I needed to prove $E((E(Y|G)|H) = E(Y|H)$, where $H$ is a sub $\sigma$-algebra of $G$.
The following is my thought on it, I feel that it's wrong, but don't know how, please help me.
If $Y$ is positive, then $E((E(Y|G)|H) - E(Y|H) = 0 \iff E(E((E(Y|G)|H) - E(Y|H)) = 0$
Then the law of total expectation proves the statement.
For general $Y$, use Jordan decomposition to get $Y = Y^+ - Y^-$, then the property follows from the linearity of expectation.
This doesn't quite work. The problem is with the $\Leftarrow$ implication. You could a priori have $E(E(Y|G)|H) > E(Y|H)$ on some $H$-measurable set of positive measure and $E(E(Y|G)|H) < E(Y|H)$ on another $H$-measurable set of positive measure, but they could balance out and give you $ E(E(E(Y|G)|H) - E(Y|H) ) = 0$.
Instead, observe that for any $A \in H$, $$ \int_A E(E(Y|G)|H) \, dP = \int_A E(Y|G) \, dP = \int_A Y \, dP $$ The first equality is from the definition of the conditional expectation $E(E(Y|G)|H)$. The second is from the definition of the conditional expectation $E(Y|G)$, together with the fact that $A \in G$ since $H$ is a sub-$\sigma$-algebra of $G$. But then $E(E(Y|G)|H)$ has the same integral as $Y$ over any $A \in H$. This is the property defining $E(Y|H)$, and the conditional expectation is unique up to a null set, so we must have $E(E(Y|G)|H) = E(Y|H)$.