Why $E[X|\mathcal{G}]=X$ if $X$ is $\mathcal{G}$-measurable?

Question

If $X$ is a $\mathcal{G}$-measurable random variable, why $E[X|\mathcal{G}] = X$? I know the intuition (basicly we're conditioning on the same informations on which $X$ is defined, $\sigma(X)$, we can't gain any different information), but I don't know how to prove it rigourously

Answer 1

To show $E[X|\mathcal{G}]$ equals some $Y$. You need to check:

1. $Y$ is $\mathcal{G}$-measurable;
2. $\int_A Y(\omega)dP(\omega)=\int_A X(\omega)dP(\omega)$ for all $A\in\mathcal{G}$.

In this case, the claim is $Y=X$ works. (1) holds by assumption and (2) holds trivially.