# Why is that a Cauchy sequence in $L^2(\mathbb{P})$, by $(1)$?

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).
• Why does $(2)$ imply $\mathbb{E}\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]\to0$? – Strictly_increasing Nov 7 '20 at 18:18
• 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? – Strictly_increasing Nov 7 '20 at 18:36
• Sorry, I was accidentally calling $\phi$ what it was suppose to be $f.$ I fixed it in the answer. – UBM Nov 7 '20 at 18:48