So I am trying to fill some deatils of the Itô's isometry for a simple function $g(t)$ that is defined in a $k-$partition of $[a,b]$ and such that $\xi_i=g(t_i)$ for $(t_i, t_{i+1}]$. I have that: $$\mathbb{E} [ \vert \int_a^b g(t) dB(t) \vert^2 ] = \sum_{0\leq i,j \leq k-1} \mathbb{E} [\xi_i \xi_j (B(t_{i+1})-B(t_i))]=\sum_{0\leq i\leq k-1} \mathbb{E} [\xi_i^2 (B(t_{i+1})-B(t_i))^2]=\sum_{0\leq i\leq k-1} \mathbb{E} [\xi_i^2] \mathbb{E}[(B(t_{i+1})-B(t_i))^2]=\mathbb{E}[\int_a^b\vert g(t) \vert ^2 dt]$$
And I want to know what happens with the mixed terms and why are they $0$?
I defined the Brownian motion for a filtration $\mathcal{F}$ with the usual conditions such that $\mathcal{F}(t_i)\subset \mathcal{F}(t_j)\subset \mathcal{F}$ if $i<j$ and so, $B(t_{j+1})-B(t_j)$ is also independent from $\mathcal{F}(t_j)$. Also, I know that $B(t_{j+1})-B(t_j)$ is independent from $\xi_i\xi_j(B(t_{i+1})-B(t_i))$ if $i<j$. Is this enough to write that: $$\mathbb{E}[\xi_i\xi_j(B(t_{i+1})-B(t_i))(B(t_{j+1})-B(t_j)]=\mathbb{E}[\xi_i\xi_j(B(t_{i+1})-B(t_i))\vert \mathcal{F(t_j)}]\mathbb{E}[(B(t_{j+1})-B(t_j)]$$
Or may I use the tower property? $$\mathbb{E}[\xi_i\xi_j(B(t_{i+1})-B(t_i))(B(t_{j+1})-B(t_j)]=\mathbb{E}[\xi_i\xi_j(B(t_{i+1})-B(t_i))\mathbb{E}[(B(t_{j+1})-B(t_j)\vert \mathcal{F}(t_j)]]$$
In any case, $\textbf{I think that I am conditioning my expectation to a coarser filter}\, \; \mathcal{F}(t_j)\subset \mathcal{F}$: $$\mathbb{E}[\xi_i\xi_j(B(t_{i+1})-B(t_i))(B(t_{j+1})-B(t_j))]=\mathbb{E}[\xi_i\xi_j(B(t_{i+1})-B(t_i))(B(t_{j+1})-B(t_j))\vert \mathcal{F}(t_j)]$$
I am not even sure if I can do so.
For the first option I think that I need $B(t_{j+1})-B(t_j)$ to be independent from $\sigma(\xi_i\xi_j(B(t_{i+1})-B(t_i)), \mathcal{F(t_j)})$ not just from each one individually and I'm not sure if $B(t_{i+1})-B(t_i)$ is independent from $\mathcal{F}(t_j)$. However, my intuition is that since $\mathcal{F}(t_i)\subset \mathcal{F}(t_j)$, then if I condition to $\mathcal{F}(t_{j})$ it is as if I know the values $B(t)$ up to the time $t_j$, so in this case as $i<j$, the values of $B(t_{i+1})-B(t_i)$ would be deterministic and hence $\mathcal{F}(t_j)-$measurable.
I will really appreciate any help, specially with the part written in bold font.
Assume that $\xi_i$ in $\mathcal{F}(t_i)$ measurable for all $i$. Also suppose $i<j$, i.e. $t_i < t_{i+1} \le t_j < t_{j+1} $. Then by tower property $$ E[\xi_i \xi_j (B(t_{i+1}) - B(t_i))(B(t_{j+1}) - B(t_j))] \\= E \Bigg\{ E \Big[\xi_i \xi_j (B(t_{i+1}) - B(t_i))(B(t_{j+1}) - B(t_j)) \mid \mathcal{F}(t_j)\Big]\Bigg\} \\= E \Bigg\{ \xi_i \xi_j (B(t_{i+1}) - B(t_i)) E \Big[(B(t_{j+1}) - B(t_j)) \mid \mathcal{F}(t_j)\Big]\Bigg\} \\ = E \Bigg\{ \xi_i \xi_j (B(t_{i+1}) - B(t_i)) E \Big[B(t_{j+1}) - B(t_j) \Big]\Bigg\} = 0. $$ The last line follows from the fact that $B(t_{j+1}) - B(t_j)$ and $\mathcal{F}(t_j)$ are independent.
So you have to modify your simple function to be $g = \sum_i \xi_i 1_{(t_i, t_{i+1}]}$.