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I'm trying to find the answer of the following question: let $\mathcal{G}$ and $\mathcal{H}$ be two sub-$\sigma$-algebras of $\mathcal{F}$. Does \begin{eqnarray} \mathbb{E}[X\vert\mathcal{G}]=\mathbb{E}[X\vert\mathcal{H}]\quad\textrm{ a.s., }\forall\:X\in L^1 \end{eqnarray} imply $\mathcal{G}=\mathcal{H}$?

Now, given any $A\subset\mathcal{G}$, the characteristic function $1_A$ is $\mathcal{G}$-measurable, hence \begin{eqnarray} 1_A=\mathbb{E}[1_A\vert\mathcal{G}]=\mathbb{E}[1_A\vert\mathcal{H}]\quad\textrm{ a.s. } \end{eqnarray} and, by definition $\mathbb{E}[1_A\vert\mathcal{H}]$ is $\mathcal{H}$-measurable. So $1_A$ is equal almost surely to an $\mathcal{H}$-measurable function. If $\mathcal{H}$ is complete (i.e. contains every subset of a null set), this implies that $1_A$ is $\mathcal{H}$-measurable and we are done. But what if it is not? Can someone come up with a way of fixing it or a counterexample in this case?

Thanks!!!

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This is the best you can get. Indeed, if $\mathcal G$ is a $\sigma$-algebra and $N$ is a set of probability $0$, defining $\mathcal H$ as the $\sigma$-algebra generated by $\mathcal G$ and $N$, we get the equal of conditional expectations but not of the $\sigma$-algebras.

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  • $\begingroup$ Simple and clear answer. Thank you! $\endgroup$
    – havmath
    Dec 14 '16 at 14:20

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