Prove that the Itô integral for elementary predictable processes builds a martingale Let


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*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$\mathbb F=(\mathcal F)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$

*$B=(B_t)_{t\ge 0}$ be an $\mathbb F$-adapted Brownian motion with respect to $\mathbb F$

*$h_i$ be a bounded, real-valued, $\mathbb F$-adapted random variable on $(\Omega,\mathcal A)$


Let $H=(H_t)_{t\ge 0}$ be of the form $$H_t(\omega)=\sum_{i=1}^nh_{i-1}(\omega)1_{(t_{i-1},t_i]}(t)\;\;\;\text{for all }\Omega\times[0,\infty)$$ and $$I_t^B(H):=\sum_{i=1}^nh_{i-1}\left(B_{t_i\wedge t}-B_{t_{i-1}\wedge t}\right)\;\;\;\text{for }t\ge 0$$

I want to show, that $\left(I_t^B(H)\right)_{t\ge 0}$ is a $\mathbb F$-martingale. Please note, that $B$ is a $\mathbb F$-martingale. Hence, $$\operatorname E\left[I_\tau^B(H)\right]=0$$ by the optional stopping theorem, for any $\mathbb F$-stopping time $\tau$:
This might help to prove the desired statement. We need to show, that $$\operatorname E\left[I_t^B(H)\mid\mathcal F_s\right]=I_s^B(H)\;\;\;\text{for all }s<t\;.$$ How can we do that?
 A: You should also mention that $h_i$ is $\mathbb{F}_{t_i}$-measurable. Since $i$ is not a time index, it is not clear what it means for $(h_i)$ to be adapted. That being said, note that linear combinations of martingales are again martingales. So then we are done if we just show that whenever $s < t$
$$E[ h_{i-1} (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_s] = h_{i-1} (B_{t_i\wedge s} - B_{t_{i-1}\wedge s})$$
First consider the case $s < t_{i-1}$. Clearly, the RHS is $0$. For the LHS use the tower property as follows
$$E[ h_{i-1} (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_s] = E[E[ h_{i-1} (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_{t_{i-1}}]\mid \mathbb{F}_s] = E[h_{i-1} E[ (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_{t_{i-1}}]\mid \mathbb{F}_s] = 0$$
To see why the last step is true note that $E[B_{t_i\wedge t} \mid \mathbb{F}_{t_{i-1}}] = B_{t_{i-1}\wedge t}$.
Now consider the case $s \geq t_{i-1}$. Doob helps a lot here. Remember that a stopped martingale is again a martingale. That is what I use below.
$$E[ h_{i-1} (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_s] = h_{i-1} E[  (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_s] = h_{i-1} E[  (B^t_{t_i} - B^t_{t_{i-1}})\mid \mathbb{F}_s] =  h_{i-1}   (B^s_{t_i} - B^s_{t_{i-1}}) = h_{i-1}(B_{t_i\wedge s} - B_{t_{i-1}\wedge s})$$
This finishes the proof I think.
