I quote Øksendal (2003).
Let $\mathcal{V}=\mathcal{V}(S,T)$ be the class of functions $f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$ such that $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$-measurable (where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $[0,\infty)$), $f(t,\omega)$ is $\mathcal{F}_t$-adapted and $\mathbb{E}\bigg[\int_{S}^T f(t,\omega)^2 dt\bigg]<\infty$. […]
Starting from a probability space $\left(\Omega,\mathbb{P},\mathcal{E}\right)$ and a Brownian motion $\left(B_t\right)_{t\ge0}$, if $\phi(t,\omega)$ is bounded and elementary, then
$$\mathbb{E}\left[\left(\int_S^T\phi(t,\omega)dB_t(\omega)\right)^2\right]=\mathbb{E}\left[\int_S^T\phi(t,\omega)^2 dt\right]\tag{1}$$ […]
If $f\in\mathcal{V}$ one can show that it is possible to choose elementary functions $\phi_n\in\mathcal{V}$ such that:
$$\mathbb{E}\left[\int_S^T|f-\phi_n|^2 dt\right]\to0\tag{2}$$
Then, define
$$\mathcal{I}\left[f\right](\omega)=\int_S^T f(t,\omega)dB_t(\omega)=\lim_{n\to\infty}\int_S^T\phi_n(t,\omega)dB_t(\omega)\tag{3}$$
The limit exists as an element of $L^2(\mathbb{P})$, since $\left\{\int_S^T\phi_n(t,\omega)dB_t(\omega)\right\}$ forms a Cauchy sequence in $L^2(\mathbb{P})$, by $(1)$.
What I cannot understand is the statement in bold above. Why is that true?
Best Answer
Note that (2) implies: $$\mathbb{E}\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]\to0. \tag{4}$$ Thus \begin{align} E \left|\int_S^T \phi_n(t)dB_t-\int_S^T \phi_m(t)dB_t \right|^2 &=E \left|\int_S^T (\phi_n(t)-\phi_m(t))dB_t \right|^2 \\ &=E \int_S^T |\phi_n(t)-\phi_m(t)|^2 dt \rightarrow 0. \end{align} The last equality follows from the Ito isometry and the convergence to zero follows follows from (4). This proves that $\{\int_S^T \phi_n(t)dB_t \}$ is a Cauchy sequence in $L^2$.
Edit: proof that (2) implies (4): \begin{align} E\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]&=E\left[\int_S^T|\phi_n(t)-f(t)+f(t)-\phi_m(t)|^2 dt\right] \\ &\leq E\left[\int_S^T 2|\phi_n(t)-f(t)|^2+2|f(t)-\phi_m(t)|^2 dt\right] \\ &= 2E\left[\int_S^T |\phi_n(t)-f(t)|^2dt\right]+2E\left[\int_S^T|f(t)-\phi_m(t)|^2 dt\right] \rightarrow 0. \end{align} The '$\leq$' follows from the inequality $(a+b)^2 \leq 2a^2 + 2b^2$ and the convergence to zero from (2).