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 Good Integrator. The Stochastic Integral $(H\cdot X)$ of $H$ w.r.t. $X$ is defined as:
$$(H\cdot X)_t:=\sum_{i=1}^{n-1} H_i(X_t^{T_{i+1}} - X_t^{T_i}).$$
I want to prove that:
$$[(H\cdot X),(H\cdot X)]=(H^2\cdot [X,X]),$$
where $[Y,Y]$ is the quadratic variation of the process $Y$, defined as: $$[X,X]:=\lim_{n\rightarrow\infty} \sum_{i=1}^{n} (X_t^{T_{i+1}} - X_t^{T_i})^2 .$$
I worked out the following but I am not sure it is correct:
First note: $(H\cdot X)_{T_{i+1}\wedge t}-(H\cdot X)_{T_i\wedge t} =H_i(X_{T_{i+1}\wedge t} - X_{T_{i}\wedge t})$,
then: $$[(H\cdot X),(H\cdot X)]_t=\lim_{n\rightarrow\infty} \sum_{i=1}^{n} ((H\cdot X)_t^{T_{i+1}} - (H\cdot X)_t^{T_i})^2=\lim_{n\rightarrow\infty} \sum_{i=1}^{n} H_i^2(X_{T_{i+1}\wedge t} - X_{T_{i}\wedge t})^2=$$ $$=(H^2\cdot [X,X])_t$$
Is this reasoning correct?