2
$\begingroup$

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.

  • 0
    @A$s$af: Let me mention that the other page deal$s$ with a va$s$tly more difficult questio$n$.2012-07-16

2 Answers 2

1

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.

4

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.

  • 0
    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?2012-07-16