# Almost identical $\sigma$-algebras and measurability

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.

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 $$g(x)\equiv y_0$$.
• 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. – triple_sec Dec 21 '19 at 18:44