Measure of $X$ an uncountable set. Let $X$ be an uncountable set and let $\mathcal{A}$ be the collection of subsets $A$ of $X$ s.t. either $A$ or $A^c$ is countable. Define $\mu(A) = 0$ if $A$ is countable and $\mu(A) = 1$ if $A$ is uncountable. Prove $\mu$ is a measure.
To prove $\mu$ is a measure I want to show 


*

*$\mu(\emptyset) = 0$

*$\mu(\cup_{i=1}^{\infty}{A_i}) = \sum_{i=1}^{\infty}{\mu(A_i)}$


Since $\emptyset$ is countable then $\mu(\emptyset) = 0$.
Considering $\mu(\cup_{i=1}^{\infty}{A_i}) \in \{0, 1\}$, then there cannot contain two sets $A_i$ and $A_j$ $i \neq j$ such that they are both uncountable. 
Is there some fact that two disjoint subsets of an uncountable set both cannot be uncountable?
 A: Hint: If $A^c$ and $B^c$ are countable then $A \cap B$ cannot be empty (because the uncountable set $A$ cannot be contained in the countable set $B^c$). So if $A_i$ is a pairwise disjoint family of members of $\cal A$, at most one of the $A_i$ is uncountable.
A: Hint: If $\{A_{i}\}_{i=1}^{\infty}$ is a collection of countable sets, then $\bigcup_{i=1}^{\infty}A_{i}$ is countable. 
A: 
Let $A, B \subseteq X$ be disjoint sets such that $A^c$ and $B^c$ are countable.
$$\emptyset = A\cap B \implies X = (A\cap B)^c = A^c\cup B^c$$
This implies that $X$ is countable as a union of two countable sets. Contradiction.

Let $(A_n)_{n=1}^\infty$ be a pairwise disjoint sequence of subsets of $X$.
If all $A_n$ are countable, then:
$$\mu\left(\underbrace{\bigcup_{n=1}^\infty A_n}_{\text{countable}}\right) = 0 = \sum_{n=1}^\infty \underbrace{\mu(A_n)}_{=0}$$
If $A_{i}^c$ is countable for some $i \in \mathbb{N}$, then the argument above implies that $A_n$ are countable for $n \ne i$:
$$\mu\left(\underbrace{\bigcup_{n=1}^\infty A_n}_{\text{uncountable}}\right) = 1 = \underbrace{\mu(A_i)}_{=1} + \sum_{n\in\mathbb{N}\setminus\{i\}}\underbrace{\mu(A_n)}_{=0}$$
Hence, $\mu$ is $\sigma$-additive.
A: If $\{A_n\}_{n\in\mathbb N}\subseteq{\cal A}$ are disjoint subsets of $X$, then $A_m\subseteq A_n^c$ for $m\neq n$.
So, if $A_k\in{\cal A}$ is uncountable for some $k\in\mathbb N$, it follows that $A_k^c$ is countable and therefore $A_j$, being a subset of $A_k^c$ ($j\neq k$), is countable.
Therefore, the family $\{A_n\}$ must have at most one uncountable subset of $X$.
