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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.

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I believe the answer is yes. First of all, either $\mathscr F=\mathscr G$ or both $\sigma$-algebras consist of sets of measure $0$ or $1$. Indeed, suppose that a set $E$ is in $\mathscr F$ but not in $\mathscr G$ and $\mathbb P(E)=1$ (otherwise, take $E^c$). For every set $A\in \mathscr F$, either $A\cap E$ or $A^c\cap E$ is not in $\mathscr G$, so $\mathbb P(A)$ is either $0$ or $1$. If $\mathscr G\not\subseteq \mathscr F$ then, by a similar argument, $\mathbb P(A)$ is either $0$ or $1$ for all $A\in \mathscr G$.

Let $\Pr:=\mathbb P\circ (x\to (x,f(x)))^{-1}$. Since $$\Pr(x\in A, y\in B)=\mathbb P(x\in A, f(x)\in B)=\mathbb P(x\in A)\mathbb P(f(x)\in B),$$ because $f^{-1}(B)\in \mathscr F$ has measure $0$ or $1$, $\Pr=\mathbb P\times \mu$ is a product measure on $\mathscr X \times \mathscr Y$ and, moreover, $\mu(B)=0$ or $1$ for all $B\in\mathscr Y$.

Given a set $A\in \mathscr X\times \mathscr Y$, let $A_x=\{y : (x,y)\in A\}$. Since $\mu(A_x)\in \{0,1\}$, the rectangle $r(A):=\{x: \mu(A_x)=1\}\times Y$ differs from $A$ by a set of measure zero. Therefore, any simple function on the product space is $\Pr$-almost surely equal to a function of $x$ and, as a result, the same is true for all measurable functions, i.e. $h(x,y)=\bar h(x)$ a.s.. This implies that, for some $y_0$, $h(x,y) = h(x,y_0)$ a.s. and we can take $g(x)\equiv y_0$.

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  • $\begingroup$ Thank you, @D_809, the first part of your answer is crystal clear! I am not sure, though, that the existence of such a set $B\in\mathscr Y$ can be guaranteed if $\mathscr Y$ is not countably generated. Let $(X,\mathscr X,\mathbb P)$ be $[0,1]$ with the Borel $\sigma$-algebra and the Lebesgue measure, $(Y,\mathscr Y)$ be $[0,1]$ with the $\sigma$-algebra of countable and co-countable sets, and $\mathscr F=\mathscr Y$. If $f$ is the identity, then your condition states that there exists a minimal co-countable subset (and so a maximal countable subset) of $[0,1]$, which is clearly not the case. $\endgroup$ – triple_sec Dec 21 '19 at 18:44
  • $\begingroup$ Yes, you are right. Let me think some more. $\endgroup$ – D_809 Dec 21 '19 at 22:59
  • $\begingroup$ I made a second attempt. Let me know if this makes sense. $\endgroup$ – D_809 Dec 22 '19 at 0:23
  • $\begingroup$ This is awesome, much appreciated! $\endgroup$ – triple_sec Dec 22 '19 at 1:35

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