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This question arrives from A. N. Shiryaev's "Probability" Page $138$ Problem $1$.

Let $\Omega:=[0,1]$, $\mathcal{A}$ be the system of sets each of which is a finite union of disjoint sets of one of the forms: $(a,b)$, $[a,b)$, $(a,b]$, $[a,b]$ with $a,b\in\Omega$. By checking that intervals of these forms from $\mathcal{A}$ are closed under complements and finite unions, we can extend the closure to any sets from $\mathcal{A}$. Hence, $\mathcal{A}$ is an algebra.

My question is:

Is $\mathcal{A}$ a $\sigma$-algebra?

Well, the construction of $\mathcal{A}$ does not guarantee the closure under countable unions, and I cannot think of a proof. But, I could not find a counterexample, either.

Normally, for example in this post https://math.stackexchange.com/questions/233702/example-of-an-algebra-which-is-not-a-σ-algebra, if $\mathcal{A}$ only contains finite disjoint unions of the following intervals $[0,a], (a,b],(b,1]$, then we can use the countable union of $(0,1-\frac{1}{n}]$ to get $(0,1)\notin\mathcal{A}$. But this counterexample does not work for our $\mathcal{A}$ since it contains intervals of the form $(a,b)$.

Any idea? Thank you!

JacobsonRadical
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1 Answers1

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No, $\mathcal A$ is not a $\sigma$-algebra.

It is easy to see that any set in $\mathcal A$ either consists of finitely many points, or it has nonempty interior. So $A=\mathbb Q\cap[0,1]$ is not in $\mathcal A$, while $A=\bigcup_n[q_n,q_n]$ if $\{q_n\}$ is an enumeration of $A$.

Alternatively, Cantor's Ternary set is a countable intersection of sets in $\mathcal A$, but it is not a finite union of intervals.

Martin Argerami
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  • Great answer. Thank you – JacobsonRadical Apr 18 '23 at 22:16
  • A follow-up question inspired by your example. If now I take $\Omega:=\mathbb{Q}\cap [0,1]$, and $\mathcal{A}$ consisting of all the finite disjoint union of sets of the forms $[a,b]\cap\mathbb{Q},(a,b]\cap\mathbb{Q},[a,b)\cap\mathbb{Q}$ and $(a,b)\cap\mathbb{Q}$. Clearly $\mathcal{A}$ is an algebra by a similar proof. In this case, is $\mathcal{A}$ even a $\sigma$-algebra? – JacobsonRadical Apr 19 '23 at 21:50
  • Taking $a=0,b=1$, isn’t $\Omega=[a,b]\cap \mathbb{Q}$? I am sorry for not making it clear. – JacobsonRadical Apr 19 '23 at 23:19
  • You are right, my bad. In this case every singleton is in $\mathcal A$, and $\Omega$ is countable, so $\mathcal A=\mathcal P(\Omega)$. – Martin Argerami Apr 19 '23 at 23:23
  • Great. Thank you very much! – JacobsonRadical Apr 19 '23 at 23:25