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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.

share|cite|improve this question… may be a thread of interest. – Asaf Karagila Jul 19 '13 at 7:32
up vote 2 down vote accepted

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.

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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 $\mathcal{F}$..."

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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 4:42
@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 4:54
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 4:59
@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 5:02
Ah, I see. Thanks a lot! The confusion is cleared. – Doubting Thomas Jul 19 '13 at 5:06

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