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Find examples where $\mathcal{F}_1 \subseteq \mathcal{F}_2 \subseteq \cdots$ are increasing $\sigma$-fields but $\bigcup_{n = 1} ^\infty \mathcal{F}_n$ is not. I have constructed an example that I believe is correct, but having trouble to show this is actually the right counter-example:

Define $\mathcal{F}_n = \sigma(\{ [0, i]: i \leq n \})$ on $\mathbb{R}$. Then I believe $\bigcup_{i = 1} ^\infty [0, i] = [0, \infty) \not\in \bigcup_{n = 1} ^\infty \mathcal{F}_n$.

However, I am having trouble to show $[0, \infty) \not\in \bigcup_{n= 1} ^\infty \mathcal{F}_n$. Any suggestions?

Partial T
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  • My earlier comment was based on the wrong assumption that your universal set was $[0,\infty)$, instead of $\mathbb R$. I have deleted the comment and posted an answer. – Kavi Rama Murthy Aug 22 '23 at 04:59

2 Answers2

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$\{1\},\{2\},..,\{n-1\}, [0,1],(1,2], (2,3],...,(n-1,n],\mathbb R \setminus [0,n]$ is a partition of the real line, and it is easy to check that $\mathcal F_n$ is also the $\sigma-$ algebra generated by this partition. This implies that every set in $\mathcal F_n$ is a union of of some sets from this collection. If $[0,\infty)$ belongs to $\mathcal F_n$ it follows that $[0,\infty)$ must be a union of sets on which is necessarily $\mathbb R \setminus [0,n]$ (since $[0,\infty)$ is unbounded). But then $(-\infty,0)\subseteq \mathbb R \setminus [0,n] \subseteq [0,\infty)$ which is a contradiction.

Remark: The $\sigma-$ algebra generated by any countable partition $A_1,A_2,...$ is the collection of all possible unions of the sets $A_i$.

Kavi Rama Murthy
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Your question isn't very clear about what the underlying sets of the ${\cal F}_n$ are. If the underlying set isn't required to be fixed, you could take $\cal{F}_n = \Bbb{P}\{1, \ldots, n\}$. Then each ${\cal F}_n$ is a $\sigma$-field. But $\cal F = \bigcup_n{\cal F}_n$ is not a $\sigma$-field over the positive integers: the empty set is in $\cal F$, but its complement is not; $\cal F$ is also not closed under countable unions.

Rob Arthan
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  • Why the unexplained down-vote? If my idea is wrong, then I'd love to have that explained. – Rob Arthan May 02 '24 at 20:43
  • Hello again! I wasn't the downvoter. There is some ambiguity in this question but I think the natural way to interpret it is "find a set $X$ and sigma-algebras $\mathcal F_1 \subseteq \dotsb$ on $X$ such that their union is not a sigma-algebra", since usually "sigma-algebra" implicitly means "sigma-algebra on some fixed set we're working in". This seems consistent with OP's attempt. With this interpretation, this answer wouldn't work. This might be the cause of the downvote (or it might just be random noise)... I think it's not a silly answer though and OP could be a bit clearer. – Izaak van Dongen May 02 '24 at 20:55
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    @IzaakvanDongen: thanks. I've edited my answer to point out the ambiguity. Maybe I've done too much work in algebra: for me terms like "group", "ring", algebra and, hence, in this case, $\sigma$-algebra don't imply a fixed universe. – Rob Arthan May 02 '24 at 21:16