I was looking for a projection measurability theorem with the least restrictions and as self-contained as possible, without the need to use capacity theory, for example.
I thought I found what I was looking for in Theorem 5.72 in the book "Research Topics in Analysis, Volume I" by Hu and Papageorgiou.
Theorem 5.72: If $(\Omega, \Sigma)$ is a complete measurable space, $Y$ is a Souslin space, and $A \in \Sigma \otimes B(Y)$, then $\pi_{\Omega}(A) \in \Sigma$.
But I ended up finding a theorem in a paper that is even more general, which left me a bit confused with some counterexamples. The paper can be found on arxiv and published here. It's theorem 5.1 in arxiv and 5.6 in the paper.
Let $(M_1,\mathcal{A}_1)$ and $(M_2,\mathcal{A}_2)$ be two measurable spaces. For any $G \in \mathcal{A}_1 \otimes \mathcal{A}_2$, its projection $\pi_{M_1}(G)$ belongs to $\hat{\mathcal{A}}_1$.
Where $\hat{\mathcal{A}}_1$ it is the universal completion of $\mathcal{A}_1$.
Typically when we think about the projection theorem, we want to see some example of when the projection is not measurable. We can find some examples here on the MSE. E1, E2 and E3.
When we deal with these theorems we want to know if the projection of a Borelian is still Borelian, so we have to be careful when we apply the theorem, because the theorem discusses the product of $\sigma$-algebras, not the borel $\sigma$-algebra of the product, but this $\sigma$-algebras are equal when it comes to second-countable space. Which I believe covers some of the examples I presented.
So, I'm wondering if I'm interpreting the theorem wrong, applying it wrongly in these counterexamples, or if there's a problem with the proof.
