A question about the space of functions the Ito integral is defined on

Question

Consider the probability space $(\Omega,\mathcal{F}\,P)$ and the filtration $\mathcal{F}_t$.

In Oksendal's book the Ito integral $I(f)(\omega)=\int_S^T f(t,\omega)dB_t(\omega)$ is defined on a space $\mathcal{V}(S,T)$. This is the space of functions such that

1. $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$ measurable where $\mathcal{B}$ is the borel sigma algebra on $[0,\infty)$.
2. $f(t,\cdot)$ is $\mathcal{F}_t$ adapted.
3. $E[\int_S^Tf^2(t,\omega)dt]<\infty$

Then the Ito integral of $f$ is defined by $$\int_S^T f(t,\omega)dB_t(\omega):=\lim_{n\to \infty}\int_S^T \phi_n (t,\omega) dB_t(\omega)\,\,in\,\, L^2(P)$$ where $\{\phi_n\}$ is sequence of elementary functions such that $$E\Big[\int_S^T (f(t,\omega)-\phi_n(t,\omega))^2 dt\Big]\to 0 \,\,as\,\,n\to\infty$$

Now if we consider the space $\mathcal{W}$ of adapted elementary functions such that $E[\int_S^T\phi^2(t,\omega)dt]<\infty$ and define the an inner product $$\langle \phi,\psi\rangle_{\mathcal{W}}=E[\int_S^T\phi \psi dt]$$. Is $\mathcal{V}$ the closure of $\mathcal{W}$ with respect to the norm defined by the above inner product?

Answer 1

Yes, $\mathcal{V}$ is the closure of $\mathcal{W}$ with respect to the norm induced by the inner product. You can find a proof for instance in the book Brownian Motion - An Introduction to Stochastic Processes by Schilling & Partzsch (Theorem 15.20 in the 2nd edition).