# Two conditional expectations equal almost everywhere

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?

No. If $$X,Y,Z$$ are independent, $$\mathcal F=\sigma (Y)$$ and $$\mathcal G =\sigma (Z)$$ then $$\mathbb E( X|\mathcal F)=\mathbb E(X|\mathcal G)=\mathbb EX$$.
• 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 – Kavi Rama Murthy Jul 23 '20 at 8:50