I am trying to understand the proof of the Ito Isometry, but somwhow do not understand the follwing two steps:
Setup: $B(t)$ is a Brownian motion and $\{\mathcal{F}_t\}_{a\leq t\leq b}$ a filtration such that:
- For each $t$, $B(t)$ is $\mathcal{F}_{t}$-measurable.
- For any $s\leq t$, the random variable $B(t)-B(s)$ is independent of the $\sigma$-field $\mathcal{F}_{s}$.
Let $\{a=t_1<t_2<\ldots<t_n=b\}$ be a partition of $[a,b]$ and $\xi_i, 0\leq i \leq n-1$, bounded random variables such that each $\xi_i$ is $\mathcal{F_{t_i}}$-measurable.
Then during the proof the author says
$E\{\xi_{i-1}(B(t_i)-B(t_{i-1}))\}=E\{E[\xi_{i-1}(B(t_i)-B(t_{i-1}))|\mathcal{F}_{t_{i-1}}]\}$.
Why is this true? I´ve looked through several properties of the conditional expectation in other books but could not find one that could explain this step.
Then later the author says (for $i<j$):
$E\{\xi_{i-1}\xi_{j-1}(B(t_i)-B(t_{i-1}))(B(t_j)-B(t_{j-1}))\}=E\{E[\xi_{i-1}\xi_{j-1}(B(t_i)-B(t_{i-1}))(B(t_j)-B(t_{j-1}))]|\mathcal{F}_{t_{j-1}}\}\\ =E\{\xi_{i-1}\xi_{j-1}(B(t_i)-B(t_{i-1}))E[(B(t_j)-B(t_{j-1}))]|\mathcal{F}_{t_{j-1}}\} $
The first equality then probably can be justified by the same explanation as above, but why is the second equality true? I know that $E[XY|\mathcal{G}]=YE[X|\mathcal{G}]$ if $Y$ is $\mathcal{G}$-measurable. But why is $\xi_{i-1}\xi_{j-1}(B(t_i)-B(t_{i-1}))$ then $\mathcal{F}_{t_{j-1}}$ measurable?
Many thanks!