Let $S$ and $T$ be stopping times with respect to the filtration $\{\mathcal{F}_n\}_{n \in \mathbb{N}}$. I want to show that for an integrable random variable $Y$, $$E(E(Y|\mathcal{F}_S)|\mathcal{F}_T) = E(Y|\mathcal{F}_{S \wedge T})$$
Clearly $\mathcal{F}_{S \wedge T} = \mathcal{F}_{T} \cap \mathcal{F}_{S}$ so that $\forall A \in \mathcal{F}_{S \wedge T}$, we have $$ E(E(E(Y|\mathcal{F}_S)|\mathcal{F}_T)1_A) = E(E(Y|\mathcal{F}_S)1_A) = E(Y1_A)$$
where the first equality follows since $A \in \mathcal{F}_T$ and the second follows from $A \in \mathcal{F}_S$ and applying the definition of conditional expectation.
Thus it suffices to show that $Z \equiv E(E(Y|\mathcal{F}_S)|\mathcal{F}_T)$ is $\mathcal{F}_{S \wedge T}$ measurable. I know that the sets $\{S<T\}$ and $\{S \ge T\}$ are both in $\mathcal{F}_S \cap \mathcal{F}_T$, so we may write for any Borel set $B$
$$\{Z \in B\} \equiv C = (C \cap \{S \ge T\}) \cup (C \cap \{S < T\})$$
Because $C \in \mathcal{F}_T$ by definition of cond. expectation, we have $C \cap \{S \ge T\} \cap \{S \wedge T \leq n\} = C \cap \{S \ge T\} \cap \{T\leq n\} \in \mathcal{F}_n \forall n$ so that $C \cap \{S \ge T\} \in \mathcal{F}_{S \wedge T}$.
Finally I just need to show $C \cap \{S < T\} \in \mathcal{F}_{S \wedge T}$ which is something I have been incapable of doing thus far. Any ideas?
(For reference, this is from Jacod and Protter exercise 24.11)
One approach to this issue is to consider the uniformly integrable martingale defined by $M_n:=E(Y\mid \mathcal F_n)$. The stopped process $K_n:=M_{n\wedge S}$ is also a UI martingale, with terminal value $K_\infty=\lim_{n\to\infty}K_n=M_S=E(Y\mid\mathcal F_S)$. And $E(K_\infty\mid\mathcal F_T)=K_T$, to finish.