Let $(\Omega, \mathcal{F}, \mathbb{P} )$ be the probability space, and {$W_t,0\leq t\leq T$} is a Brownian motion and $\mathcal{F_t}^W$ is the canonical filtration.
For the $f(t)\in L^2([0, T])$(a deterministic function), we can define the stochastic integral $\int_s^tf(u)dW_u$,where $0<s<t<T$.
So I am wondering if the $\int_s^tf(u)dW_u$ is independent with the $\mathcal{F_s}$?
Because by the definition of the stochastic integral, $\int_s^tf(u)dW_u$ is the $L^2$-limit of the stochastic integral of simple process, which could have the form like $\sum_{i=0}^{n-1}f(u_i)(W_{u_{i+1}}-W_{u_i})$, where $\{u_{i}\}_{i=1}^n$ is partition of the interval $[s, t]$: $s=u_0<u_1<...<u_n=t$.
For $\sum_{i=0}^{n-1}f(u_i)(W_{u_{i+1}}-W_{u_i})$, we know that it is independent with the $\mathcal{F_s}$. So I just want to know if the independence still holds between the $L^2$-limit and the $\sigma$-algebra.
Thank you!
Of course. In order to see this, just write that for $A\in F_s$,
\begin{align} E\left[ 1_A \int_s^t fdW \right] &= \lim E\left[1_A \sum f(u_i)(W_{u_{i+1}} - W_{u_i}) \right] \\&= \lim E\left[1_A\right] E\left[ \sum f(u_i)(W_{u_{i+1}} - W_{u_i})\right] \\&= E\left[ 1_A\right] E\left[\int_s^t fdW\right] \end{align}