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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.)

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

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