I completed the elementary part of @Joseph Van Name's answer:

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 $f[U] \mathrel{\delta_\mathcal{B}} f[V]$.

I completed the elementary part of @Joseph Van Name's answer:

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 $f[U] \mathrel{\delta_\mathcal{B}} f[V]$.

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 $f[U] \mathop{\delta}_\mathcal{B} f[V]$.

I completed the elementary part of @Joseph Van Name's answer:

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 $f[U] \mathrel{\delta_\mathcal{B}} f[V]$.