I'm trying to prove the next fact using in a proof of quadratic variation of martigales:
For every $0\leq r<s,$ and for every bounded $\mathcal{F}_{r}-$measurable variable $Z,$ the process $N_{t}=Z(M_{s\land t}-M_{r\land t})$ is a martingale, where $M$ is a martingale.
I'm stuck proving it. I would like to prove the above because I'm interested in proving that martingale transform $$X_{t}^{n}=\sum_{i=1}^{p_{n}}M_{t_{i-1}^{n}}(M_{t_{i}^{n}\land t}-M_{t_{i-1}^{n}\land t})$$ is a martingale.
I'm able to prove that martingale transform of a martinagle $M$ is martingale too, but when is a stopped martingale I don't know how to proceed.
Any kind of help is thanked in advanced.
Note that for $t \leq r$, $N_t = 0$ and for $t \geq r$ $N_t$ is certainly $\mathcal{F}_t$ measurable since $Z, M_{s \wedge t}, M_{r \wedge t}$ all are. Therefore $N$ is adapted.
Also $N_t \in L^1$, since $Z$ is bounded so $N_t \leq C(M_{s \wedge t} - M_{r \wedge t})$ for some constant $C$ and $M_{\cdot \wedge t} \in L^1$.
Finally we need to calculate the conditional expectation $\mathbb{E}[N_t \mid \mathcal{F}_{t-1}]$. Here we split in to cases.
If $t > s$ then $N_t = Z(M_s - M_r)$. Then $t-1 \geq s$ so $\mathbb{E}[N_t \mid \mathcal{F}_{t-1}] = Z(M_s - M_r) = N_{t-1}$ since all the terms are $\mathcal{F}_{t-1}$-measurable.
If $t \in (r,s]$ then $N_t = Z(M_t - M_r)$. Then $t-1 \in [r,s]$ so $\mathbb{E}[N_t \mid \mathcal{F}_{t-1}] = Z \mathbb{E}[M_t \mid \mathcal{F}_{t-1}] - Z M_r = Z(M_{t-1} - M_r) = N_{t-1}$.
Finally if $t < r$, $N_t = 0$ so $\mathbb{E}[N_t \mid \mathcal{F}_{t-1}] = 0 = N_{t-1}$.
Therefore $\mathbb{E}[N_t \mid \mathcal{F}_{t-1}] = N_{t-1}$ so $N$ is a martingale as desired.