Doob-Dynkin factorization through the sigma-algebra generated by a function

If $g=h\circ f$ for Borel $h$, then for each Borel $C\subseteq \bar{\mathbb{R}}$, $g^{-1}(C)=f^{-1}(h^{-1}(C))\in \sigma (f)$ by [L1].

Conversely suppose $g$ is $\sigma (f)$-measurable. For every rational $q$, [L1] and [L2] say that the family of Borel $A\subseteq \mathbb{R}$ with $f^{-1}(A)=\{g\le q\}$ is nonempty. By [L3], $\mathbb{Q}$ is countable; apply the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ to choose one $A_q$ for every $q$. Define $B_q={\bigcap }_{r\in \mathbb{Q},r>q}{A}_r$. These sets are Borel and increasing in $q$, and $f^{-1}(B_q)={\bigcap }_{r>q}\{g\le r\}=\{g\le q\}$, including when $g$ takes an infinite value.

The family also satisfies $B_q={\bigcap }_{r\in \mathbb{Q},r>q}{B}_r$. Indeed the right side expands to the intersection of all $A_s$ with rational $s>r>q$; every rational $s>q$ has a rational $r$ strictly between $q$ and $s$, and every $s>r>q$ also has $s>q$. Thus the two collections of $A_s$ have the same intersection.

For $y\in \mathbb{R}$ put $h(y)=\inf \{q\in \mathbb{Q}:y\in B_q\}$ in $\bar{\mathbb{R}}$, with empty infimum $+\mathrm{\infty }$. For rational $q$, monotonicity and step 2.1 give $\{h\le q\}={\bigcap }_{r\in \mathbb{Q},r>q}{B}_r=B_q$: if the infimum is at most $q$, some index below each $r>q$ belongs to the defining set, and conversely membership in every $B_r$ forces the infimum at most $q$. The threshold criterion [L2] therefore makes $h$ Borel measurable.

For every $x\in X$ and rational $q$, steps 1.2 and 3.1 give $h(f(x))\le q$ exactly when $g(x)\le q$. Rational thresholds distinguish all points of $\bar{\mathbb{R}}$, including both infinities, hence $h(f(x))=g(x)$. Together with step 1.1 this proves the equivalence.