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

Sets $U \subset X$ and $V \subset X$ in a measure space $(X, \mathcal{A})$ are called *$\mathcal{A}$-proximal*, written $U \mathop{\delta}_\mathcal{A} V$, 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
$U \mathop{\delta}_\mathcal{A} V \Rightarrow f(U) \mathop{\delta}_\mathcal{B} f(V)$ 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 $U \mathop{\delta}_\mathcal{A} V$ 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 $U \mathop{\delta}_\mathcal{A} V$, therefore $f[U] \mathop{\delta}_\mathcal{B} f[V]$.