Show that $$ I_t^W(H) := \sum_{i=1}^n h_{i-1}\big(W(t_i \wedge t) - W(t_{i-1}\wedge t)\big), \qquad t \ge 0 $$ where $(W_t)_{t \ge 0}$ is the Wiener process and $h_{i-1}$ an $\mathcal F_{t_{i-1}}$-measurable and bounded random variable, is a martingale.
This is what I managed to do: we need to consider, for $s<t$, $$ \boldsymbol{\mathrm E} \left [ I_t^W(H) \mid \mathcal{F}_s \right] = \sum_{i=1}^n \boldsymbol{\mathrm E} \left [h_{i-1}(W(t_i \wedge t) - W(t_{i-1}\wedge t)) \mid \mathcal{F}_s \right] $$ First note that since Brownian motion is a martingale, for any $s < t$ $$ \boldsymbol{\mathrm E} \left [W(t) \mid \mathcal{F}_s \right] = W(s). $$ In particular if $s < t_i \wedge t $ then $$ \boldsymbol{\mathrm E} \left [W(t_i \wedge t) \mid \mathcal{F}_s \right] = W(s) \underbrace{= W(t_i \wedge s)}_{s < t_i \implies s = t_i \wedge s} $$ Further, since $W(t)$ is adapted w.r.t. $\mathcal{F}_t$, if $t_i \wedge t < s$ then $$ \boldsymbol{\mathrm E} \left [W(t_i \wedge t) \mid \mathcal{F}_s \right] = W(t_i \wedge t) = \underbrace{W(t_i \wedge s)}_{t_i \wedge t < s \text{ and } s<t \implies t_i \wedge t = t_i \wedge s}. $$ And we conclude tht for any $t_i,$ and andy $s<t$ $$ \boldsymbol{\mathrm E} \left [W(t_i \wedge t) \mid \mathcal{F}_s \right] = W(s) = W(t_i \wedge s) $$ Thus what is left is to consider what happens with $h_{i-1}$. There is two cases; when $t_{i-1} < s$ then $h_{i-1}$ is $\mathcal{F}_s$-measurable and \begin{multline*} \boldsymbol{\mathrm E} \left [h_{i-1}(W(t_i \wedge t) - W(t_{i-1}\wedge t)) \mid \mathcal{F}_s \right] = h_{i-1} \boldsymbol{\mathrm E} \left [W(t_i \wedge t) - W(t_{i-1}\wedge t) \mid \mathcal{F}_s \right] \\ = h_{i-1} \left (W(t_i \wedge s) - W(t_{i-1}\wedge s) \right ) =I_s^W(H); \end{multline*}
My question is: what happens when $t_{i-1} > s$? since then $h_{i-1}$ isn't necessarily $\mathcal{F}_s$-measurable.
If $t_{i-1} > s$, you can use the tower property of conditional expectation: \begin{align*}\mathbb{E}[h_{i-1}(W(t_i\wedge t) - W(t_{I-1}\wedge t))|\mathcal{F}_s] &= \mathbb{E}[\mathbb{E}[h_{i-1}(W(t_i\wedge t) - W(t_{I-1}\wedge t))|\mathcal{F}_{t_{i-1}}]|\mathcal{F}_s] \\ &=\mathbb{E}[h_{i-1}\mathbb{E}[(W(t_i\wedge t) - W(t_{I-1}\wedge t))|\mathcal{F}_{t_{i-1}}]|\mathcal{F}_s] \\ &=0 \end{align*}