$\sigma$-algebra produced by a subclass of a class. im studying the book 'probability & measure' by Patrick Billingsley. in chapter 2 there's an exercise 2.9 say's: show that: If $B\in\sigma(A)$, then there exists a countable subclass $A_B$ of $A$ such that $B\in\sigma(A_B)$.

This is - as so often - done by the good sets principle. Let $$ \def\A{\mathscr A} \A' := \left\\{B\in \sigma(\A) : \exists \A_B \subseteq \A, \A_B \text{ countable }, B \in \sigma(\A_B) \right\\} $$ Now we show that $\A'$ is a $\sigma$-algebra: $\emptyset\in \A'$, as $\emptyset \in \sigma(\emptyset)$ and $\emptyset$ is countable. If $B\in \A'$, then there is $\A_B$ with $B \in \sigma(\A_B)$, hence $B^c \in \sigma(\A_B)$, so $B^c \in \A'$. If $B_i \in \A'$ for $i \in \mathbf N$, then choose $\A_i$ countable with $B_i \in \sigma(\A_i)$, let $\A_B := \bigcup_{i} \A_i$. Then $\A_B$ is a countable subset of $\A$ and $B_i \in \sigma(\A_B)$ for all $i$. Hence $\bigcup_i B_i \in \sigma(\A_B)$.

As $\A'$ contains $\A$, we conclude that $\sigma(\A) \subseteq \A'$ and are done.