I don't have much experience with measure spaces, so I am asking for verification of some proofs. Thank you.
Let $(X, \Sigma)$ be a standard Borel space. Consider the product space $(Z, \Sigma') = (X, \Sigma) \times (X, \Sigma)$, where $\Sigma' = \sigma\{ B_1 \times B_2 : B_1, B_2 \in \Sigma\}$.
I want to prove the following:
- The diagonal $\Delta = \{(x,x) : x \in X\}$ is measurable in $(Z, \Sigma')$.
- Let $\pi: Z \to X$ be the projection $(x,y) \mapsto x$. The restriction $\pi|_\Delta$ is an isomorphism between $(\Delta, \Sigma'|_\Delta)$ and $(X,\Sigma)$.
As I understood, the first point is stated here: the product $\sigma$-algebra of $(X, \Sigma) \times (X, \Sigma)$ is the same as the Borel $\sigma$-algebra of the underlying Polish spaces. Since $\Delta$ is closed in the product, it is also measurable.
As for the second point, let $E \in \Sigma$, then $E \times X \in \Sigma'$ and $\pi_\Delta^{-1}(E) = (E \times X) \cap \Delta \in \Sigma'|_\Delta$. Therefore, $\Sigma \subseteq \pi_\Delta(\Sigma')$. Having that $\pi_\Delta(\Sigma')$ is a measurable bijection between standard Borel spaces, the inverse mapping is also measurable by Suslin's theorem.
Further notes:
- The first point is not true in general.
- The first part of the proof of the second point should be always true once we have the measurability of $\Delta$, right?
- The second part really uses the special property of standard Borel spaces, is there any milder sufficient condition?
