I completed the elementary part of @[Joseph Van Name](https://mathoverflow.net/users/22277/joseph-van-name)'s [answer](https://mathoverflow.net/a/453530): 

Sets $U \subset X$ and $V \subset X$ in a measure space $(X, \mathcal{A})$ are called *$\mathcal{A}$-proximal*, written $U \mathrel{\delta_\mathcal{A}} V$, if there is no $A \in \mathcal{A}$ such that $U \subset A$ and $V  \subset A^c$. 

A map $f\colon X \to Y$ between measure spaces $(X, \mathcal{A})$ and  $(Y, \mathcal{B})$ is a called *proximal map* if
$U \mathrel{\delta_\mathcal{A}} V \Rightarrow f[U] \mathrel{\delta_\mathcal{B}} f[V]$ for all $U, V \subset X$.


*Statement:*
* Then $f$ is $\mathcal{A}$–$\mathcal{B}$-measurable if and only if $f$ is a proximal map.

*Proof:*

"If."
Take $B \in \mathcal{B}$. From rules for images and preimages, $f[f^{-1}[B]] \subset B$, $f[(f^{-1}[B])^c] = f[f^{-1}[B^c]] \subset B^c$.
To avoid contradiction with proximality of $f$, there is $A \in \mathcal{A}$ such that 
$f^{-1}[B] \subset A$ as well as  $(f^{-1}[B])^c \subset A^c$ or equivalently $f^{-1}[B]\supset A$. Thus  $f^{-1}[B] \in \mathcal{A}$.


"Only if." Let $f$ be $\mathcal{A}$–$\mathcal{B}$ measurable and take $U, V \subset X$ with $U \mathrel{\delta_\mathcal{A}} V$ and assume that there is $B\in \mathcal{B}$ such that $f[U] \subset B$, $f[V] \subset B^c$. By measurability of $f$, $ f^{-1}[B] \in \mathcal{A}$.   
This time 
$ f^{-1}[B]  \supset f^{-1}[f[U]] \supset U$ and
$ f^{-1}[B]  \supset f^{-1}[f[V]] \supset V$. This contradicts $U \mathrel{\delta_\mathcal{A}} V$, therefore $f[U] \mathrel{\delta_\mathcal{B}} f[V]$.