I want to show the following claim:
Let $B$ be a one-dimensional Brownian motion and let $$I(\phi):=\int_0^1 \phi(s) \text{d}B_s.$$ Show that $\mathbb{E}(I(\phi))=0$ and $\mathbb{V}(I(\phi))=\int_0^1(\phi(s))^2\text{d}s$.
I want to use characteristic functions to prove this.
I already showed that $$\int_0^1 \phi(s) \text{d}B_s = -\int_0^1 B_s(\omega) \text{d}\phi(s),$$ where the right hand integral is constructed as a Riemann-Stieltjes integral:$$\int_0^1B_s(\omega) \text{d}\phi(s) = \lim_\limits{|\mathscr{P}|\rightarrow 0} \sum_{[u,v) \in \mathscr{P}} B_{\xi(u,v)}(\omega)\phi_{u,v}.$$
So, $$\mathbb{E}\exp(itI(\phi))= \mathbb{E}\exp \bigg[-it \big(\lim_\limits{|\mathscr{P}|\rightarrow 0} \sum_{[u,v) \in \mathscr{P}} B_{\xi(u,v)}(\omega)\phi_{u,v} \big) \bigg].$$
Next step will probably be to prove that the limit can be taken out of the expectation but how do I proceed from then on? Additionally, I know that the summands are Gaussian and hence the sum is again gaussian.
Thanks in advance!