Suppose $\mathcal{A}$ is a $\sigma$-algebra generated by $\mathcal{E}$. Then $\mathcal{A}$ is the union of the $\sigma$-algebras generated by $\mathcal{F}$ where $\mathcal{F}$ ranges over all countable subsets of $\mathcal{E}$

I am kind of confused when the propostion says "ranges over all countable ... "

Am I asked to show that

$$ \mathcal{A} = \bigcup \{ \mathcal{M} : \mathcal{M} = \sigma( \mathcal{F}), \mathcal{F} \; \; \text{countable subset of $\mathcal{E}$} \} \text{ ??}$$

  • 1
    $\begingroup$ Yes, you have correctly interpreted the question. $\endgroup$ – Nate Eldredge Jul 23 '15 at 5:51
  • $\begingroup$ But what is the proof? $\endgroup$ – Ali Jul 23 '15 at 6:09

To show the claim, define $$\mathscr{G} = \{B \in \mathscr{A}: B \in \mathscr{F}, \mathscr{F} \text{ is a countable subclass of } \mathscr{E}\}.$$

It is not hard to check that $\mathscr{G}$ is a $\sigma$-algebra that contains $\mathscr{E}$. Therefore $\mathscr{A} = \sigma({\mathscr{E}}) \subset \mathscr{G} \subset \mathscr{A}$, hence the result follows.


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