# Measurability with respect to equivalent sub-$\sigma$-algebras for mappings into countably generated spaces

Let $$(\Omega,\mathcal{F},\mathbb{P})$$ be a probability space, and let $$\mathcal{G}_1$$ and $$\mathcal{G}_2$$ be sub-$$\sigma$$-algebras of $$\mathcal{F}$$ that are equivalent to each other in the sense that $$\sigma(\mathcal{G}_1 \cup \{\mathbb{P}\textrm{-null sets}\}) \ = \ \sigma(\mathcal{G}_2 \cup \{\mathbb{P}\textrm{-null sets}\}).$$ Let $$(X,\Sigma)$$ be a countably generated measurable space.

Given a $$(\mathcal{G}_1,\Sigma)$$-measurable function $$g_1 \colon \Omega \to X$$, does there necessarily exist a $$(\mathcal{G}_2,\Sigma)$$-measurable function $$g_2 \colon \Omega \to X$$ such that $$\,g_1 \! \overset{\mathbb{P}\textrm{-a.s.}}{=} \! g_2\,$$?

(I know that the answer is yes if $$(X,\Sigma)$$ is a standard Borel space, since if we regard $$X$$ as a Borel subset of $$\mathbb{R}$$, then taking $$g_2$$ to be a suitable version of $$\mathbb{E}[g_1|\mathcal{G}_2]$$ gives the result. But I am curious about the more general scenario that $$(X,\Sigma)$$ is a countably generated measurable space.)

• +1 I like your questions :) – mathworker21 Feb 21 '20 at 20:55
• Why doesn't conditioning work in more general spaces? – aduh Sep 14 '20 at 0:10