Prove that $\mathcal{A}$ is an $\sigma$-algebra.

Question

Let $S$ be a set and let $\mathcal{A} = \{A\subseteq S: \text{A is countable or $A^c$ is countable}\}$.

Exercise: Prove that $\mathcal{A}$ is an $\sigma$-algebra.

I know that $\mathcal{A}$ is called an $\sigma$-algebra if the following properties hold:

i) $\emptyset, S \in \mathcal{A}$.

ii) $A\in \mathcal{A} \Rightarrow A^c \in \mathcal{A}$.

iii) $A_1, A_2, ..., \in \mathcal{A} \Rightarrow \bigcup\limits_{i = 1}^{\infty}A_i \in \mathcal{A}.$

My approach:

i) Quite straightforward: if $A = \emptyset$ then $A \in \mathcal{A}$, because $\emptyset$ is countable, so $\emptyset \in \mathcal{A}$. If $A = S$ we have that $A^c = S\backslash S = \emptyset$ is countable so we know that $S\in \mathcal{A}$.

ii) Pick an arbitrary set $A \in \mathcal{A}$. We know that $A$ is countable or $A^c$ is countable. Assume that $A$ is countable. $A^c = S\backslash A$ is in $\mathcal{A}$, because $(S\backslash A)^c = A$ is countable. Now assume that $A^c$ is countable. $A \in \mathcal{A}$ because $A^c$ is countable.

iii) I'm not sure what to do here. I feel like the solution might be rather obvious because we know that one of the $A_i = S$. Hence $\bigcup\limits_{i = 1}^{\infty}A_i = S \in \mathcal{A}$.

Question: How do I solve part iii)?

Answer 1

If all $A_n$ are countable, then $\bigcup_{n=1}^\infty A_n$ is also countable as a countable union of countable sets. So, $\bigcup_{n=1}^\infty A_n \in \mathcal{A}$ since it is countable.

If at least one $A_{n_0}$ is such that $A_{n_0}^c$ is countable, then:

$$\left(\bigcup_{n=1}^\infty A_n\right)^c = \bigcap_{n=1}^\infty A_n^c \subseteq A_{n_0}^c$$

So, $\left(\bigcup_{n=1}^\infty A_n\right)^c$ is countable as a subset of a coutable set. Hence, $\bigcup_{n=1}^\infty A_n \in \mathcal{A}$ since its complement is countable.

Answer 2

All $A_i$'s are countable then countable union of countable set is countable. If suppose $A_m$ is not countable then its complement will be countable. Now complement of union of given sets is intersection of complement of those sets. So complement of ${A_m}$ is countable and the complement of union of $A_i$'s will be subset of complement of ${A_m}$ which will be countable hence it belings to sigma algebra.

Answer 3

You don't know that one of the $A_i =S.$ For example, $S=\mathbb{R}$ and $A_i =\{ i \} .$

You have to use the knowledge that all the $A_i$:s are countable, or otherwise one of them is not.