Sigma algebra question

If $\mathcal{F}_1 \subset \mathcal{F}_2 \subset \dotsb$ are sigma algebras, what is wrong with claiming that $\cup_i\mathcal{F}_i$ is a sigma algebra?

It seems closed under complement since for all $x$ in the union, $x$ has to belong to some $\mathcal{F}_i$, and so must its complement.

It seems closed under countable union, since for any countable unions of $x_i$ within it, each of the $x_i$ must be in some $\mathcal{F}_j$, and so we can stop the sequence at any point and take the highest $j$ and we know that all the $x_i$'s up to that point are in $\mathcal{F}_j$, and thus so must be their union. There must be some counterexample, but I don't see it.

-
I don't know a counterexample off hand, but a problem with your argument is that you have to do that "stopping" infinitely many times. If you take $x_i\in\mathcal{F}_i\setminus\mathcal{F}_{i-1}$, then there is no $i_0$ such that $x_i$ is in $\mathcal{F}_{i_0}$ for all $i$. –  Jonas Meyer Sep 21 '10 at 4:42
Heh. I assigned this as a homework problem last week. –  Nate Eldredge Sep 21 '10 at 17:40
Not to me :) But, even if I were one of your students, I don't think this question goes beyond what students would discuss between themselves. –  Neil G Sep 21 '10 at 19:54
I wasn't objecting, just amused. Anyway, my students handed in that problem a week earlier. –  Nate Eldredge Sep 29 '10 at 20:08
This will be true if your $i$'s go upto $\omega_1$. –  hot_queen Jan 9 at 1:47

The problem arises in the countable union; your argument is correct as far as it goes, but from the fact that $\cup_{i=1}^n x_i\in \cup_{i=1}^{\infty}F_i$ for each $n$ you cannot conclude that $\cup_{i=1}^{\infty} x_i$ lies in $\cup_{i=1}^{\infty} F_i$: the full union must be in one of the $F_j$ in order to be in $\cup_{i=1}^{\infty}F_i$.

For an explicit example, take $X=\mathbb{N}$; let $F_n$ be the sigma algebra that consists of all subsets of $\{1,\ldots,n\}$ and their complements in $X$. Now let $x_i=\{2i\}$. Then each $x_i$ is in $\cup F_i$, but the union does not lie in any of the $F_k$, hence does not lie in $\cup F_i$.

Added: In this example, $\cup_{i=1}^{\infty}F_n$ is the algebra of subsets of $X$ consisting of all subsets that are either finite or cofinite, so any infinite subset with infinite complement will not lie in the union, and such a set can always be expressed as a countable union of elements of $\cup F_i$.

-

Let $\Omega=[0,1]$, $A_{0}= \{\emptyset, \Omega \}$ and $A_{k}=\sigma \{[0,\frac{1}{2^k}],[\frac{1}{2^k},\frac{2}{2^k}],[\frac{2}{2^k},\frac{3}{2^k}],.....,[\frac{2^k-1}{2^k},1]\}$

pick irrational number $x\in(0,1)$ and sequence $s_{1},s_{2},...$ converging to $x$ from the left (binary representation allows to find such sequence from $A_{i}$'s). Then $(x,1]=\cap_{i=1}^{\infty}(s_{i},1] \in \cup_{i=0}^{\infty} A_{i}$. Then $x \in\cup_{i=0}^{\infty} A_{i}$ But for all fixed k $A_{k}$ contains only rational numbers and intervals.

-

If $x_1\in F_1$, $x_2\in F_2$, and so forth, then the infinite union $x_1 \cup x_2 \cup \cdots$ does not lie in any one $F_j$, and therefore does not lie in the union of the $F_j$'s.

-
You mean to say also that $x_2\notin F_1$, $x_3\notin F_2$, etc., presumably, and this doesn't prove that the result, but only shows the problem with the argument. –  Jonas Meyer Sep 21 '10 at 4:44
The union of the increasing sequence of $sigma$-algebras is only an algebra, unless there is some $n$ so that $F_n$ fails to increase after $n$. –  ncmathsadist Jul 12 '11 at 1:33
By continuity, the set of $\mathcal{F}_{i}$'s converge to the set $\cup_{i=1}^{\infty}\mathcal{F}_i$. The limiting set is a $\sigma$-algebra by definition of the problem. What is invalid about this argument?
What kind of "continuity" do you believe applies here? And what "definition of the problem" do you believe asserts that the union is a $\sigma$-algebra? –  Arturo Magidin Oct 7 '10 at 3:51
Yes! I meant to post it as a comment/question. So as $n \to \infty$, doesn't $\mathcal{F}_n$ converge to the infinite union? –  Naga Oct 7 '10 at 4:41