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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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Comic Book Guy
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2 Answers2

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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 $n$.

Davide Giraudo
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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 $F$.

Brian M. Scott
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Michael Hardy
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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. – Comic Book Guy Feb 11 '12 at 22:15
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    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 00:38
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    @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
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    @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. – Kashif Sep 08 '13 at 02:32
  • @Glassjawed : My answer states that that sequence is in every $F_n$, i.e. I said that point is in their intersection. Then I said that the set containing only that one point---the sequence of $1$s---is not a member of $F$. The reason it's not a member of $F$ is that it's not a member of any $F_n$. – Michael Hardy Sep 08 '13 at 03:39
  • How is this set not a member of F? I thought that the definition of A being a subset of B was that all elements of A are in B. Wouldn't the sets F_n where the first n terms are fixed at 1 include SUBSETS where EVERY element is 1? I just don't see the difference between subset inclusion and element inclusion. – Kashif Sep 08 '13 at 03:49
  • To see why it's not a member of $F$, read the definition of $F$. It a union of other sets. A set is a member of the union if and only if it's a member of at least one of those sets. Which one could it be a member of? – Michael Hardy Sep 08 '13 at 04:13
  • Why would it not be a member of every F_n? Generally could the element of the set be in every F_n but the set itself not be a subset? – Kashif Sep 08 '13 at 10:28
  • It's not in any $F_n$. Membership in $F_n$ is decided by the first $n$ components. There's no condition you can impose on the first $n$ components that will force all of the components beyond that to be equal to $1$. – Michael Hardy Sep 08 '13 at 15:41
  • Do you need to force all the components beyond the nth to equal 1 though? I thought you could take the condition that the first n components are 1, use that as our subset of $F_n$, and then say that the sequence of infinite 1's is a subset of this. Then our set is a subset of $F_n.$ Where am I going wrong here? – Kashif Sep 08 '13 at 16:42
  • I understand your explanation--I just wonder where my logic is wrong. – Kashif Sep 08 '13 at 18:29
  • I'm not sure what your reasoning is. – Michael Hardy Sep 08 '13 at 20:15
  • I'm saying that the sequence of infinite ones is an element of the set of sequences with the first $n$ components equal to 1, which is a subset of $F_n$. Therefore, the set consisting solely of the sequence of infinite ones is a subset of $F_n$. I just don't understand what's wrong with that argument because that just follows from the statement that $A\subseteq B$ iff $x\in A \rightarrow x\in B$ for all $x\in A.$ – Kashif Sep 08 '13 at 21:01
  • 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". – Michael Hardy Sep 08 '13 at 21:35