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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 $B\in\sigma(A_B)$.

Augustin
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A.sarabi
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1 Answers1

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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.

Daniel Fischer
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martini
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    Why is $\mathcal{A} \subset \mathcal{A}'$ ? – no lemon no melon Sep 25 '23 at 06:31
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    @nolemonnomelon for any set $B\in\mathcal{A}$, there exists a countable subset of $\mathcal{A}$ containing $B$. Then, $B$ naturally also belongs to the sigma field generated by the countable subset. Hence, $B\in\mathcal{A}'$ (The good set can contain sets not in $\mathcal{A}$). – reyna Feb 18 '24 at 17:53
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    @zaira when you say "there exists a countable subset of $\mathcal{A}$ containing $B$", I guess an obvious set would just be ${B}$? and $B \in \sigma\left(\mathcal{\left{B\right}}\right)$ XD, This was probably just an obvious fact that was too obvious and I brushed it off hastily, hence why I got confused! Thnx! – no lemon no melon Feb 19 '24 at 07:33