# Wanted: example of an increasing sequence of $\sigma$-fields whose union is not a $\sigma$-field [duplicate]

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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## marked as duplicate by Asaf Karagila, Henning Makholm, Arturo Magidin, Brian M. Scott, Byron SchmulandFeb 12 '12 at 0:13

Duplicity of a previous question aside, you should really try and give meaningful titles (this is not the first bad title you give). "Searching for a Counter example" could mean anything. In fact if I wouldn't know I could have guessed the tag was wrong as well. Giving a title which hints on the topic of the question and on some occasion might help you find similar posts via the automatic suggestions when asking your question. – Asaf Karagila Feb 11 '12 at 23:02
It is clearly not a duplicate of the previous question. It’s a consequence, but a much easier result that can very usefully be answered independently. – Brian M. Scott Feb 11 '12 at 23:28
– Arturo Magidin Feb 11 '12 at 23:47
@AsafKaragila Thanks mate i shall keep your feedback in mind when thinking of future titles. – Comic Book Guy Feb 11 '12 at 23:48
@Arturo: Your answer there is a much better fit; I’m now voting to close. – Brian M. Scott Feb 11 '12 at 23:57

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 $F$.
@MichaelHardy Why is $\{\omega \in \Omega: \omega_k = 1, \quad \forall k \in \mathbb{N}\}$ not a member of $F$? – Elements Oct 28 '12 at 20:59
@Elements : $F$ is defined as the union of a certain family of sets. A set is a member of the union if and only if it is a member of at least one member of that family. For this set to be a member of $F_1\cup F_2\cup F_3\cup\cdots$, there would have to be some positive integer $n$ such that this set is a member of $F_n$. In order to be a member of $F_n$, it would have to be characterized by a condition on the first $n$ terms. – Michael Hardy Oct 28 '12 at 23:23
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 $[2^{-(n+1)},2^{-n})\in\mathcal F_n$ for all $n$, but $(0,1)\notin \mathcal F_n$ for all $n$.