# A finitely generated $\sigma$-algebra is finite.

I am trying to solve Problem 3.5 (ii) of Schilling book on measure theory for which it must be proven that a finitely generated $$\sigma$$-algebra is finite. I can show it, given the following two statements, which I have problems proving.

1. Let $$\{B_1,...,B_N\}$$ be all the atoms of $$\sigma$$-algebra $$\mathcal A$$ of a set $$X$$ (where atom means non-empty set that contains no other non-empty set in $$\mathcal A$$). Then any set in $$\mathcal A$$ can be written as a (countable or finite) union of the $$B_j$$. In particular $$X = \bigcup B_j$$.

Idea: It is easy to show that the $$B_j$$ are disjoint. Otherwise $$\mathcal A \ni B_j \cap B_k \subset B_j$$ proper for some $$k \neq j$$. However I can not show $$X = \bigcup B_j$$. Clearly this is equivalent to $$C := \bigcap B_j^c = \emptyset$$. My idea was to assume $$C$$ non-empty. This implies that $$C$$ must contain a proper non-empty subset $$\mathcal A \ni C_1 \subset C$$, since otherwise $$C$$ would be another atom, contrary to hypothesis. The same argument can be applied to $$C_1$$ and so on, giving an infinite sequence of non-empty sets $$(C_j)_{j \in \mathbb N} \subset \mathcal A$$ such that $$C \supset C_1 \supset C_2 \supset ...$$ with all inclusions proper. Then $$\bigcap C_j \in \mathcal A$$ must be an atom? But we may have $$\bigcap C_j = \emptyset$$ (right?) so this does not appear to give the desired contradiction.

1. $$\sigma(\{ A_1,...,A_N \})$$ has at most $$2^N$$ atoms.

Idea: By induction on $$N$$. $$N=1$$ is obvious. Assume the statement is true $$N$$ and let $$\{ B_1, ..., B_M\}$$, $$M \leq 2^N$$, be the atoms of $$\sigma(\{ A_1,...,A_N \})$$. Then I presume that $$C_j := B_j \cap A_{N+1}$$ and $$D_j := B_j/C_j$$ are all the atoms of $$\sigma(\{ A_1,...,A_{N+1} \})$$ (eventually dropping some $$\emptyset$$'s) and since $$\# (\{ C_j \} \cup \{D_j \}) \leq 2^{N+1}$$ this proves the theorem. However I do not see how to show that $$\{ C_j \} \cup \{D_j \}$$ are indeed all the atoms of $$\sigma(\{ A_1,...,A_{N+1} \})$$.

This is probably very easy, but I am completely new to the subject. Also, I know that there exists solutions to this, but they do not explain the above steps.

• Why is $C_1$ an atom? Commented Feb 3, 2022 at 20:09
• I do not claim that it is an atom. I say that if it is not there must exist a $C_2$ such that $C_2 \subset C_1$ etc. Commented Feb 3, 2022 at 21:15
• You had the phrase "Then I presume that ..." which shows that for you all $C_i$ are atoms. You are not only sure these are all the atoms. Commented Feb 3, 2022 at 21:18
• Ah sorry. I was thinking you mean $C_1$ in statement 1. In fact I am not sure if the $C_j$ in statement 2. are atoms (so my formulation is kind of wrong). With presume I mean that it intuitively makes sense to me, but I fail to prove it, i.e. it might also be wrong. Commented Feb 3, 2022 at 21:23

Let $$A_1,\dots, A_n$$ be subsets of universal set $$X$$ and let $$\mathcal A$$ denote the algebra generated by these sets.

Now let $$\mathcal B$$ denote the collection of sets of form $$B_1\cap\cdots\cap B_n$$ where $$B_i\in\{ A_i, A_i^c\}$$ for $$i=1,\dots,n$$.

Evidently $$\mathcal B\subseteq\mathcal A$$ and the elements of $$\mathcal B$$ are mutually disjoint and cover $$X$$.

The collection contains at most $$2^n$$ elements (some might be empty).

Now let $$\mathcal C$$ be the collection characterized by: $$C\in\mathcal C\iff C\text{ is a union of elements of }\mathcal B$$ Here also the empty union is accepted so that $$\varnothing\in\mathcal C$$.

Then $$\mathcal C$$ is a finite collection (having at most $$2^{2^n}$$ elements) and again $$\mathcal C\subseteq \mathcal A$$. But also $$\mathcal C$$ is evidently an algebra.

So we are allowed to conclude that $$\mathcal A=\mathcal C$$ which means that $$\mathcal A$$ is a finite collection.

Finally every finite algebra is automatically a $$\sigma$$-algebra so that $$\mathcal A$$ can also be classified as the $$\sigma$$-algebra generated by $$A_1,\dots,A_n$$.

• Thanks for your answer. It is not obvious to me why $\mathcal C$ is an algebra (what I assume to mean that it is closed under arbitrary intersections, unions and complements?) and in fact, the smallest algebra containing $A_1,...,A_N$. Can you spell that out please? Commented Feb 3, 2022 at 21:44
• A collection $\mathcal A$ is an algebra if for every $U,V\in\mathcal A$ we also have $U^c\in\mathcal A$ and $U\cup V\in\mathcal A$. Do you agree with that? Commented Feb 4, 2022 at 7:17
• I get it now. Thanks Commented Feb 4, 2022 at 9:21

Let $$\Sigma$$ be the $$\sigma$$-algebra generated by sets of $$A_1,...,A_n$$. Then every element $$X\in\Sigma$$ is obtained by countably many complements, intersections and unions. The de Morgan laws imply that we can take all complements, add them to our collection: $$A_1,...,A_n, A_1',...,A_n'$$ and then take all unions and then all intersections of them. In every such countable union, only $$2n$$ terms are different, so there are only finitely many unions. Add them to the collection: $$A_1,...,A_n, A_1',...A_n', U_1,...,U_k$$. Now the number of countable intersections of sets from this collection is also finite. Hence $$\Sigma$$ is finite.

• Thanks for your answer. It is not obvious to me why $A_1,...,A_n,A_1',...,A_n',U_1,...,U_k$ and all countable intersections of them is a $\sigma$-algebra and in fact the smallest $\sigma$-algebra containing $A_1,...,A_n$. Can you spell that out please? Commented Feb 3, 2022 at 21:44
• That is De Morgan law(s): the union of intersections is the intersection of the unions, complement to the union is the intersection of the complements and so on. Commented Feb 3, 2022 at 22:18