Itô Isometry proof Let $\{\phi_n\}$ a sequencie of functions in the space $\mathcal{V}(S,T)$ of those functions that verify:

*

*$f(t,\omega)$ is $\mathcal{B} \times\mathcal{F}$ mesurable.

*$\mathbb{E}[\int_S^T|f(t,\omega)|^2 dt] < \infty$.



*$ f$ is $\mathcal{F}_t$-adapted.
Moreover, for each $n$, $E[(\int_{S}^{T} \phi_n(t,\omega) dB_t)^2] = E[\int_{S}^{T}\phi_n^2(t,\omega) dt]$.
Finally,let $f \in \mathcal{V}$ and supose $\lim_n E[\int_S^T(f(t,\omega)-\phi_n(t,\omega))^2 ds] = 0.$
I want to prove:
$E[(\int_{S}^{T} f(t,\omega) dB_t)^2] = E[\int_{S}^{T}f^2(t,\omega) dt]$.

 A: We can use almost surely $\phi $ instead of $f(t,w)$ ,and$\phi $ is bounded and elementary function. let's define $\phi(t,w)=\Sigma \phi_i$
for $E[(\int_S^T\phi(t,w)dB_t)^2]$ first discreitize , define $\Delta B_i=B_{t_{i+1}}-B_{t_i},\Delta B_j=B_{t_{j+1}}-B_{t_j}$ over intervals $[t_{j},t_{j+1}],[t_{i},t_{i+1}]$
so $$E[\phi_i\phi_j\Delta B_i\Delta B_j]=\\\begin{cases}0 & i<j\\E[\phi_i^2(\Delta B_i)^2] & i=j\\0  &i>j\end{cases}\\=\begin{cases}0 & i<j\\E[\phi_i^2(\Delta t_i)] & i=j\\0  &i>j\end{cases}\\
=\begin{cases}0 & i<j\\E[\phi_i^2]\Delta t_i & i=j\\0  &i>j\end{cases}$$ with respect that $$[t_{j},t_{j+1}],[t_{i},t_{i+1}]$$ have not cross secton when $ i\neq j$  now
$$E[(\int_{S}^{T}\phi^2(t,w) dB_t)^2]=\\\Sigma_{i}\Sigma_{j}E[\phi_i\phi_j\Delta B_i\Delta B_j]=\\\Sigma_{i=j}E[\phi_i\phi_j\Delta B_i\Delta B_j]+\underbrace{\Sigma_{i\neq j}E[\phi_i\phi_j\Delta B_i\Delta B_j]}_{0}\\
= \Sigma_{i=j}E[\phi_i\phi_i\Delta B_i\Delta B_i]\\=\Sigma_{i=j}E[(\phi_i)^2(\Delta B_i)^2]\\=
\Sigma_{i=j}E[(\phi_i)^2(\Delta t_i)]\\=
\Sigma_{i=j}E[(\phi_i)^2]\Delta t_i\\=E[(\int_S^T(\phi(t,w))^2dt]$$
Remark $(\Delta B_t)^2\to \Delta t$  and $\Delta B_i,\Delta B_j$ are independent when $i\neq j$
