TITLE: Existence of minimal $\sigma$-algebra and transfinite induction
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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 $\mathcal{F}$..."