TITLE: Proof of the tower property for conditional expectations
QUESTION [10 upvotes]: Let $Z$ be a $\mathfrak{F}$-measurable random variable with $\mathbb E(|Z|)<\infty$ and let $\mathfrak{H}\subset \mathfrak{G}\subset \mathfrak{F}$.
Show that then $\mathbb E(\mathbb E(Z|\mathfrak{G})|\mathfrak{H})=\mathbb E(Z|\mathfrak{H})=\mathbb E(\mathbb E(Z|\mathfrak{H})|\mathfrak{G})$.
The second equality is clear to me since the random variable $E(Z|\mathfrak{H})$ is $\mathfrak{H}$-measurable, hence its also $\mathfrak{G}$-measurable, so we can take it out. 
If I want to prove the first equality I have to show that $E(Z|\mathfrak{G})$ and $Z$ do agree on $\mathfrak{H}$-measurable sets, right? Because I have to show that $Z$ and $\mathbb E(Z|\mathfrak{G})$ are equal if I condition both with the sigma algebra $\mathfrak{H}$.
Is this argumentation correct?
For the proof: $\int_\Omega\mathbb E(Z|\mathfrak{G})\mathbb 1_HdP=\int_\Omega Z\mathbb 1_H$ just because $E(Z|\mathfrak{G})$ is $\mathfrak{H}$ measurable, hence "I can remove one $\mathbb E$ and $\mathfrak{G}$ and still have equality".
I know I wrought much for a simple task, but I just want to understand it correctly, thanks :)

REPLY [3 votes]: Prove of $\mathbb E(\mathbb E(Z|\mathcal{G})|\mathcal{H})=\mathbb E(Z|\mathcal{H})$:
$\mathbb E(\mathbb E(Z|\mathcal{G})|\mathcal{H})$ is a conditional expectation of $E(Z|\mathcal{G})$, hence $\int_{H}\mathbb E(\mathbb E(Z|\mathcal{G})|\mathcal{H})d P=\int_H\mathbb E(Z|\mathcal{G})dP$ for all $H\in \mathcal{H}$.
$\mathbb E(Z|\mathcal{G})$ is a conditional expectation of $Z$, hence $\int_G\mathbb E(Z|\mathcal{G})dP=\int_GZdP$ for all $G\in \mathcal{G}$.
Combining both equalities and using that $H\in \mathcal{H}$ yields: $\int_{H}\mathbb E(\mathbb E(Z|\mathcal{G})|\mathcal{H})d P=\int_HZdP$ as desired.