# Show that $P$ is a probability measure with a hint.

Let $$\Omega=\mathbb{R},$$ $$$$\mathcal{F} = \{A \subset \mathbb{R}: \text{either } A \text{ or } A^c \text{ is countable}\}$$$$ where $$P(A)=0$$ if $$A$$ is countable and $$P(A)=1$$ if $$A$$ is uncountable.

Show that $$P$$ is a probability measure. [Hint: You need to show that if $$\{A_n\}$$ is any countable collection of pairwise disjoint sets in $$\mathcal{F}$$ then it can contain at most one uncountable set.]

Here's my attempt (I have already proved $$\mathcal{F}$$ is $$\sigma-$$algebra):

(i) Since $$\Omega = \mathbb{R},$$ we've got $$P(\Omega)=1.$$

(ii) Suppose $$A_1,A_2,\ldots \in \mathcal{F}$$ where $$A_i \cap A_j = \varnothing \, \forall i \neq j.$$

For (ii), I divided for 2 cases;

(a) If $$A_1,A_2,\ldots$$ are all countable, that is, $$\cup_{n=1}^{\infty} A_n$$ is countable. So, $$P(\cup_{n=1}^{\infty} A_n) = 0 = \sum_{n=1}^{\infty} 0 = \sum_{n=1}^{\infty} P(A_n).$$

(b) If there exists $$m \in \mathbb{N}$$ such that $$A_m$$ is uncountable. Since $$A_m \in \mathcal{F}$$, it means $$A_m^c$$ is countable. Because $$A_j \cap A_m = \varnothing$$, so $$A_j \subset A_m^c$$ where $$j \neq m$$. So, $$A_j$$ is countable $$\forall j \neq m.$$

Since $$A_m$$ is uncountable and $$A_m \subseteq \cup_{n=1}^{\infty}A_n,$$ that is, $$\cup_{n=1}^{\infty}A_n$$ is uncountable.

Thus, $$P(\cup_{n=1}^{\infty} A_n) = 1 = P(A_m) + \sum_{j=1}^{\infty} P(A_j) = \sum_{j=1}^{\infty} P(A_j).$$

But I have no clue about "contain at most one uncountable set". How could I show in that way. Thank you in advance.

This should be an application of De Morgan's laws. Towards a contradiction, suppose that there are $$i,j \in \mathbb{N}$$ with $$i \neq j$$ such that $$A_i^c$$ and $$A_j^c$$ are countable. As $$A_i \cap A_j = \emptyset$$, we have \begin{align*} \mathbb{R} = \emptyset^c = (A_i \cap A_j)^c =A_i^c \cup A_j^c, \end{align*} showing that $$\mathbb{R}$$ is countable. As this a contradiction, $$A_m$$ is the only set within $$(A_i)_{i \in \mathbb{N}}$$ having a countable complement. Hence, \begin{align*} \sum_{i = 1}^\infty P(A_i) = P(A_m) = 1. \end{align*}