Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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

5 Answers 5

up vote 12 down vote accepted

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 $\cup F_i$.

share|improve this answer

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.

share|improve this answer

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".

share|improve this answer

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.

share|improve this answer
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?

share|improve this answer
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
The fact that a countable nested union of sigma algebras need not be a sigma algebra is nicely demonstrated in Arturo Magidin's answer on this page. I don't see anything mathematically precise in your answer, and I wouldn't call it an argument. –  Jonas Meyer Oct 7 '10 at 3:55
@Jonas: given the reputation of the user, I suspect (s)he would have posted it as a comment rather than an answer had (s)he been able to. Of course, it would have been better to post it as a question (with suitable references). –  Arturo Magidin Oct 7 '10 at 4:06
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
Naga, The problem is that your question doesn't make sense unless you define "converge" in this context. If convergence is defined in such a way that an increasing sequence of sets converges to the union, then convergence will not preserve the property of being a sigma algebra, as explained in Arturo's answer above. –  Jonas Meyer Oct 7 '10 at 4:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.