# Rosenthal's proof of the tower property for conditional expectation

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 $1$, $$E(E(Y|\mathcal{G}_2)|\mathcal{G}_1) = E(Y|\mathcal{G}_1).$$

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 $$E(X|\mathcal{G}_1) = Z.$$ So I must show that (i) $Z$ is a $\mathcal{G}_1$ measurable function and (ii) that $$E(Z\mathbb{1}_G) = E(X\mathbb{1}_G) \qquad \forall G \in \mathcal{G}_1.$$

(2) But in the textbook Rosenthal shows that (i) $E(X|\mathcal{G}_1)$ is a $\mathcal{G}_1$ measurable function and (ii) that $$E(E(X|\mathcal{G}_1)\mathbb{1}_G) = E(Y\mathbb{1}_G) \qquad \forall G \in \mathcal{G}_1.$$

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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.
$E(Y\mathbb{1}_G)=E(X\mathbb{1}_G)$ by the definition of conditional expectation.