For stochastic integral $I$ of simple process $X$, $0 \le s < t< \infty$, show $E[I_t(X) | \mathscr{F}_s] = I_s(X)$ a.s.
This is Karatzas + Shreve 2nd Edition, Chapter 3.2.B, equation (2.13) page 137.
Quick Definitions A simple process $X$ is defined such that there exists a sequence of real numbers $x_i \uparrow \infty$ and some corresponding sequence of random variables $\xi_i$ such that for each $t$ there is some $n$ such that $t_n < t \le t_{n+1}$ and $X_t = \xi_n$. The stochastic integral of this process is defined as follows relative to another martingale process $M$:
\begin{align*} I_t(X) &\triangleq \sum_{i=0}^{n-1} \xi_i (M_{t_{i+1}} - M_{t_i}) + \xi_n (M_t - M_{t_n}) \\ &= \sum_{i=0}^\infty \xi_i (M_{t \land t_{i+1}} - M_{t \land t_i}), \quad 0 \le t < \infty \\ \end{align*}
We want to show $E[I_t(X) | \mathscr{F}_s] = I_s(X)$. This assumes a filtration that both $M$ and $X$ are adapted to. It will be sufficient to show that:
\begin{align*} E[\xi_i (M_{t \land t_{i+1}} - M_{t \land t_i}) | \mathscr{F}_s] &= \xi_i (M_{s \land t_{i+1}} - M_{s \land t_i}) \quad \text{a.s.} \, P \tag{A}\\ \end{align*}
The textbook says that this above equation can be verified:
this can be verified separately for each of the three cases $s \le t_i$, $t_i < s < t_{i+1}$, and $t_{i+1} < s$ by using the $\mathscr{F}_{t_i}$ measurability of $\xi_i$.
So, if $t_i \le s$ this is easy: then $\mathscr{F}_{t_i} \subset \mathscr{F}_s$ so that $\xi_i$ is also $\mathscr{F}_s$ measurable, then:
\begin{align*} E[\xi_i (M_{t \land t_{i+1}} - M_{t \land t_i}) | \mathscr{F}_s] = \xi_i E[(M_{t \land t_{i+1}} - M_{t \land t_i}) | \mathscr{F}_s] \end{align*}
and then the above equation (A) follows from the martingale property of $M$.
When $s < t_i$, I don't see how we can get (A).
When $s \le t_i$, then $s \land t_i = s \land t_{i+1} = s$, so the right hand side is zero. For the left hand side, we start with the tower property, since $\xi_i$ is $\mathscr{F}_{t_i}$ measurable it comes out of the conditional expectation, then with the martingale property, the inner conditional expectation is zero:
\begin{align*} E[\xi_i (M_{t \land t_{i+1}} - M_{t \land t_i}) | \mathscr{F}_s] &= E \left[ E \left(\xi_i (M_{t \land t_{i+1}} - M_{t \land t_i}) | \mathscr{F}_{t_i}\right) | \mathscr{F}_s \right] \\ &= E \left[ \xi_i E \left( M_{t \land t_{i+1}} - M_{t \land t_i} | \mathscr{F}_{t_i}\right) | \mathscr{F}_s \right] = 0 \\ \end{align*}