_n$ is as well, and hence $E\in\mathcal F$. If not, there exists a positive integer $m$ for which $E_m$ is co-countable. Since $$ E^c:=\left( \bigcup_{n=1}^\infty E_n\right)^c = \bigcap_{n=1}^\infty E_n^c \subset E_n^c, $$ it follows that $E$ is co-countable, and hence $E\in\mathcal F$.
For $\mathbb P$ to be a probability measure, we must show that it is a (positive) measure, and that $\mathbb P(\Omega)=1$. By definition, $\mathbb P(E)\in\{0,1\}$ for any $E\in\mathcal F$, and in particular $\varnothing$ is countable, so $\mathbb P(\varnothing)=0$. If $(E_n)_n$ is a sequence of pairwise disjoint sets in $\mathcal F$, then either each $E_n$ is countable, so that $$ \mathbb P\left(\bigcup_{n=1}^\infty E_n\right) = 0 = \sum_{n=1}^\infty \mathbb P(E_n), $$ or there exists a positive integer $m$ for which $E_m$ is co-countable. Then for each positive integer $n\ne m$, we have $E_n\subset E_m^c$, so that $E_n$ is countable and hence $\mathbb P(E_n)=0$. This implies that $\sum_{n=1}^\infty \mathbb P(E_n) = \mathbb P(E_m) = 1$. Again, as $E^c\subset E_m^c$, we see that $E$ is co-countable, so that $\mathbb P(E)=1$ - together, we have $$ \mathbb P\left(\bigcup_{n=1}^\infty E_n\right) = 1 = \sum_{n=1}^\infty \mathbb P(E_n). $$ (Note that these are the only two cases we need consider, as if $E$ and $F$ were co-countable subsets of $\Omega$, then they could not be disjoint. Feel free to prove this as an exercise.) Finally, since the complement of $\Omega=\mathbb R$ is the empty set which is clearly countable, we see that $\mathbb P(\Omega)=1$, as desired.