How to prove that two conditional expectations are equivalent?

Question

I have the following question: what is claimed when one say that two conditional expectations, with respect to different sigma algebras are equivalent? That is to verify $E[1_F|\mathcal {F } ]=E[1_F|\mathcal {G}]$ do I need to show that $E[1_F|\mathcal {F }]$ is a version of $E[1_F|\mathcal {G} ]$ and then that $E[1_F|\mathcal {G } ]$ is a version of $E[1_F|\mathcal {F }]$? Or is it sufficient to show just that one of them is a version of the other?

The specific contex is that in a proof it is claimed that $E[1_F|\sigma(X_n)]=E[1_F|\sigma(X_1,X_2,...X_n)]$ and to verify it the author sets about to show that $E[1_F|\sigma(X_n)]$ is a version of $E[1_F|\sigma(X_1,X_2,...X_n)]$ but not the converse. Why is this sufficient?

Would it be different for the cases $E[X|\mathcal {F}]=E[Y|\mathcal {F}]$ and $E[X|\mathcal {F}]=E[Y|\mathcal {G}]$ (for $X \neq Y $)?

Thanks in advance!

Answer 1

Here it would be sufficient because $\sigma(X_n)$ is contained in $\sigma(X_1,\dots,X_n)$.

Let $X$ be an integrable random variable and let $\mathcal F$, $\mathcal G$ be two $\sigma$-algebras such that $\mathcal F\subset \mathcal G$. In order to check that $\mathbb E\left[X\mid \mathcal F\right]=\mathbb E\left[X\mid \mathcal G\right]$, it suffices to check that $\mathbb E\left[X\mid \mathcal F\right]$ is a version of $\mathbb E\left[X\mid \mathcal G\right]$. Since $\mathcal F\subset \mathcal G$ we already know that $\mathbb E\left[X\mid \mathcal F\right]$ is $\mathcal G$-measurable so everything reduces to prove that for all $G\in\mathcal G$, $$ \mathbb E\left[\mathbb E\left[X\mid \mathcal F\right]\mathbf 1_G\right]=\mathbb E\left[X\mathbf 1_G\right].$$