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Possible Duplicate:
Sigma algebra question
The union of a strictly increasing sequence of $\sigma$-algebras is not a $\sigma$-algebra

If $F_n$ is an increasing sequence of sigma fields then $F = \bigcup_{n=1}^\infty F_n$ is a field. Please help me find a counter-example to show that $F$ may not be a sigma-field.

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    @Arturo: Your answer there is a much better fit; I’m now voting to close. – 2012-02-11

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Let $\Omega$ be the set of all infinite sequences in which each term is $0$ or $1$. Let $F_n$ be $\lbrace A\subseteq\Omega : \forall\omega\in\Omega\ (\omega \in A\iff \text{some condition on the first $n$ terms of }\omega\text{ holds}) \rbrace. $ Let $F$ be the union. Let $\omega_k$ be the $k$th term in the seqence $\omega$. Then $ \begin{align} & \{ \omega\in\Omega : \omega_1 = 1 \} \in F_1 \subseteq F \\ & \{ \omega\in\Omega : \omega_1 = \omega_2 = 1 \} \in F_2 \subseteq F \\ & \{ \omega\in\Omega : \omega_1 = \omega_2= \omega_3 = 1 \} \in F_3 \subseteq F \\ & \cdots\cdots \end{align} $ But the intersection of these sets contains only the sequence in which every term is $1$, and that is not a member of $F$.

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    The set of all sequences whose first $n$ terms are $1$ is a member of $F_n$. This sequence is a member of that set. But the set whose _only_ member is this sequence is not a member of $F_n$, for reasons I explained above. Your statement beginning with "Therefore" simply has nothing to support it. Also, you shouldn't say "infinite ones" if you mean "infinitely many ones". – 2013-09-08
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Take $\Omega=[0,1)$ and $\mathcal F_n$ the $\sigma$-algebra generated by intervals of the form $\left[k2^{-n},(k+1)2^{-n}\right)$, $k\in\{0, \ldots,2^n-1\}$ . Then $(0,1)=\bigcup_{n\in\mathbb N}[2^{-(n+1)},2^{-n})$ and $[2^{-(n+1)},2^{-n})\in\mathcal F_{n+1} $ for all $n$, but $(0,1)\notin \mathcal F_n$ for all $n$.