Proving martingale property of $N_t = Z(M_{t\wedge s} - M_{t \wedge r})$ for martingale $M$ (Stochastic calculus and Brownian motion, LeGall, page 80). 

Suppose $M = (M_t)$ is a martingale. Also, let $Z$ be a bounded random variable which is $\mathcal{F}_r$ adapted. Then we like to show that for any $0 \leq r < s$, 
  $$N_t = Z(M_{t\wedge s} - M_{t \wedge r})$$ 
  is a martingale. 

My attempt:
$Z(M_{t\wedge s} - M_{t \wedge r}) \in L_1$, should be fine since both $Z$ and $M$ are bounded. I am not sure how to show it is adapted.
Finally we need to show the martingale identity. Suppose $v \geq r$, then we have 
$$\mathbb{E} \{Z(M_{t\wedge s} - M_{t \wedge r}) \mid \mathcal{F}_v \} = Z \mathbb{E} \{(M_{t\wedge s} - M_{t \wedge r}) \mid \mathcal{F}_v \} = Z (M_{v\wedge s} - M_{v \wedge r})$$
where in the first equality we use the fact that $Z \in \mathcal{F}_r$, and $v \geq r$. This proves the result for this case. However, I don't know how to get the result for general $v$.
Thanks for you helps in advance.
 A: First of all, we note that
$$N_t = \begin{cases} 0, & t \leq r, \\ Z (M_t-M_r) & t \in (r,s), \\ Z (M_s-M_r), & t \geq s. \end{cases} \tag{1}$$
Since $Z$ is, by assumption, $\mathcal{F}_r$-measurable and $(M_t)_{t \geq 0}$ is $(\mathcal{F}_t)_{t \geq 0}$-adapted, this implies, in particular, that $(N_t)_{t \geq 0}$ is $(\mathcal{F}_t)_{t \geq 0}$-adapted. Moreover, $M_t \in L^1$ together with the boundedness of $Z$ entails $N_t \in L^1$ for all $t \geq 0$. It remains to check that
$$\mathbb{E}(N_t \mid \mathcal{F}_u) = N_u \quad \text{for all $u \leq t$}. \tag{2}$$
We consider two cases separately.  If $t \geq u  \geq r$ then it follows from the $\mathcal{F}_r$-measurability of $Z$ that
$$\mathbb{E}(N_t \mid \mathcal{F}_u) = Z \mathbb{E}(M_{t \wedge s}-M_{t \wedge r} \mid \mathcal{F}_u).$$
Because of the martingal property of $(M_t)_{t \geq 0}$ we have
$$ \mathbb{E}(M_{t \wedge s}-M_{t \wedge r} \mid \mathcal{F}_u) = M_{t \wedge s \wedge u} - M_{t \wedge r \wedge u} \stackrel{t \geq u}{=} M_{u \wedge s} - M_{u \wedge r},$$
and so
$$\mathbb{E}(N_t \mid \mathcal{F}_u) = N_u, \qquad r \leq u \leq t. $$
In particular,
$$\mathbb{E}(N_t \mid \mathcal{F}_r) = N_r \stackrel{(1)}{=} 0. \tag{3}$$
It remains to consider the case $u < r$. If $t \leq r$ then, by $(1)$, $N_t = 0$ and so
$$\mathbb{E}(N_t \mid \mathcal{F}_u)=0=N_u.$$
Suppose now that $t \geq r$. Since the tower property gives
$$\mathbb{E}(N_t \mid \mathcal{F}_u) = \mathbb{E} \bigg[ \mathbb{E}(N_t \mid \mathcal{F}_r) \mid \mathcal{F}_u \bigg], $$
we conclude from $(3)$ that
$$\mathbb{E}(N_t \mid \mathcal{F}_u) = 0 = N_u.$$
This finishes the proof of $(2)$.
