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 Answer 1


This follows from two standard facts both of which can be found in Halmos' book.

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 $P(A\Delta B)<\epsilon$.

2) The algebra generated by a countable collection of sets is countable.

  • $\begingroup$ Halmos actually proved the second for ring, will the proof work for algebra? $\endgroup$
    – The R
    Commented Apr 8, 2019 at 3:04
  • 1
    $\begingroup$ @TheR Yes the arguments in Halmos' book can be used for algebras too. $\endgroup$ Commented Apr 8, 2019 at 5:17

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