Wanted: example of an increasing sequence of $\sigma$-fields whose union is not a $\sigma$-field 
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. 
 A: 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 $n$. 
A: 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 $F$.
