Show that $E[ E[Z\lvert \mathcal{F}_{T}\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]$ Let $T$ and $S$ be stopping times and $Z$ some integrable random variable. I already know that:
$\mathcal{F}_{T\land S}=\mathcal{F}_{T}\cap\mathcal{F}_{S}$ and $\{T>S\},\;\{T\leq S\},\; \{T<S\},\;\{T\geq S\} \in \mathcal{F}_{T\land S}$. And furthermore I have shown that:
$E[Z\lvert \mathcal{F}_{T}]=E[Z\lvert \mathcal{F}_{S\land T}]\; \; \mathbb P$-a.s. on $\{T\leq S\} \; (*)$
I want to show that:
$E[ E[Z\lvert \mathcal{F}_{T}]\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]\; \; \mathbb P$-a.s.
I guess we have to use $(*)$ but I get stuck:
$E[ E[Z\lvert \mathcal{F}_{T}]\lvert \mathcal{F}_{S}]=E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T\leq S\}}\lvert \mathcal{F}_{S}]+E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T> S\}}\lvert \mathcal{F}_{S}]= E[Z\lvert \mathcal{F}_{T\land S}]1_{\{T\leq S\}}+E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T> S\}}\lvert \mathcal{F}_{S}]$
How can I go about showing that the second term
$$ E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T> S\}}\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]1_{\{T>S\}}$$ or am I going about this the wrong way?
Reference to Problem 2.17 from Karatzas and Shreve p. 9 Brownian Motion and Stochastic Calculus
 A: Let $Y$ be the random variable $Y=E[Z| \mathcal{F}_{T}]$. I am assuming we work with a complete filtration $(\mathcal F_t)$ on some space $(\Omega, \mathcal F,\Bbb P)$.
Then $(*)$ gives:
$$
\begin{aligned}
E[Z| \mathcal{F}_{T}]
&=E[Z |\mathcal{F}_{T\land S}] \qquad\text{ on } A:=\{T\le S\}\ ,\\
E[Y| \mathcal{F}_{S}]
&=E[Y |\mathcal{F}_{T\land S}] \qquad\text{ on } B:=\{S\le T\}\ ,\text{ so }
\\[2mm]
E[Z\cdot 1_A| \mathcal{F}_{T}]
&=E[Z\cdot 1_A |\mathcal{F}_{T\land S}] \qquad(1)
\\
E[Y\cdot 1_{\bar A}| \mathcal{F}_{S}]
&=E[Y\cdot 1_{\bar A} |\mathcal{F}_{T\land S}]
\qquad(2)\qquad\text{ so }
\\[2mm]
E[\ E[Z\cdot 1_A| \mathcal{F}_{T}]\ | \mathcal{F}_{S}]
&= E[\ E[Z\cdot 1_A |\mathcal{F}_{T\land S}] \ |\mathcal F_S]
   \qquad\text{ by }(1)\\
&= E[Z\cdot 1_A |\mathcal{F}_{T\land S}]\\
&= E[Z |\mathcal{F}_{T\land S}]\cdot 1_A
\\[2mm]
E[\ E[Z\cdot 1_{\bar A}| \mathcal{F}_{T}]\ | \mathcal{F}_{S}]
&=E[Y\cdot 1_{\bar A}| \mathcal{F}_{S}]\\
&=E[Y\cdot 1_{\bar A}| \mathcal{F}_{T\land S}]
   \qquad\text{ by }(2)\\
&=E[Y| \mathcal{F}_{T\land S}]\cdot 1_{\bar A}\\
&=E[\ E[Z| \mathcal{F}_{T}]\ | \mathcal{F}_{T\land S}]\cdot 1_{\bar A}\\
&=E[Z | \mathcal{F}_{T\land S}]\cdot 1_{\bar A}\\
\end{aligned}
$$
And we add.
(We have used $\bar A\subseteq B$.)
$\square$
A: What seems to be missing here is the following fact:

If $Y$ is $\mathcal{F}_{S}$-measurable, then $Y 1_{\{S < T\}\}}$ is
$\mathcal{F}_{S \wedge T}$-measurable.

(You already seem to have noticed that $Y 1_{\{S \leq T\}}$ is $\mathcal{F}_{S \wedge T}$-measurable.)
Now notice that if $A \in \mathcal{F}_{S \wedge T}$, then $A \cap \{S < T\} \in \mathcal{F}_{S \wedge T}$ and, thus,
\begin{align*}
E[E[Z\mid\mathcal{F}_{S}] 1_{\{S < T\}} : A] &= E[Z : A \cap \{S < T\}] \\
&= E[E[Z \mid \mathcal{F}_{S \wedge T}] 1_{\{S < T\}} : A].
\end{align*}
Thus, since $E[Z \mid \mathcal{F}_{S}]1_{\{S < T\}}$ is $\mathcal{F}_{S \wedge T}$-measurable by the observation above, the arbitrariness of $A$ implies that
\begin{equation*}
E[Z \mid \mathcal{F}_{S}] 1_{\{S < T\}} = E[Z \mid \mathcal{F}_{S \wedge T}] 1_{\{S < T\}}.
\end{equation*}
Finally, notice that if $A \in \mathcal{F}_{S}$, then our observation gives $A \cap \{S < T\} \in \mathcal{F}_{S \wedge T}$ so we can write
\begin{equation*}
E[E[Z\mid\mathcal{F}_{T}] : A \cap \{S  < T\}] = E[Z: A \cap \{S < T\}] = E[Z 1_{\{S < T\}} : A].
\end{equation*}
By the arbitrariness of $A$ and the inclusion $\{S < T\} \in \mathcal{F}_{S}$, this implies
\begin{equation*}
E[E[Z\mid \mathcal{F}_{T}] \mid \mathcal{F}_{S}] 1_{\{S < T\}} = E[E[Z \mid \mathcal{F}_{T}] 1_{\{S < T\}} \mid \mathcal{F}_{S}] = E[Z \mid \mathcal{F}_{S}] 1_{\{S  <T\}}.
\end{equation*}
Combining the two identities obtained above, we find the equation we needed:
\begin{equation*}
E[E[Z\mid \mathcal{F}_{T}] 1_{\{S < T\}} \mid \mathcal{F}_{S}] = E[E[Z\mid \mathcal{F}_{T}] \mid \mathcal{F}_{S}] 1_{\{S < T\}} = E[Z \mid \mathcal{F}_{S \wedge T}] 1_{\{S  <T\}}.
\end{equation*}
