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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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
@AsafKaragila Thanks mate i shall keep your feedback in mind when thinking of future titles. – Hardy 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

2 Answers 2

up vote 2 down vote accepted

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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Actually if it is not too much to ask i would really love u provided reasoning behind various choices you made in constructing your brilliant counter example. I would genuinely great full to be honest i do n't care so much for what the final answer ended up being this is not for school assignment, i want to learn how to think like you and build my own. Your step by step reasoning such as why did we choose omega to be infinite sequences of 0 or 1 and in the sigma algebra const "if some condition on the first n terms of ω holds". Your help would be very much appreciated. – Hardy Feb 11 '12 at 22:15
I was just trying to find what seemed like the simplest possible example. The first sigma-algebra just splits the space in half; the second splits those halves in half; the third splits all four of those parts in half, and so on. – Michael Hardy Feb 12 '12 at 0:38
@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
But how is the sequence {1,1,1,...} NOT in every F_n? It satisfies the conditions for each F_n that the first n elements are equal to 1. – Clark Kent Sep 8 '13 at 2:32

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 $n$.

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