Itô integral of a step on a stopping-time interval

First verify that $H_s=\xi 1_{(\tau ,\sigma ]}(s)$ is predictable pointwise. For $s>0$ and Borel $C\subseteq \mathbb{R}$, put $A_C=\{\xi \in C\}\in {\mathcal{F}}_{\tau }$. Then $A_C\cap \{\tau <s\}={\bigcup }_{q\in \mathbb{Q}\cap [0,s)}{A}_C\cap \{\tau \le q\}\in {\mathcal{F}}_s,$ and $\{\sigma \ge s\}=\{\sigma <s{\}}^c\in {\mathcal{F}}_s$, since $\{\sigma <s\}={\bigcup }_{q\in \mathbb{Q}\cap [0,s)}\{\sigma \le q\}$. With $E_s=\{\tau <s\}\cap \{\sigma \ge s\}$, both $E_s$ and $A_C\cap E_s$ belong to ${\mathcal{F}}_s$. The preimage $\{H_s\in C\}$ is $A_C\cap E_s$ if $0\mathrm{\notin }C$, and $(A_C\cap E_s)\cup E_s^c$ if $0\in C$; thus it belongs to ${\mathcal{F}}_s$. Thus $H$ is adapted. Its paths are left-continuous and $H_0=0$. On the nested deterministic grids $t_j^{(r)}=jT{2}^{-r}$, define $P_0^{(r)}=0$ and $P_s^{(r)}=H_{t_j^{(r)}}$ on each cell $(t_j^{(r)},t_{j+1}^{(r)}]$. Each $P^{(r)}$ is predictable by [F2]; the left endpoints of the cells containing any fixed $s>0$ increase to $s$ from below, so left-continuity gives $P_s^{(r)}\to H_s$. Hence $H$ is a pointwise limit of predictable processes and is predictable. Now let ${\delta }_n=T{2}^{-n}$, and round each stopping time upward to the next point of this deterministic grid: ${\tau }_n={\delta }_n\lceil \tau \mathrm{/}{\delta }_n\rceil$ and ${\sigma }_n={\delta }_n\lceil \sigma \mathrm{/}{\delta }_n\rceil$. The stopping-time test shows these are finite-valued stopping times, ${\tau }_n\le {\sigma }_n$, ${\tau }_n\downarrow \tau$, and ${\sigma }_n\downarrow \sigma$. On a grid cell $(t_k,t_{k+1}]$, the process $H^n=\xi 1_{({\tau }_n,{\sigma }_n]}$ has coefficient $\xi 1_{\{{\tau }_n\le t_k<{\sigma }_n\}}$. This coefficient is ${\mathcal{F}}_{t_k}$-measurable: the active event is in ${\mathcal{F}}_{t_k}$, and $\xi$ restricted to $\{{\tau }_n\le t_k\}=\{\tau \le t_k\}$ is ${\mathcal{F}}_{t_k}$-measurable by [F2]. Thus $H^n$ is a finite predictable step process.

The intervals $({\tau }_n,{\sigma }_n]$ and $(\tau ,\sigma ]$ differ in Lebesgue measure by at most $2{\delta }_n$ on each path. Hence $\mathbb{E}{\int }_0^T\mathrm{\mid }{H}_s^n-H_s{\mathrm{\mid }}^{2}ds\le 2{\delta }_n\mathbb{E}\mathrm{\mid }\xi {\mathrm{\mid }}^2⟶0.$ Also $\mathbb{E}{\int }_0^T{H}_s^{2}ds=\mathbb{E}[\mathrm{\mid }\xi {\mathrm{\mid }}^2(\sigma -\tau )]\le T\mathbb{E}\mathrm{\mid }\xi {\mathrm{\mid }}^2<\mathrm{\infty }$. The predictable $L^2$ space is closed under almost-everywhere limits: since the displayed squared errors are summable in $n$, Tonelli implies $H^n\to H$ for $dt\otimes P$-almost every $(s,\omega )$; therefore $H$ has a predictable representative.

Apply [F1] to the deterministic-grid representation of $H^n$. The active cells are consecutive, so their finite sum telescopes and gives ${\int }_0^t{H}_s^{n}d{W}_s=\xi (W_{t\wedge {\sigma }_n}-W_{t\wedge {\tau }_n})$ almost surely for every fixed $t$. By the isometry in [F1] and step 1.2 the left side converges in $L^2$ to ${\int }_0^t{H}_{s}d{W}_s$. By [F3], path continuity and the endpoint convergence imply that the right side converges almost surely to $\xi (W_{t\wedge \sigma }-W_{t\wedge \tau })$. Uniqueness of limits in probability identifies the two limits almost surely.

Apply step 2.1 at the countable dense set of rational times in $[0,T]$ and intersect those measurable probability-one events with the continuity events from [F3] and [F4]. The two continuous processes then agree at every time on that one event. If $\tau =\sigma$, both sides vanish; $\tau =0$, $\sigma =T$, and $t=0$ are included by the same formula. AC is used only through [A1].