# Union of increasing $\sigma$-algebras which is no $\sigma$-Algebra [duplicate]

Give an example for an increasing series of $\sigma$ algebras $$\mathcal{A}_1\subset\mathcal{A}_2\subset\ldots$$ so that $$\bigcup_{i=1}^{\infty}\mathcal{A}_i$$ is no $\sigma$-algebra.

Could you pls give me a hint how to find such an example?

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## marked as duplicate by Nate Eldredge, Prahlad Vaidyanathan, Daniel Fischer♦, Did, Dennis GulkoOct 22 '13 at 15:58

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## 1 Answer

Hint: If I remember correctly, almost any example works, as long as $\mathcal A_i \subsetneq \mathcal A_{i+1}$, to give a concrete example think on finite $\sigma$-algebras $\mathcal A_i$ on $\mathbb N$, such that $\bigcup \mathcal A_i$ contains all singletons.

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Would $F_n:=\left\{\left\{1\right\},\left\{2\right\},\ldots,\left\{n\right\}\right\}$, and then $\sigma(F_n)$ be an appropriate example? – math12 Oct 22 '13 at 15:36
Exactly this was the example i was thinking of. – martini Oct 22 '13 at 17:10