# Does a sub-$\sigma$-algebra of a countably generated $\sigma$-algebra have to be countably generated?

Assume that $A \subseteq B$ are $\sigma-$algebras and $B$ is a countably generated (separable) $\sigma-$algebra. Now my question:

Is it possible that $A$ is not countably generated?

I'm testing many examples but I can't find any example.

## 2 Answers

Yes, it is possible. For instance, let $B$ be the Borel $\sigma$-algebra on $\mathbb{R}$ and let $A\subset B$ be the $\sigma$-algebra consisting of all sets which are either countable or cocountable. Then $B$ is countably generated (by the intervals with rational endpoints, say), but $A$ is not. Indeed, if you have countably many elements $\{X_n\}$ of $A$, you may assume they are all countable subsets of $\mathbb{R}$ (if not, replace them by their complements), and then the $\sigma$-algebra they generate is contained in the algebra of sets $Y$ such that either $Y$ or $\mathbb{R}\setminus Y$ is contained in the countable set $X=\bigcup X_n$.

Hint: Consider $\mathcal{B} := \mathcal{B}(\mathbb{R})$, the Borel-$\sigma$-algebra on $\mathbb{R}$, and

$$\mathcal{A} := \{A \subseteq \mathbb{R}; \text{A or A^c is countable}\},$$

the countable-cocountable $\sigma$-algebra.