$\exists$ countably generated $\mathcal F$, s.t. $\sigma(\{ \{\omega \}: \omega\in\Omega \}) \subsetneqq \mathcal F \subsetneqq \mathcal B(\Omega)$?

Question

Does there exist a countably generated $\sigma$-field $\mathcal F$ on a second countable space $\Omega$ such that \begin{equation*} \sigma(\{ \{\omega \}: \omega\in\Omega \}) \subsetneqq \mathcal F \subsetneqq \mathcal B(\Omega)? \end{equation*}

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Here is the motivation, also some clues.

One the one hand, since $\Omega$ is second countable, its Borel $\sigma$-field $\mathcal B(\Omega)$ is clearly countably generated (by a countable topological base). But the $\sigma$-field generated by singletons $\sigma(\{ \{\omega \}: \omega\in\Omega \})$ is just the countable co-countable $\sigma$-field, which is not countably generated in general, say e.g., when $\Omega=\mathbb R$.

On the other hand, $\sigma(\{ \{\omega \}: \omega\in\Omega \})$ is countably generated if and only if $\Omega$ is itself a countable set. In this case, $\sigma(\{ \{\omega \}: \omega\in\Omega \})$ coincides with $\mathcal B(\Omega)$, and there is no such intermediate $\mathcal F$.

So does such intermediate $\mathcal F$ exist in some general cases? Or it definitely does not exist whenever the second countable space $\Omega$ is?

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Any comments or hints will be appreciated. TIA...

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EDIT: There should be some appropriate examples for the intermediate $\mathcal F$, as shown in the comment by @bof and the answer by @Henno Brandsma. But what happens if we force $\Omega$ to be a Polish space, which is stronger and more commonly used in measure theory than second countable space? I think then there will be no such intermediate $\mathcal F$, but I don't know how to prove it...

Answer 1

It is a theorem of Blackwell

Blackwell, David, On a class of probability spaces, Proc. 3rd Berkeley Sympos. Math. Statist. Probability 2, 1-6 (1956). ZBL0073.12301.

Let $(A,\mathscr A)$ be an analytic measurable space, and let $\mathscr A_0$ be a countably-generated sub-$\sigma$-algebra of $\mathscr A$. Then a subset of $A$ belongs to $\mathscr A_0$ if and only if it belongs to $\mathscr A$ and is the union of a family of atoms of $\mathscr A_0$.

A special case of an analytic measurable space is a Polish space with its Borel sets. If $\{x\} \in \mathcal A_0$ for all $x$, then the condition "is the union of a family of atoms of $\mathscr A_0$" holds for all sets. So in that case $\mathscr A_0 = \mathscr A$.

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This is Theorem 8.6.7 on page 291 of the text

Cohn, Donald L., Measure theory, Boston, Basel, Stuttgart: Birkhäuser (1980). ZBL0436.28001.

Answer 2

A candidate is taking $\Omega$ to be the irrationals (as a subspace of the reals) and as $\mathcal{F}$ the $\sigma$-algebra generated by the compact subsets of $\Omega$. This is clearly an in-between $\sigma$-field (finite sets are compact and compact sets are closed so Borel) but I think it might not be countably generated ($\Omega$ and its open subsets are all not $\sigma$-compact).. Throwing it out as my first impulse... Maybe I'll get better ideas later. Similar and probably not countably generated either: the Borel zero-sets (and their complements) for a given Borel measure, or the first category Borel sets (and their complements) in a Baire space (to avoid degeneration)..