Question about Kallenberg's proof of Doob-Dynkin Lemma

Question

The following is the proof for the Doob Dynkin Lemma from Kallenberg's Foundations of Modern Probability. In the theorem, $(S,\mathscr{S})$ is assumed to be Borel, i.e. Borel isomorphic to a Borel subset of $[0,1]$. Thus, it is natural to assume that $S \in \mathscr{B}([0,1])$. But how can we modify $h$, to further reduce to the case when $S = [0,1]$?

Answer 1

As you noticed, there exists a measurable function $\phi:S\to [0,1]$ s.t. $\phi^{-1}$ is Borel. So if you find measurable $h':T\to[0,1]$ s.t. $$ \phi \circ f=h'\circ g, $$ then $f=h\circ g$, where $h=\phi^{-1}\circ h'$ is the required mapping.

Answer 2

There are two types of Borel spaces (also called standard Borel spaces by some authors):

1. countable ones, for which the $\sigma$-algebra is necessarily discrete, and
2. uncountable ones.

If $S$ is countable, then the conclusion of Lemma 1.13 can be proved directly. If $S$ is uncountable, then by a standard result (*) about Borel spaces, for every uncountable Polish space (i.e., a separable and completely metrizable topological space) $X$, there is a Borel isomorphism $\phi : (S, \mathcal{S}) \to (X, \mathcal{B}(X))$. In particular, $X = [0, 1]$ is one such uncountable Polish space. Thus, without loss of generality one may assume $S = [0, 1]$ since they are Borel isomorphic anyway.

(*) See e.g. Theorem 7.1 in these notes.