# Show that for a countably generated $\sigma$-algebra, there exists a countable dense collection of subsets.

Countably generated means it is a $$\sigma$$-algebra generated by a countable collection of sets, say $$\{A_{n}\}$$. I want to show that for a countably generated $$\sigma$$-algebra $$\mathscr{F}$$, there exists a countable collection $$\{F_{n}\}\subset \mathscr{F}$$ s.t. $$\forall \varepsilon>0, \forall B\in\mathscr{F}$$, $$\exists F_{m}\in\{F_{n}\}$$ s.t. $$P\left(F_{m} \Delta B\right)<\varepsilon$$. $$P$$ is a probability measure but I think this works for any measure space?

My idea is to show that if there is no such countable dense collection, then it contradicts the fact that $$\mathscr{F}$$ is the smallest $$\sigma$$-algebra generated by a countable collection, but I am really not sure about it.

1). If an algebra $$\mathcal A$$ generates a sigma algebra $$\mathcal F$$ and $$P$$ is a finite measure on $$\mathcal F$$ then, for any $$\epsilon>0$$ and any $$A\in \mathcal F$$, there exists $$B \in \mathcal A$$ such that $$P(A\Delta B)<\epsilon$$.