Prove projection of a measurable set from product space is measurable

[Lagrange fanboy]

TL;DR Summary

Given a set from a product sigma algebra, how do I prove that it's projection is measurable?

I was reading page 33 of https://staff.fnwi.uva.nl/p.j.c.spreij/onderwijs/TI/mtpTI.pdf when I saw this claim:
Given measurable spaces $(\Omega_1,\Sigma_1), (\Omega_2,\Sigma_2)$ and the product space $(\Omega_1\times \Omega_2, \Sigma)$ where $\Sigma$ is the product sigma algebra, the embedding $e_b:\Omega_1\rightarrow \Omega_1\times\Omega_2, x\mapsto (x,b)$ is measurable.
After some reduction, I notice that $e^{-1}_b(S)=\emptyset$ if $S\not = \{(x,b)|x\in A\}$ for some $A\subseteq \Omega_1$, otherwise $e^{-1}_b(S)=\pi_1(S)$, the projection of $S$ onto the first component. Here I'm stuck. I can't show that if $S\in \Sigma$ then $\pi_1(S)\in \Sigma_1$. I thought about using structural induction, taking advantage of the fact that $\Sigma$ is the smallest sigma algebra generated, but I think for structural induction to work I need to have a well-founded/well-ordered relation, which I cannot find.

 

[wrobel]

I think you need a lemma from p.289
S Lang

 

[mathwonk]

Just an ignorant suggestion without doing the work: a measurable product set is presumably a union of rectangles with both sides measurable, and projection of a rectangle is a side. so projection of a measurable set is apparently a union of measurable sets, hence measurable. does something like this work?

 

[Hawkeye18]

You need the following simple fact/lemma:

Let a function $f:X\to Y$ be fixed, and let $\Sigma$ be a sigma-algebra on $X$, and $\mathcal C$ be some collection of subsets in $Y$. Assume that for any $A\in\mathcal C$ the inverse image $f^{-1}(A)$ belongs to $\Sigma$. Then $f^{-1}(B)\in\Sigma$ for all $B$ in $\Sigma(A)$, where $\Sigma(A)$ is the sigma-algebra generated by $\mathcal C$.

One of the corollaries of the above lemma is the fact that if for a real-valued function $f$ the sets $f^{-1}([a, \infty))$ are measurable for all $a\in\mathbb R$, then $f^{-1}(B)$ is measurable for any Borel set $B\subset\mathbb R$. (This is also true for rays of form $(a, \infty)$, $(-\infty, a]$, $(-\infty, a)$).

The proof of the lemma is quite elementary: consider the collection $\mathcal B$ of all $B\subset Y$ such that $f^{-1}(B)\in\Sigma$. It is easy to show, using the fact that $\Sigma$ is a sigma-algebra, that $\mathcal B$ is also a sigma-algebra. Therefore $\Sigma(\mathcal C) \subset\mathcal B$, which proves the statement.

Returning to the original question, it is easy to see that if $A\in\Sigma_1$, $B\in\Sigma_2$, then $\pi_b^{-1}(A\times B)$ is measurable (it is $A$ if $b\in B$, and $\varnothing$ othervise). But such sets $A\times B$ generate $\Sigma$, so the embedding $\pi_b$ is indeed measurable.

Finally, projections of measurable sets are not always measurable. For example, a coordinate projection of a Borel measurable set in $\mathbb R^2$ is not necessarily Borel. It is a so-called Souslin (a.k.a. analytic) set, but generally it is not Borel.