Let $h:[0,\infty) \to \mathbb{R}$ be a measurable, square integrable function on $[0,t]$, for all $t \geq 0$.
I want to show that if $H_t = \int_0^t h(s)\;dB_s$, where $(B_t)_{t\geq0}$ is a standard Brownian Motion, then $\operatorname{Cov}(H_s,H_t)=\int_0^sh^2(u)\;du,\text{ for }s
Using a characteristic funtion of $\mathcal{N}$ and Ito's lemma I have shown that
$H_t \sim \mathcal{N(0,\int_0^th^2(u)\;du)}$
So I am only left to show that $\int_s^t h(u)\;dB_u$ is independent of $\mathcal{F}_s$. ($\mathcal{F}_{t\geq 0}$ is the natural filtration of $B_t$)
The problem here is that $\int_s^t h(u)\;dB_u$ is a limit of the functions of a form $\int_0^{\infty}\displaystyle\sum_{k=1}^{N_k}a_k\mathbf{1}_{(s_k,t_k]}dB_u$ in $\displaystyle \sup_{t\geq 0}||\cdot||_2$ norm, where a_k's are deterministic constants in this case and $s_1 > s$
The stochastic integrals of simple processes (as above) are definitely independent of $\mathcal{F}_s$, but how can I deduce that the independence is preserved after taking the limits in the $\displaystyle \sup_{t\geq 0}||\cdot||_2$ norm?