I'm currently reading some stuff on the existence of a measurable inverse, or measurable choice theorems from Bogachev's Measure Theory book, Volume II, where I'm trying to connect these two theorems on measurable choice:
The first one is (see P.34): see below, the third screenshot for the definition of Souslin sets, that're also called the Analytic Sets
and the second one is (see P. 268):
I understand that:
Definition (Souslin set): A Souslin set is a continuous image of a complete, separable (= having a countable dense subset) metric space that's also Hausdorff (see. P. 19 in the same book)
Now given the definition, I think it's clear that Borel sigma algebra is strictly contained in the signma algebra generated by analytic/Souslin sets, but not vice versa. But then I wonder why, on P.268 of the book above, the author writes: "Theorem 9.1.3 is an immediate corollary of Theorem 6.9.1"? It's clear that a selection or choice or right inverse of $F$ is measurable w.r.t. to the sigma algebra of analytic sets, so inverse images by $F$ are analytic, but why must they be Borel?
The resulting function will, in general, not be Borel measurable, and Bogachev doesn't write that it is. Theorem 9.1.3. is about the completion (!) of (finite) Borel measures on $Y$. That is, if you have a measure defined on all Borel sets and look at the $\sigma$-algebra on $Y$ generated by all Borel sets and subsets of Borel sets of measure zero, then $g$ is measurable with respect to the resulting $\sigma$-algebra. That this works follows from Theorem 7.4.1.