I've been re-reading one passage from the book of Bernt Øksendal for a whole week and still can't understand one moment in proof of that theorem:
Lemma 6.2.7.: $\hat{X_t} = E[X_t] + \int_{0}^{t}\frac{\partial}{\partial s} E[X_tR_s]\, dR_s $
My question about this moment:
$E[X_t\int_{0}^{t}f(s)\, dR_s] = E[\hat{X_t}\int_{0}^{t}f(s)\, dR_s] = > E[\int_{0}^{t}g(s)\,dR_s \int_{0}^{t}f(s)\, dR_s] = > E[\int_{0}^{t}g(s)f(s)\, ds]$
for all $f$ from $L^2[0,t]$
I understand, that $E[\int_{0}^{t}g(s)f(s)\, ds]$ they got from $E[\int_{0}^{t} g(s)f(s) \, dRs]$ by Ito's isometry.
But how did they get this $E[\int_{0}^{t} g(s)f(s) \,dRs]$ from $E[\int_{0}^{t} g(s)\,dRs \int_{0}^{t} f(s)\, dRs]$? I cant understand only this step here.
definitions:
$dR_s = \frac{1}{D(t)} dN_t(\omega); t \ge0, R_0=0,N_t - $ innovation process $N_t$: $N_t = Z_t - \int_{0}^{t} G(s)\hat{X_s} \,ds$
$G(t),D(t) \in \mathbb{R}$ - functions from representation of linear observations:
$dZ_t = G(t)X_tdt + D(t)dV_t$, $V_t$ - Brownian motion
$\hat{X_t}$ is projection of $X_t$ on $L(R,t) $space