2

$\mathcal{F_{1}}$ is some sub-algebra and $\mathcal{F_{n+1}}$ is class of all sets that can be represented as a countable union or intersection of sets $\mathcal{F_{n}}$

Prove that $\bigcup_{n \in \mathbb{N}} \mathcal{F_{n}}$ is not necessarily $\sigma$-algebra

I thought that this example can be reduced to some solved exercise like in this question but its definitely other task

So now I have no idea what to do with that. I've tried to take $\mathcal{F_{i}}$ like interval (a;b] with rational ends but it also doesn't work. May be here can be some counter-example?

Ryuk
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  • Suggestion: take ${\cal F}_1$ to be the $\sigma$-field over $\Bbb{N}$ comprising the finite and cofinite sets. – Rob Arthan May 02 '24 at 20:38
  • @RobArthan, then $\mathcal F_2$ is already $\mathcal P(\Bbb N)$, isn't it? (which is a sigma-algebra). – Izaak van Dongen May 02 '24 at 20:42
  • @IzaakvanDongen: yes, you're right. My suggestion isn't helpful. Would you be kind enough to have a look to the answer I added to the question https://math.stackexchange.com/questions/4756598/prove-that-the-union-of-increasing-sigma-field-is-not-a-sigma-field that Ryuk links to and check whether I was having a brainstorm about that too. – Rob Arthan May 02 '24 at 20:46
  • The typical example here is the Borel $\sigma$-algebra, with an $\mathcal F_1$ like the one you suggest. This fact is a nontrivial result from descriptive set theory, though. I'm not sure if there's a simpler example. – Izaak van Dongen May 03 '24 at 12:31
  • @IzaakvanDongen, I've just found very similar question https://math.stackexchange.com/questions/3024183/why-it-mustnt-be-a-sigma-algebra but I cant understand why F must contain all Borel sets. May be you can understand and explain it? Because right now I dont see that this prove is right – Ryuk May 03 '24 at 19:50
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    That part is correct. Do you know the definition of the Borel $\sigma$-algebra? Do you know how you can show that another $\sigma$-algebra contains the Borel algebra? (There is a slight mistake in the definition of $\mathcal F_1$ in the post (it's not an algebra), but that is not a critical error). – Izaak van Dongen May 03 '24 at 20:00
  • @IzaakvanDongen, but in this example it looks like I show only that it's not Borel $\sigma$-algebra. But non-Borel $\sigma$-algebra can still be $\sigma$-algebra, isn't it? – Ryuk May 03 '24 at 20:04
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    The point is: "there is some Borel set $A$ which does not lie in any $\mathcal F_n$. If $\bigcup_n \mathcal F_n$ was a $\sigma$-algebra, it would contain all the Borel sets, including $A$. But that's absurd". Please take some time to check the relevant definitions and reflect on this argument if you're still not getting it right away. – Izaak van Dongen May 03 '24 at 20:19
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    @IzaakvanDongen, ok, I think I understood it, thank you so much! – Ryuk May 03 '24 at 20:22

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