Before presenting my problem I will introduce some notation. Time index $t\in [0,T]$.
$$C_t = \begin{cases} Z_n = B_{t_{n-1}}, & \text{if $t=T$} \\[2ex] Z_i = B_{t_{i-1}} , & \text{if $t_{i-1}\leq t\leq t_i$} \end{cases}$$
where $C_t$ is a simple process and $B_t$ is brownian motion at time $t$.
The Ito integral of $C_t$ on $[0,t]$ with the partition defined by $t_{k-1}\leq t \leq t_k$, is given by $$ I_t(C)=\int_0^t C_sdB_s :=\sum_{i=1}^{k-1}Z_i(B_i-B_{i-1}) + Z_k(B_t-B_{t_{k-1}}).$$
In addition, $I_t(C)$ is a martingale with respect to the natural brownian filtration $(\mathcal{F}_t, \, t\in [0,T]).$
I am trying to check that $E[I_t(C)|\mathcal{F}_s]= I_s(C)$ when $s\in[t_{l-1},t_l]$ and $t\in[t_{k-1}, t_k]$ with $l<k$. To do this, I want to use the following decomposition
$$I_t(C)=I_s(C)+R(s,t)$$
with
$$R(s,t)=Z_l(B_{t_l}-B_s)+\sum_{l+1}^{k-1}Z_i(B_i-B_{i-1})+Z_k(B_t-B_{t_{k-1}}).$$
So I need to show that $E[R(s,t)|\mathcal{F}_s]= 0$.
In my attempt I tackle each of the 3 terms of $R(s,t)$, one at a time.
First, since $Z_l$ is function of $B$ up to time s and $(B_{t_l}-B_s)$ is independent of $\mathcal{F}_s$:
$$E[Z_l(B_{t_l}-B_s)|\mathcal{F}_s]=Z_lE[(B_{t_l}-B_s)|\mathcal{F}_s]=Z_lE[(B_{t_l}-B_s)]=0.$$
Next, since \begin{align} E[Z_i(B_{t_i}-B_{t_{i-1}})|\mathcal{F}_s] = E[B_{t_{i-1}}B_{t_i}|\mathcal{F}_s]-E[B_{t_{i-1}}^2|\mathcal{F}_s] &= E[B_{t_{i-1}}B_{t_i}]-(t_{i-1}-s+B_s^2)\\&=t_{i-1}-(t_{i-1}-s+B_s^2). \end{align} I find $$E\left[\sum_{l+1}^{k-1}Z_i(B_{t_i}-B_{t_{i-1}})\bigg|\mathcal{F}_s\right] = \sum_{l+1}^{k-1}E[B_{t_{i-1}}(B_{t_i}-B_{t_{i-1}})|\mathcal{F}_s] = (s-B_s^2)\sum_{l+1}^{k-1}1 = (s-B_s^2)(k-1-l).$$
And finally
$$E[Z_k(B_{t_k}-B_{t_{k-1}})|\mathcal{F}_s]=(s-B_s^2).$$
Summing up every result I obtain $$0+(s-B_s^2)(k-l-1)+(s-B_s^2)=(s-B_s^2)(k-l).$$
And this is not zero. So I do not understand what I am doing wrong. Can you help me?
Thank you in advance.
Your mistake is in rewriting $\mathbb{E}\left\lbrack B_{t_i} B_{t_{i-1}} \middle| \mathcal{F}_s\right\rbrack$ as $\mathbb{E}\left\lbrack B_{t_i} B_{t_{i-1}} \right\rbrack$.
In making that step in your working, you're implicitly assuming that $B_{t_i}B_{t_{i-1}}$ is independent of $\mathcal{F}_s$; this isn't the case for any time indices $t_i$ & $t_{i-1}$ - intuitively, knowing $(B_r, 0 \leq r \leq s)$ tells you something about the value of $B_{t_i} B_{t_{i-1}}$. There's an often-used trick to evaluate conditional expectations like this, which is to write
$B_{t_i}B_{t_{i-1}} = ((B_{t_i} - B_s) + B_s)((B_{t_{i-1}} - B_s) + B_s)$
and expand the outer brackets.