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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.

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    $\begingroup$ 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 $i$. $\endgroup$ – Jonas Meyer Sep 21 '10 at 4:42
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    $\begingroup$ Heh. I assigned this as a homework problem last week. $\endgroup$ – Nate Eldredge Sep 21 '10 at 17:40
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    $\begingroup$ 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. $\endgroup$ – Neil G Sep 21 '10 at 19:54
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    $\begingroup$ I wasn't objecting, just amused. Anyway, my students handed in that problem a week earlier. $\endgroup$ – Nate Eldredge Sep 29 '10 at 20:08
  • $\begingroup$ This will be true if your $i$'s go upto $\omega_1$. $\endgroup$ – hot_queen Jan 9 '14 at 1:47
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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 $\cup F_i$.

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  • $\begingroup$ With $F_{2n}$ we can have the subset $\{2,4,\dots,2n \} \in F_{2n} = \cup_{j=1}^{n} x_i$, what is wrong? $\endgroup$ – Olba12 Nov 2 '16 at 20:01
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    $\begingroup$ @Olba12 That's a union of finitely many $x_i$. The point is the union of all the $x_i$, that is, the set of all even positive integers. That lies in no $F_k$, and hence not in $\bigcup F_k$. $\endgroup$ – Daniel Fischer Dec 14 '16 at 14:31
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Something more drastic is true: If $\langle \mathcal{F}_n: n \geq 1\rangle$ is a strictly increasing sequence of sigma algebras over some set $X$ then $ \bigcup_{n \geq 1} \mathcal{F_n}$ is not a sigma algebra. As a corollary, there is no countably infinite sigma algebra. See, for example, "A comment on unions of sigma fields, A. Broughton, B. Huff, American mathematical monthly, 1977 Vol. 84 No. 7, pp 553-54".

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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.

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