Let $M$ be a right continuous $L^2$ martingale, show that $(\int I_{[0,t]}XdM)_{t \geq 0}$ is a right continuous $L^2$ martingale where $X=\sum_{i=1}^n c_i I_{(s_i,t_i]\times G_i}$, $c_i\in \mathbb{R}, 0\leq s_i<t_i$ and $G_i$ is $\mathfrak{F}_{s_i}$-measurable.
It suffices to show that it holds of the case $X=I_{(a,b]\times G}$, where $a<b$ and $G$ is $\mathfrak{F}_a$ measurable. Note that
$$\int I_{[0,t]}XdM=I_{G}(M_{b\wedge t}-M_{a\wedge t}).$$ Since $M$ is right continuous and $L^2$, it is obvious that $I_{G}(M_{b\wedge t}-M_{a\wedge t})$ also right continuous and $L^2$. Finally, we are showing that if $s<t$, then
$$
E[I_{G}(M_{b\wedge t}-M_{a\wedge t})| \mathfrak{F}_{s}]=I_{G}(M_{b\wedge s}-M_{a\wedge s}). (\star)
$$
I tried solve it case by case, and I have no problem of the cases if $s\geq a.$ But I got stcuk of the cases:$s<a$.
For example: $t\geq b$, $s<a,$ I wrote the left side of $(\star)$ by
$$
E[I_{G}(M_{b\wedge t}-M_{a\wedge t})| \mathfrak{F}_{s}]=E[I_GM_{b}|\mathfrak{F_s}]-E[I_{G}M_a|\mathfrak{F}_s] .(\star\star)
$$
Since right hand side of $(\star)$ will be
$$
I_{G}(M_s-M_s)=0.
$$
But, I have no idea how to claim the $(\star\star)$ is zero.
Let $t \geq b > a > s$. Since $\mathcal{F}_s \subseteq \mathcal{F}_a$, we have $$\mathbb{E}[I_G(M_b - M_a)|\mathcal{F}_s] = \mathbb{E}[\mathbb{E}[I_G(M_b - M_a)|\mathcal{F}_a]|\mathcal{F}_s] = \mathbb{E}[I_G\mathbb{E}[M_b - M_a|\mathcal{F}_a]|\mathcal{F}_s] = 0.$$