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Suppose $X$ is a continuous random variable. If $\mathbb{E}[X\,|\,\mathcal{F}]=\mathbb{E}[X\,|\,\mathcal{G}]$ almost everywhere for two sub-sigma algebra $\mathcal{F}$ and $\mathcal{G}$, does this imply $\mathcal{F}$ and $\mathcal{G}$ are set theoretically identical?

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No. If $X,Y,Z$ are independent, $\mathcal F=\sigma (Y)$ and $\mathcal G =\sigma (Z)$ then $\mathbb E( X|\mathcal F)=\mathbb E(X|\mathcal G)=\mathbb EX$.

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  • $\begingroup$ What if $\mathcal F$ is properly contained in $\mathcal G$ and $X$ is measurable w.r.t. $\mathcal F$. In this case both the conditional expectaions are equal to $X$. @ZiyuanWang $\endgroup$ – Kavi Rama Murthy Jul 23 '20 at 8:50

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