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$X=(-\infty,\infty)$, $\mathcal{F}_n$ is the $\sigma$-field generated by $[0,1),[1,2),...,[n-1,n)$. Prove $\mathcal{F}_n\subset \mathcal{F}_{n+1}$ and $\cup_{n=1}^\infty\mathcal{F}_n$ is not a $\sigma$-field.

My solution:

$\mathcal{F}_{n+1}=\mathcal{F}_{n}\cup[n,n+1)\Rightarrow \mathcal{F}_n\subset \mathcal{F}_{n+1}$.

Proof by contradiction. If $\cup_{n=1}^\infty\mathcal{F}_n=[0,\infty)$ is a $\sigma$-field, then $\emptyset\in\cup_{n=1}^\infty\mathcal{F}_n \Rightarrow X=(-\infty,\infty)\subset\cup_{n=1}^\infty\mathcal{F}_n=[0,\infty)$, which is a contradiction. So, $\cup_{n=1}^\infty\mathcal{F}_n$ is not a $\sigma$-field.

I wonder if my proof is right or not. Thanks in advance.

The above is wrong. Let me try again.

$\mathcal{F}_{n}=\sigma([0,1),...,[n-1,n)),\mathcal{F}_{n+1}=\sigma([0,1),...,[n-1,n),[n,n+1))$.

Proof by contradiction. If $\mathcal{F}_n\supset \mathcal{F}_{n+1}\Rightarrow \mathcal{F}_n\supset \mathcal{F}_{n+1}\supset\{[0,1),...,[n-1,n),[n,n+1)\}\supset\{[0,1),...,[n-1,n)\}\Rightarrow \mathcal{F}_{n}$ is not the smallest $\sigma$-field containing $\{[0,1),...,[n-1,n)\}$, which is a contradiction.

Proof by contradiction. If $\cup_{n=1}^\infty\mathcal{F}_n$ is a $\sigma$-field, then $\cup_{n=1}^\infty\mathcal{F}_n=\cup_{n=1}^\infty\sigma([0,1),...,[n-1,n))\supset\cup_{n=1}^\infty\{[0,1),...,[n-1,n)\}=[0,\infty)$, i.e., $\cup_{n=1}^\infty\mathcal{F}_n$ is the smallest $\sigma$-field containing $[0,\infty)$. Since $\cup_{n=1}^\infty\mathcal{F}_n$ is a $\sigma$-field, $\emptyset\in\cup_{n=1}^\infty\mathcal{F}_n \Rightarrow X\in\cup_{n=1}^\infty\mathcal{F}_n\Rightarrow X=(-\infty,\infty)$ is also the smallest $\sigma$-field containing $[0,\infty)$, which is not true. So, $\cup_{n=1}^\infty\mathcal{F}_n$ is not a $\sigma$-field.

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It is not true that $\mathcal{F}_{n+1}=\mathcal{F}_{n}\cup[n,n+1)$.What you want is that relation on the generating sets implies that $\mathcal{F}_n\subseteq \mathcal{F}_{n+1}$. To add more, it seems as if you are looking at the problem wrong. The $\mathcal{F}_n$ are not subsets of $\mathbb{R}$ but sets of subsets of $\mathbb{R}$. Therefore $X$ cannot be a subset of $\bigcup_{n=1}^\infty$ as that union is a set of subsets of $\mathbb{B}$ and not a subset of $\mathbb{R}$. Therefore your proof is not right. I have not yet thought about how to prove it though. –  user45150 Dec 7 '12 at 6:13
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Note that $\mathcal F_n$ is a subset of $2^X$ hence $\mathcal F_n$ contains no real number hence $\mathcal F_{n}\cup[n,n+1)$ has nothing to do with anything. –  Did Dec 7 '12 at 6:15
    
Thank you, user45150 and did. Let me keep thinking. –  Sam Dec 7 '12 at 6:19
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I think you are taking mistake, because if $F_{n}$ be $\sigma$-field is generated with those intervals in $X$ so they should have complements of those intervals in $X$ too, so for examples we can make $(-n,n)$ on $F_{n}$ so also $\mathbb{R}$ is in $\cup_{n=1}^{\infty}F_{n}$ too and your reason for non-$\sigma$-field is wrong! –  AmirHosein SadeghiManesh Dec 7 '12 at 6:50
    
In my above comment I should correct that $F_{n}$ contains it's element's complement and also $\emptyset$ and $\mathbb{R}$ , but not $(-n,n)$, I'm sorry, but look below proofs. and be careful about using nomads and symbols, $(-\infty,\infty)$ is not $\sigma$-field itself , a $\sigma$-field is a subset of powerset of something with some conditions! –  AmirHosein SadeghiManesh Dec 8 '12 at 3:31
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2 Answers

up vote 4 down vote accepted

The following are hints for two different ways of showing that $\mathcal{F} = \bigcup_{n=1}^\infty \mathcal{F}_n$ is not a $\sigma$-field. In each, I will denote by $\mathcal{G}$ the $\sigma$-field generated by $\mathcal{F}$.

Hint 1: Note that $| \mathcal{F}_n | = 2^{n+1}$ for $n \geq 1$, and therefore $| \mathcal{F} | = \aleph_0$. Using the fact that $\{ [ n-1 , n ) : n \geq 1 \} \subseteq \mathcal{F} \subseteq \mathcal{G}$, show that $| \mathcal{G} | \geq 2^{\aleph_0}$.

Hint 2: Given $n \geq 1$, note that each $A \in \mathcal{F}_n$ has the property that either $[ n , + \infty ) \subseteq A$ or $A \cap [ n , + \infty ) = \emptyset$. Find a set in $\mathcal{G}$ for which this property fails for all $n \geq 1$ (and thus cannot belong to any $\mathcal{F}_n$).

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@AmirHoseinSadeghiManesh: A few points. (1) Regarding the first line of your comment: what does $[n,n-1)$ mean? (This would usually be interpreted as $\varnothing$.) (2) Writing out the generating set for $\mathcal{F}_n$ as $[0,1) , [1,2) , \ldots , [n-1,n)$ (as was done in the question), note that the first one has right endpoint $1$, the second has right endpoint $2$, ... the $n$th one has right endpoint $n$, and this is the last one, meaning that there are $n$ elements of this set. (cont...) –  Arthur Fischer Dec 7 '12 at 15:27
    
(...inued) (3) I never claimed that $| \mathcal{F}_n | = n+1$; I claimed that $| \mathcal{F}_n | = 2^{n+1}$. (4) While I appreciate being told the definition of a $\sigma$-field, I must admit I cannot find a single spot where I misused this concept in my answer. Please enlighten me. –  Arthur Fischer Dec 7 '12 at 15:27
    
Thank @Arthur for your helpful hints. I am trying to follow hint 2 but I still cannot find such a set in $\mathcal{G}$ s.t. it fails that property. –  Sam Dec 7 '12 at 16:51
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@Sam: (re: first comment) There is a bit of a connection between my first hint and my second. In particular, the first hint gives you a subset of the $\sigma$-field $\mathcal{G}$ which you should be able to use to construct the set required. –  Arthur Fischer Dec 7 '12 at 18:43
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@Sam: (re: second comment) The fact that $\bigcup_{n=1}^N \mathcal{F}_n$ is a $\sigma$-field relies quite strongly on the fact that $\bigcup_{n=1}^N \mathcal{F}_n = \mathcal{F}_N$, which we know is a $\sigma$-field. When we move to the infinite union we no longer have a known $\sigma$-field which equals this union. (It is also very rare that the union of even two arbitrary $\sigma$-fields is a $\sigma$-field.) –  Arthur Fischer Dec 7 '12 at 19:13
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I'm sorry for my previous mistake and thank Mr. Arthur for his clever notation. Because every $F_{n}$ is a $\sigma$-field so $\forall n\in\mathbb{N}\; :\;\emptyset,\mathbb{R}\in F_{n}$ so it is brightly that $\emptyset,\mathbb{R}\in F:=\cup_{n=1}^{\infty}F_{n}$. First condition is firmed.

Let $A\in F=\cup_{n=1}^{\infty}F_{n}$ so there is one of elements of the union like $F_{m}$ for a number $m\in\mathbb{N}$ such that $A\in F_{m}$, as $F_{m}$ is a $\sigma$-field so $A^{c}\in F_{m}$ and thus $A^{c}\in F=\cup_{n=1}^{\infty}F_{n}$. Hence second condition is correct too.

The main problem is in the Third condition. If $\{A_{i}\}_{i=1}^{m}\subset F=\cup_{n=1}^{\infty}F_{n}$ for a natural number $m$, then it means $\forall i\in\{1,2,...,m\}\; :\; \exists n_{i}\in\mathbb{N}\; such\; that\; A_{i}\in F_{n_{i}}$ and as $\forall n\in\mathbb{N}\; :\; F_{n}\subset F_{n+1}$ so we can conclude $\forall i\in\{1,2,...,m\}\; :\; A_{i}\in F_{max\{n_{1},n_{2},...,n_{m}\}}$, let $n_{0}:=max\{n_{1},n_{2},...,n_{m}\}$, because $F_{n_{0}}$ is a $\sigma$-field so $\cup_{i=1}^{m}A_{i}\in F_{n_{0}}$ and thus $\cup_{i=1}^{m}A_{i}\in F=\cup_{n=1}^{\infty}F_{n}$. So $F$ is a field but it is not a $\sigma$-field because if $\{A_{i}\}_{i=1}^{\infty}\subset F=\cup_{n=1}^{\infty}F_{n}$ we can not do same work, as here there is not any guaranty that there will be one contains all $A_{i}$.

Get every $A_{i}$ as $[0,i)$, it will be in $F_{i}$ because $[0,i)=\cup_{k=1}^{i-1}[k-1,k)$ and we know that $[0,1),[1,2),...,[i,i-1)$ are in $F_{i}$. now $\cup_{i=1}^{\infty}A_{i}=\cup_{i=1}^{\infty}[0,i)=[0,\infty)$ but $[0,\infty)$ is not belong to any $F_{n}$, so it is not in $F=\cup_{n=1}^{\infty}F_{n}$ that don't let $F$ be a $\sigma$-field.

Now why $[0,\infty)$ does not belong to any $F_{n}$? Because let $n$ be constant natural number, then every subset of $\mathbb{R}$ that is in $F_{n}$ should contains $(-\infty,0)\cup[n,\infty)$ or be contained in $[0,n)$.

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Hello, AmirHosein, please check my updated version. –  Sam Dec 7 '12 at 7:23
    
You are right. The proof is not rigorous. I am trying to follow @Arthur's hints. –  Sam Dec 7 '12 at 15:43
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Note that if $A \in \mathcal{F} = \bigcup_{n=1}^\infty \mathcal{F}_n$ it does not mean that $A = \bigcup_{n=1}^\infty A_n$ where $A_n \in \mathcal{F}_n$ for all $n$. It only means that there is an $n$ such that $A \in \mathcal{F}_n$. –  Arthur Fischer Dec 7 '12 at 17:26
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@AmirHosein I like your rigorous proof. Thank you very much. –  Sam Dec 9 '12 at 18:01
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@AmirHoseinSadeghiManesh: +1 for the attempt. –  B. S. Dec 15 '12 at 14:30
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