I know that for one-dimensional case, $$ E \left[ \left(\int^T_S f(t,\omega)dB_t \right)^2 \right] = E\left[ \int^T_S f^2(t,\omega) \, dt \right]$$ for adapted, measurable f that satisfies that are in $L^2(dt \times dP)$.
For $f = [f_{ij}(t,\omega)]$ and $n$-dimensional Brownian motion, we have $$ E\left[ \left|\int^T_S f(t,\omega)dB_t \right|^2 \right] = E \left[ \int^T_S |f(t,\omega)|^2 \, dt \right]?$$
where $|f|^2 = \sum f_{ij}^2$. I think to derive this, we need (for $j \neq k$)
$$E \left( \int^T_S f_{ij}\, dB_j \int^T_S f_{ik} \, dB_k \right) =0 ?$$
But I am not sure about the last assertion (by indep of the Brownian motions? but the $f$'s are not independent?). Thanks.
For any $f \in L^2(\lambda \otimes \mathbb{P})$, we know that
$$\int_0^t f(s) \, dB_s$$
is $\mathcal{F}^B_t$-measurable. Here $\mathcal{F}_t^B := \sigma\{B_s; s \leq t\}$ denotes the $\sigma$-algebra generated by the Brownian motion $B$.
If $(W_t)_{t \geq 0}$ and $(B_t)_{t \geq 0}$ are independent Brownian motions, this implies in particular that the random variables
$$\int_0^t f(s) \, dB_s$$
and
$$\int_0^t g(s) \, dW_s$$
are independent for any $t \geq 0$ and $f,g \in L^2(\lambda \otimes \mathbb{P})$ which are progressively measurable with respect to the corresponding (canonical) filtration.