Countable Sets, sigma-algebra and probability measure The problem I'm trying to solve is the following:
Consider Ω uncountable and F = {A ⊆ Ω: A is countable or $A^c$ is countable}.

*

*Show that F is a σ-algebra of Ω


*Consider P: F → [0,1]. If A is uncountable P(A) = 1 and if A is countable P(A) = 0. Is P a probability measure?
I think that for this problem I just have to prove that the conditions for a σ-algebra is valid for 1) and Kolmogorov axioms are valid for 2).
However, I'm having a lot of trouble with countable (and uncountable) sets properties. For example, if A is countable, $A^c$ may be countable or uncountable, right?
At the moment I have the following:
1)
i) F is non-empty. This is true, since A is in F
ii) F is Closed under complementation. I'm not sure on how to work A and $A^c$ in this
iii) F is Closed under countable unions.
2)
i) P>= 0. This is true, it can only be 0 or 1.
ii) P(Ω) = 1. Also true, do I need to prove this?
iii) P(Union of all the sets) = ∑ P(each set) if they are disjoint. I'm not sure on how to prove this one.
Sorry for all the questions. I'm not a mathematician and moving areas has been very hard for me.
 A: Hints for statement (ii) in Part 1:
If $A \in \mathcal{F}$ then $A$ is countable or $A^c$ is countable. If $A^c$ is countable, then $A^c \in \mathcal{F}$. If $A = (A^c)^c$ is countable, then $A^c \in \mathcal{F}$. Thus, in any case $A^c \in \mathcal{F}$.
Hints for statement (iii) in Part 1:
Any countable union of countable sets is necessarily countable (assuming the Axiom of Choice). If $\{A_n \}_{n \in \mathbb{N}}$ is any countable subcollection of $\mathcal{F}$, consider the case where there exists an uncountable $A_n$ separately from the case where all $A_n$ are countable.
Hints for statement (iii) in Part 2:
If $ \{ A_n \}_{n \in \mathbb{N}}$ is any countable collection of pairwise disjoint sets in $\mathcal{F}$, then it can contain at most one uncountable set (because if $A_n, A_m$ are two disjoint uncountable sets, then $A_n^c, A_m^c$ are countable, so $A_n^c \cup A_m^c$ is countable, and thus $A_n \cap A_m = (A_n^c \cup A_m^c)^c$ is uncountable and thus non-empty, which is a contradiction).
