I'm struggling with an exercise from a martingales theory book:
Let $M$ be an $F$-martingale and $Z$ an adapted (bounded) continuous process. Prove that: $$ \mathbb{}E\left(M_t\int_s^tZ_u \, du \mid \mathcal{F}_s \right)=\mathbb{}E\left(\int_s^t M_u Z_u \, du \mid \mathcal{F}_s\right) $$
It was given just after the Doob-Meyer decomposition theorem.
I started from the right term interverting integral and expectation signs: $$ \mathbb{}E\left(\int_s^t M_u Z_u \, du \mid \mathcal{F}_s\right)=\mathbb{}\int_s^tE\left( M_u Z_u \mid \mathcal{F}_s\right) \, du $$ but I didn't see where it could lead.
Any hint (not a solution) would be appreciated
Since $M$ is a martingale, we have
$$M_u Z_u = \mathbb{E}(M_t \mid \mathcal{F}_u) Z_u = \mathbb{E}(M_t Z_u \mid \mathcal{F}_u)$$
for any $u \leq t$. Now integrate both sides over $u \in [s,t]$ and take the conditional expectation with respect to $\mathcal{F}_s$