Ito's Isometry and integrands in $\lambda^2_{\text{loc}}$

Question

Does the Ito isometry $$E\left[\left(\int_0^t\phi_s \, dB_s\right)^2\right]=E\left[\int_0^t(\phi_s)^2 \, ds\right]$$ hold also when $\phi$ in $\lambda^2_{\text loc}$ (namely the integrand $\phi$ is progressive and for every $t$, $\int_0^t \phi_s^2\,ds< \infty$) (the larger class of integrands)?

Answer 1

Yes, Itô's isometry holds for locally square integrable functions. Take a progressively measurable function $\phi$ such that $\int_0^t \phi(s)^2 \, ds < \infty$ and define stopping times by

$$\sigma_n := \inf\{t>0; \int_0^t \phi(s)^2 \, ds \geq n\}.$$

From $$\sigma_n(\omega)>t \iff \int_0^t \phi(s,\omega)^2 \, ds < n$$ we find that $\phi_n(t) := \phi (t) \cdot 1_{[0,\sigma_n)}(t)$ satisfies $$\mathbb{E} \int_0^t \phi_n(s)^2 \, ds \leq n$$ for all $t \geq 0$. Hence, $\phi_n \in L^2$ and, by Itô's isometry, $$\mathbb{E} \left| \int_0^t \phi_n(s) \, dB_s \right|^2 = \mathbb{E} \int_0^t \phi_n(s)^2 \, ds<\infty.\tag{1}$$ Moreover, by the definition of the Itô integral for locally square integrable functions, the identity $$\int_0^t \phi(s) \, dB_s = \int_0^t \phi_n(s) \, dB_s$$ holds on $\{t \leq \sigma_n\}$. Since $\sigma_n \uparrow \infty$, this means that $$\int_0^t \phi(s) \, dB_s = \sup_{n \in \mathbb{N}} \int_0^t \phi_n \, dB_s. \tag{2}$$ Applying the monotone convergence theorem (MCT), we get

\begin{align*} \mathbb{E} \left| \int_0^t \phi(s) \, dB_s \right|^2& \stackrel{\text{MCT}}{=} \sup_{n \in \mathbb{N}} \mathbb{E} \left| \int_0^t \phi_n(s) \, dB_s \right|^2 \\ &\stackrel{(1)}{=} \mathbb{E} \int_0^t \phi_n(s)^2 \, ds \\ &\stackrel{\text{MCT}}{=} \mathbb{E} \int_0^t \phi(s)^2 \,ds, \end{align*} where we used in the last step that $\phi_n \uparrow \phi$ as $n \to \infty$.