Itô Isometry proof

Question

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

1. $f(t,\omega)$ is $\mathcal{B} \times\mathcal{F}$ mesurable.
2. $\mathbb{E}[\int_S^T|f(t,\omega)|^2 dt] < \infty$.

3. $ 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: $E[(\int_{S}^{T} f(t,\omega) dB_t)^2] = E[\int_{S}^{T}f^2(t,\omega) dt]$.

Answer 1

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 $i\neq j$