I have a question about a proof of the tower property of conditional expectation that I found in Rosenthal's "A First Look at Rigorous Probability".
Proposition: Let $Y$ be a random variable with finite mean, and $\mathcal{G}_1 \subseteq \mathcal{G}_2$ be two sub-$\sigma$-algebras. The with probability $1$,
\begin{equation}
E(E(Y|\mathcal{G}_2)|\mathcal{G}_1) = E(Y|\mathcal{G}_1).
\end{equation}
Are both of the following approaches sufficient to prove the statement?
(1) Suppose I define $X = E(Y|\mathcal{G}_2)$ and $Z = E(Y|\mathcal{G}_1)$. Then I must show under the given assumptions that
\begin{equation}
E(X|\mathcal{G}_1) = Z.
\end{equation}
So I must show that (i) $Z$ is a $\mathcal{G}_1$ measurable function and (ii) that
\begin{equation}
E(Z\mathbb{1}_G) = E(X\mathbb{1}_G) \qquad \forall G \in \mathcal{G}_1.
\end{equation}
(2) But in the textbook Rosenthal shows that (i) $E(X|\mathcal{G}_1)$ is a $\mathcal{G}_1$ measurable function and (ii) that
\begin{equation}
E(E(X|\mathcal{G}_1)\mathbb{1}_G) = E(Y\mathbb{1}_G) \qquad \forall G \in \mathcal{G}_1.
\end{equation}
Best Answer
Yes, both are sufficient. In (1) you think of computing $E[X \mid \mathcal{G}_1]$, and you show it is given by $E[Y \mid \mathcal{G}_1]$. In (2) you go the other way: you think of computing $E[Y \mid \mathcal{G}_1]$, and you show it is given by $E[X \mid \mathcal{G}_1]$. Either way you get the desired equality.