Let $(X,\mathscr X,\mathbb P)$ be a probability space, $(Y,\mathscr Y)$ a measurable space, and $h:X\times Y\to\mathbb R$ a real-valued function measurable with respect to the product $\sigma$-algebra $\mathscr X\otimes\mathscr Y$ (where $\mathbb R$ is endowed with the Borel $\sigma$-algebra).

Moreover, let $\mathscr F$ and $\mathscr G$ be two $\sigma$-subalgebras of $\mathscr X$ such that if $E$ is in one of $\mathscr F$ or $\mathscr G$, but not in the other, then $\mathbb P(E)$ is either $0$ or $1$.

Conjecture:For every function $f:X\to Y$ that is $\mathscr F/\mathscr Y$-measurable, there exists a function $g:X\to Y$ that is $\mathscr G/\mathscr Y$-measurable such that $$h(x,f(x))=h(x,g(x))\quad\text{$\mathbb P$-a.s.}$$

The idea is that the $\sigma$-algebras $\mathscr F$ and $\mathscr G$ “almost coincide,” so that a function measurable with respect to one of them can be “slightly” modified into a function measurable with respect to the other.

**Remark:** A more natural version of the conjecture would require that $f(x)=g(x)$ $\mathbb P$-a.s., but without further assumptions on $(Y,\mathscr Y)$, it may happen that the set $\{x\in X\,|\,f(x)=g(x)\}$ is not even $\mathscr X$-measurable. I believe this modified conjecture holds if, for instance, $Y$ is a separable and metrizable topological space and $\mathscr Y$ is the Borel $\sigma$-algebra on it. That said, I am curious about whether the more general conjecture holds without assuming any kind of topological structure on $(Y,\mathscr Y)$. This explains why I use the function $h$ to make sense of the idea that $f$ and $g$ “almost coincide.”

This conjecture strikes me as something that “should” be true, but I am concerned that a counterexample involving weird $\sigma$-algebras on very large sets may ruin it. I would be grateful for any thoughts, remarks, or references.