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It is well-known that:

Given a set $X$ and a collection $\cal S$ of subsets of $X$, there exists a $\sigma$-algebra $\cal B$ containing $\cal S$, such that $\cal B$ is the smallest $\sigma$-algebra satisfying this condition.

Certain texts, Lieb and Loss, Analysis, for instance, state that the proof of this assertion requires transfinite induction. On the other hand, one can define $\mathcal B$ to be the intersection of all $\sigma$-algebras containing $\cal S$. Which statement is correct? Or, is there a hidden transfinite induction contained somewhere?

I must confess here that I have only vague ideaos of the rigorous set-theoretic foundations of mathematics.

Doubting Thomas
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  • http://math.stackexchange.com/questions/54172/the-sigma-algebra-of-subsets-of-x-generated-by-a-set-mathcala-is-the-s/ may be a thread of interest. – Asaf Karagila Jul 19 '13 at 07:32

2 Answers2

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The intersection of all $\sigma$-algebras that include $\mathcal S$ is a perfectly good way to get the smallest such $\sigma$-algebra, and the proof that it works requires no transfinite induction.

Andreas Blass
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  • Hello! Can you explain why this intersection of infinitely many sets exists and that it is indeed the smallest $\sigma$-algebra? – youthdoo Apr 12 '25 at 12:03
  • @youthdoo For any nonempty set $S$ of sets, the intersection $\bigcap S={x:(\forall a\in S),x\in a}$ exists by the axiom of separation, since it's a subset of any member of $S$. If the elements of $S$ are $\sigma$-algebras, then so is $\bigcap S$; each of the required closure properties for $\bigcap S$ follows immediately from the corresponding property of the members of $S$. For example, given countably many elements $x_n\in\bigcap S$, we have, for every $a\in S$, that $x_n\in a$ and therefore $\bigcup x_n\in a$ That is, $\bigcup x_n\in\bigcap S$. – Andreas Blass Apr 12 '25 at 19:30
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Lieb and Loss say no such thing, as far as I can tell. In fact on page 4, it reads

"Consider all the sigma-algebras that contain $\mathcal{F}$ and take their intersection, which we call $\sum$, i.e., a subset $A \subset \Omega$ is in $\sum$ if and only if $A$ is in every sigma-algebra containing $\mathcal{F}$. It is easy to check that $\sum$ is indeed a sigma-algebra. Indeed it is the smallest sigma-algebra containing $\mathcal{F}$..."

Ink
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  • It goes through transfinite induction somewhere shortly afterwards.. It's unfortunate, I don't have a physical copy with me and I can't get an e-copy. – Doubting Thomas Jul 19 '13 at 04:42
  • @DoubtingThomas It mentions transfinite induction in the context of passing an algebra to a $\sigma$-algebra. This is irrelevant to the existence of a minimal $\sigma$-algebra (Given an algebra $\mathcal{A}$, taking countable unions yields a collection $\mathcal{A_1}$, which is not closed under countable intersections. So taking countable intersections of $\mathcal{A_1}$ yields $A_2$..and so on. Proceeding like this you get a the minimal $\sigma$-algebra of $A$). – Ink Jul 19 '13 at 04:54
  • So, why does not the same intersection argument work in the case of passing from algebra to $\sigma$-algebra ? After all, $\cal A$ is just another set and there exists a minimal $\sigma$-algebra containing it. If that proof is transfinite-induction free, it should work here too; no? – Doubting Thomas Jul 19 '13 at 04:59
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    @DoubtingThomas It does work. The book was talking more about the process of building a minimal $\sigma$-algebra of an algebra. The existence part is clear, because you can just intersect all the $\sigma$-algebras containing $S$. – Ink Jul 19 '13 at 05:02
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    Ah, I see. Thanks a lot! The confusion is cleared. – Doubting Thomas Jul 19 '13 at 05:06