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?


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).

  • $\begingroup$ Why does $(2)$ imply $\mathbb{E}\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]\to0$? $\endgroup$ – Strictly_increasing Nov 7 '20 at 18:18
  • $\begingroup$ @Strictly_increasing: I edited my previous answer, hope it helps. $\endgroup$ – UBM Nov 7 '20 at 18:34
  • $\begingroup$ But $(2)$ establishes that $\mathbb{E}\left[\int_S^T|f-\phi_n|^2 dt\right]\to0$ for $f\in\mathcal{V}$, and not that $\mathbb{E}\left[\int_S^T|\phi-\phi_n|^2 dt\right]\to0$, doesn't it? $\endgroup$ – Strictly_increasing Nov 7 '20 at 18:36
  • 1
    $\begingroup$ Sorry, I was accidentally calling $\phi$ what it was suppose to be $f.$ I fixed it in the answer. $\endgroup$ – UBM Nov 7 '20 at 18:48

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