A square-integrable predictable step Itô integral is its finite sum

For every Borel set $C\subseteq \mathbb{R}$, $H^{-1}(C)$ is the finite union of the predictable rectangles $(t_k,t_{k+1}]\times \{{\xi }_k\in C\}$, together with $\{0\}\times \mathrm{\Omega }$ when $0\in C$; hence [F2] makes $H$ predictable. The intervals partition $(0,T]$, and [F3] makes the time-zero endpoint irrelevant to the product integral, so $E{\int }_0^T{H}_s^{2}ds={\sum }_{k=0}^{N-1}(t_{k+1}-t_k)E\mathrm{\mid }{\xi }_k{\mathrm{\mid }}^2<\mathrm{\infty }$.

For $n\ge 1$, clip each coefficient by ${\xi }_{k,n}=(-n)\vee ({\xi }_k\wedge n)$ and put $H_s^n={\sum }_k{\xi }_{k,n}{1}_{(t_k,t_{k+1}]}(s)$. Each $H^n$ is bounded and elementary by [F3]. Since $\mathrm{\mid }{\xi }_{k,n}-{\xi }_k{\mathrm{\mid }}^2\le \mathrm{\mid }{\xi }_k{\mathrm{\mid }}^2{1}_{\{\mathrm{\mid }{\xi }_k\mathrm{\mid }>n\}}$ and $E\mathrm{\mid }{\xi }_k{\mathrm{\mid }}^2<\mathrm{\infty }$, [F7] applied to $\mathrm{\mid }{\xi }_k{\mathrm{\mid }}^2{1}_{\{\mathrm{\mid }{\xi }_k\mathrm{\mid }\le n\}}\uparrow \mathrm{\mid }{\xi }_k{\mathrm{\mid }}^2$ gives $E\mathrm{\mid }{\xi }_{k,n}-{\xi }_k{\mathrm{\mid }}^2\to 0$. The finite partition then gives $\mathrm{\parallel }{H}^n-H{\mathrm{\parallel }}_{L^2(dt\otimes P)}^2={\sum }_k(t_{k+1}-t_k)E\mathrm{\mid }{\xi }_{k,n}-{\xi }_k{\mathrm{\mid }}^2\to 0$.

For each fixed $t\in [0,T]$, [F5] defines ${\int }_0^t{H}_{s}d{W}_s$ as an $L^2(P)$ limit of the elementary integrals of $H^n{1}_{[0,t]}$. By [F4] those approximants are $S_t^n={\sum }_k{\xi }_{k,n}(W_{t\wedge t_{k+1}}-W_{t\wedge t_k})$. Hence $S_t^n\to {\int }_0^t{H}_{s}d{W}_s$ in $L^2(P)$.

Put ${\delta }_k(t)=(t\wedge t_{k+1})-(t\wedge t_k)\ge 0$. When ${\delta }_k(t)=0$ the increment in the $k$th summand vanishes. When ${\delta }_k(t)>0$, [F1] gives its $N(0,{\delta }_k(t))$ law, [F11] gives second moment ${\delta }_k(t)$, and [A1] makes it independent of ${\xi }_{k,n}-{\xi }_k$. Thus the $L^2$ norm of that summand's difference between $S_t^n$ and $S_t$ is $\sqrt{{\delta }_k(t)}\mathrm{\parallel }{\xi }_{k,n}-{\xi }_k{\mathrm{\parallel }}_2$. The triangle inequality [F10] and the finite number of intervals imply $S_t^n\to S_t$ in $L^2(P)$. Since also $S_t^n\to {\int }_0^t{H}_{s}d{W}_s$ by step 2.1, the triangle inequality gives $\mathrm{\parallel }{S}_t-{\int }_0^t{H}_{s}d{W}_s{\mathrm{\parallel }}_2=0$, and separation [F10] makes these classes equal almost surely.

Extend $H$ by zero after $T$; [F9] makes this extension predictable and its energy stays finite on every finite horizon. The process $S$ has continuous paths on the common Brownian continuity event [F12], because it is a finite sum of fixed random coefficients times stopped Brownian increments. Let $M$ be the continuous integral version supplied by [F6]. Step 3.1 gives $M_t=S_t$ almost surely at each $t$ in the countable dense set of [F8]; intersecting those measurable probability-one events with the continuity events and using continuity gives equality for every $t\in [0,T]$ on one measurable probability-one event. If the probability space is complete, the complement of the full equality set is a subset of the measurable null complement of this event, hence is measurable and null; the versions are then indistinguishable. AC is inherited through [F6] as recorded in [A2].