Take $H\in S$ to be a simple process defined as: $$H_t:=\sum_{i=1}^{n-1} H_i1_{(T_i,T_{i+1}]}(t),\ \ H_i\in \mathcal{F}_{T_i}, \ (T_1\leq...\leq T_n \ stopping\ times),$$
and $X$ a Martingale. I want to prove that the Stochastic Integral $(H\cdot X)$ defined as:
$$(H\cdot X)_t:=\sum_{i=1}^{n-1} H_i(X_t^{T_{i+1}} - X_t^{T_i})$$
is a Martingale.
I have to show that $\mathbb{E}[(H\cdot X)_t|\mathcal{F_s}]=(H\cdot X)_s $, given that $\mathbb{E}[X_t|\mathcal{F_s}]=X_s$.
Without loss of generality, let's assume: $$H = H_11_{(T_1,T_{2}]}+H_21_{(T_2,T_{3}]}$$
then, $$(H\cdot X)_t:= H_1(X_t^{T_{2}} - X_t^{T_1})+H_2(X_t^{T_{3}} - X_t^{T_2})$$ with $T_1\leq s\leq T_2 \leq t \leq T_3$.
Then: $$\mathbb{E}[(H\cdot X)_t|\mathcal{F_s}]=\mathbb{E}[H_1(X_t^{T_{2}} - X_t^{T_1})|\mathcal{F_s}]+\mathbb{E}[H_2(X_t^{T_{3}} - X_t^{T_2})|\mathcal{F_s}]=$$ $$H_1(X_s^{T_{2}} - X_s^{T_1})+\mathbb{E}[H_2(X_t - X_{T_2})|\mathcal{F_s}]. (*)$$
I get stuck here because $H_2$ is $\mathcal{F_{T_2}}$-measurable and since $s\leq T_2$ I can not bring $H_2$ outside the expectation and apply the martingale property on $X$.
Any hint?
EDIT: This should work
Consider just the second addend in (*):
$$\mathbb{E}[H_2(X_t - X_{T_2})|\mathcal{F_s}]=\mathbb{E}[\mathbb{E}[H_2(X_t - X_{T_2})|\mathcal{F}_{T_2}]|\mathcal{F_s}]=\mathbb{E}[H_2\mathbb{E}[(X_t - X_{T_2})|\mathcal{F}_{T_2}]|\mathcal{F_s}]= 0.$$
Therefore:
$$\mathbb{E}[(H\cdot X)_t|\mathcal{F_s}]=H_1(X_s^{T_{2}} - X_s^{T_1})=(H\cdot X)_s,$$
since $$(H\cdot X)_s=H_1(X_s^{T_{2}} - X_s^{T_1})+H_2(X_s^{T_{3}} - X_s^{T_2})=$$ $$H_1(X_s^{T_{2}} - X_s^{T_1})+H_2(X_s - X_s)=$$ $$H_1(X_s^{T_{2}} - X_s^{T_1}).$$