Let $\mathcal{E} = \{F \subset \mathbb{R} \ \text{such} \ \text{that} \ F \ \text{is} \ \text{finite} \}$. I am trying to find $\mathcal{M} (\mathcal{E})$, the $\sigma$-algebra generated by $\mathcal{E}$. After some experimentations I found the set $\mathcal{C} = \{C \subset \mathbb{R}: C \ \text{or} \ \mathbb{R}-C \ \text{is} \ \text{countable} \}$ is the set I am looking for.
Now I have managed to show that $\mathcal{C}$ is a $\sigma$-algebra, but I stuck on proving that $\mathcal{C} = \mathcal{M}(\mathcal{E})$. I appreciate idea and hint.
Best Answer
Proof.
Assume there is a smaller $\sigma$-algebra $\mathcal C'\subset \mathcal C$ which contains $\mathcal E$ but also lacks the countable set $C\in \mathcal C$. Because $C$ is countable there is an enumeration $\{c_n\}$ of all the elements of $C$. Because $\mathcal E$ contains all finite sets, $\{c_n\}$ belongs to $\mathcal C'$ for all $n$. But then $\mathcal C'$ must contain
$$\bigcup_n \{c_n\}=C.$$
Contradiction. Note that if some co-countable set $\bar C$ is missing from $\mathcal C'$, so is the countable set $C:=\Bbb R-\bar C$. $\square$