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Let $\Omega = \mathbb{N}$ be the natural numbers and $\mathcal{E}_n = \{\{1\},\{2\},\ldots,\{n\}\} \subset \Omega$. $\mathcal{A}_n = \sigma (\mathcal{E}_n)$ shall be the $\sigma$-algebra induced by $\mathcal{E}_n$. Clearly $\mathcal{A}_n \subset \mathcal{A}_{n+1}$ holds. I am told that $$\bigcup_n \mathcal{A}_n $$ is no $\sigma$-algebra. But I don't see why. Shouldn't this be the power set of $\mathbb{N}$? At least the $\sigma$-algebra induced by all one element subsets of $\mathbb{N}$ is the power set? I am confused.

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Just observe that the union is the set of finite subsets of $\mathbb{N}$. It contains no infinite subset. Even it does not contain $\Omega = \mathbb{N}$ itself! – Sangchul Lee Jul 16 '12 at 12:15
Every $\mathcal{A}_n$ must contain $\Omega$ itself, as that is a property of being a $\sigma$-algebra. However, there are other infinite sets that no $\mathcal{A}_n$ will contain. – Zev Chonoles Jul 16 '12 at 12:18
@Zev: thanks, I misread that $\mathcal A_n$ is a $\sigma$-algebra on $1,\dots,n$ – Ilya Jul 16 '12 at 12:22
Also related:… – Asaf Karagila Jul 16 '12 at 12:23
I think this can go on the list of "freshman dreams" (alongside with $(a+b)^n=a^n+b^n$). – Asaf Karagila Jul 16 '12 at 12:36
up vote 1 down vote accepted

Let $\mathcal A:=\bigcup_{n\geq 1}\mathcal A_n$. $\mathcal A$ contains the finite subsets of $\Bbb N$ and their complements. Indeed, if $A$ is finite then $A\subset \{1,\dots,n\}$ for some $n$ and $A\in\mathcal A_n$ and if $A^c$ is finite, $A^c\in \mathcal A_n$ hence $A\in\mathcal A_n$. Conversely, if $A\in\mathcal A$, $A\in\mathcal A_n$ for some $n$, $A$ has the form $A'\cup\{n+1,\dots,\}$ or $A'$, where $A'\subset \{1,\dots,n\}$. Indeed, the collection of elements of the form $$A\cup\{n+1,\dots\}\mbox{ and }A,$$ where $A\subset \{1,\dots,n\}$, is a $\sigma$-algebra which contains $\mathcal E_n$.

Hence $\mathcal A$ consists of the subsets of $\Bbb N$ which are finite or have a finite complement. As Zev Chenoles counter-example shows, it's not stable by countable unions.

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For example, the countable union $$\bigcup_{n\in\mathbb{N}}\{2n\}\notin\bigcup_{n}\mathcal{A}_n$$ even though each $\{2n\}\in \mathcal{A}_{N}$ for all $N\geq 2n$, certainly.

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Yep. +1. More generally, every infinite co-infinite subset. – Did Jul 16 '12 at 12:19
Can you give some more detail? I am still confused. Because for all $N \geq 2n$, it will be $\{2n\} \in \mathcal{E}_N$, therefore $\bigcup_{i=1}^n \{2i\} \in \mathcal{A}_N$, and $\bigcup_{i=1}^n \{2i\} \subset \bigcup_{i=1}^{2n} \mathcal{A}_i$. Does it matter that the one Union has to "grow" twice as fast in the limit? – Haatschii Jul 16 '12 at 12:44

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