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Let $A_n$ be the set of all the subsets of $\{1,2,...,n\}$. $\mathcal{F}_n$ is the corresponding sigma algebra generated from $A_n$. We then define a set $B=\{2,4,6,... \}$. In the textbook, it states that $B \notin \cup_{n=1}^{\infty} \mathcal{F}_n$.

I don't understand the above statement. I think that $B$ should belong to $\cup_{n=1}^{\infty} \mathcal{F}_n$ when $n \rightarrow \infty$. I know it may be easy for you, but I am only a beginner.

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    $\begingroup$ If $A_n$ is the powerset of $\{1,\ldots,n\}$, then $\mathcal{F}_n=A_n$, unless I've misunderstood something. Moreover, if $B\in \cup_n \mathcal{F}_n$, then there exists an $n$ such that $B \in \mathcal{F}_n$, which leads to a contradiction. $\endgroup$
    – parsiad
    Nov 29 '16 at 14:08
  • $\begingroup$ @parsiad : Thank you for a quick reply. Yes. $A_n$ is the power set. But when $n = \infty$, is $B \in \mathcal{F}_n$?I think that I am misunderstanding something. $\endgroup$ Nov 29 '16 at 14:14
  • $\begingroup$ I think your confusion might come from being unfamiliar with the definition of $\cup_{n=1}^\infty X_n$. In particular, $x\in \cup_{n=1}^\infty X_n$ if and only if $x \in X_n$ for some $n$. $\endgroup$
    – parsiad
    Nov 29 '16 at 14:17
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    $\begingroup$ When you say "$\mathcal{F}_n$ as $n\rightarrow \infty$", this statement has no meaning. Make sure to use the definition of infinite union above instead. $\endgroup$
    – parsiad
    Nov 29 '16 at 14:20
  • $\begingroup$ @parsiad ok! It is clear now. Thank you! $\endgroup$ Nov 29 '16 at 14:32
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As pointed out by parsiad, the statement "$B\in\bigcup_{n=1}^{\infty} \mathcal{F}_n$ as $n\to \infty$" has no meaning; but $B\in\bigcup_{n=1}^{\infty} \mathcal{F}_n$ has.

It is true that $B$ belongs to the $\sigma$-algebra generated by $\bigcup_{n=1}^{\infty} \mathcal{F}_n$ because $\{2j\}$ belongs to $\sigma\left( \bigcup_{n=1}^{\infty} \mathcal{F}_n\right)$ and so will do $\bigcup_{j\in \mathbf N} \{2j\}$, which is $B$.

However, for any positive integer $n$, $B$ does not belong to $\mathcal F_n$. Indeed, the elements of $\mathcal F_n$ have the following form: these one are either subset of $\left\{1,\dots, n\right\}$ or can be written as $A \cup \left(\mathbb N\setminus\left\{1,\dots,n\right\}\right)$, where $A$ is a subset of $\left\{1,\dots, n\right\}$. The set $B$ is not of this form.

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