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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.

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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 $X=\bigcup X_n$.

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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.

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