Ito isometry for elementary integrands

Refining the partition of $H$ if necessary so that $s$ is a partition point, write the increments of the defining sum between the deterministic times $s\le t$ as $I_t(H)-I_s(H)={\sum }_k{\xi }_k(B_{t\wedge u_{k+1}}-B_{t\wedge u_k})-{\sum }_k{\xi }_k(B_{s\wedge u_{k+1}}-B_{s\wedge u_k})$ over the refined partition $0=u_0<\cdots <u_n=T$; this is a finite rearrangement and does not change the values by the definition of the sums. Term by term, a block with $u_{k+1}\le s$ contributes $0$, a block with $u_k\ge t$ contributes $0$, and every remaining block contributes ${\xi }_k(B_{a_k}-B_{b_k})$ with $b_k=\max (s,u_k)$ and $a_k=t\wedge u_{k+1}$, so that $s\le b_k\le a_k\le T$ and ${\xi }_k$ is ${\mathcal{F}}_{u_k}$-measurable with ${\mathcal{F}}_{u_k}\subseteq {\mathcal{F}}_{b_k}$.

For each such block, $B_{a_k}-B_{b_k}$ is independent of ${\mathcal{F}}_{b_k}$ with mean $0$ and second moment $a_k-b_k$ by (H): $E[B_{a_k}-B_{b_k}\mid {\mathcal{F}}_{b_k}]=0$ and $E[(B_{a_k}-B_{b_k}{)}^2\mid {\mathcal{F}}_{b_k}]=a_k-b_k$ almost surely.

For every remaining block, $E[{\xi }_k(B_{a_k}-B_{b_k})\mid {\mathcal{F}}_s]=E[{\xi }_{k}E[B_{a_k}-B_{b_k}\mid {\mathcal{F}}_{b_k}]\mid {\mathcal{F}}_s]=0$ almost surely, because ${\mathcal{F}}_s\subseteq {\mathcal{F}}_{b_k}$, ${\xi }_k$ is bounded and ${\mathcal{F}}_{b_k}$-measurable, and the inner conditional expectation vanishes by step 1.2; summing the finitely many blocks gives $E[I_t(H)-I_s(H)\mid {\mathcal{F}}_s]=0$, so $E[I_t(H)\mid {\mathcal{F}}_s]=I_s(H)$ almost surely by linearity of conditional expectation and the ${\mathcal{F}}_s$-measurability of $I_s(H)$.

For the variance, write ${\mathrm{\Delta }}_k^t:=B_{t\wedge t_{k+1}}-B_{t\wedge t_k}$ for the blocks of the original partition, so that $I_t(H)={\sum }_k{\xi }_k{\mathrm{\Delta }}_k^t$. For $k<l$ the random variable ${\mathrm{\Delta }}_k^t$ is ${\mathcal{F}}_{t_l}$-measurable, because $t\wedge t_{k+1}\le t_{k+1}\le t_l$, and ${\xi }_k,{\xi }_l\in {\mathcal{F}}_{t_l}$; Each increment is in $L^2$, and $2\mathrm{\mid }ab\mathrm{\mid }\le a^2+b^2$ shows that a product of two increments is integrable; bounded coefficients preserve these bounds. In the off-diagonal use of [F4], take $W={\mathrm{\Delta }}_l^t$ and $Z={\xi }_k{\mathrm{\Delta }}_k^t{\xi }_l$; $W\in L^1$, $ZW\in L^1$, and $ZE[W\mid {\mathcal{F}}_{t_l}]=0$ is integrable. In the diagonal use, $W=({\mathrm{\Delta }}_k^t{)}^2\in L^1$ and $Z={\xi }_k^2$ is bounded. By (H) applied to the increment over the interval $(t_l,t\wedge t_{l+1}]$ (which is empty, hence contributes $0$, when $t\le t_l$), $E[{\mathrm{\Delta }}_l^t\mid {\mathcal{F}}_{t_l}]=0$ almost surely, so the tower property and taking out what is known give $E[{\xi }_k{\mathrm{\Delta }}_k^t{\xi }_l{\mathrm{\Delta }}_l^t]=E[{\xi }_k{\mathrm{\Delta }}_k^t{\xi }_{l}E[{\mathrm{\Delta }}_l^t\mid {\mathcal{F}}_{t_l}]]=0$. For the diagonal terms, the same identity gives $E[{\xi }_k^2({\mathrm{\Delta }}_k^t{)}^2]=E[{\xi }_k^{2}E[({\mathrm{\Delta }}_k^t{)}^2\mid {\mathcal{F}}_{t_k}]]=E[{\xi }_k^2](t\wedge t_{k+1}-t\wedge t_k)$, where the block contributes $0$ when $t\le t_k$.

Summing the diagonal terms of step 2.2 and using $H_s^2={\sum }_k{\xi }_k^2{1}_{(t_k,t_{k+1}]}(s)$ gives $E{I}_t(H{)}^2={\sum }_{k=0}^{m-1}E[{\xi }_k^2](t\wedge t_{k+1}-t\wedge t_k)=E{\int }_0^t{H}_s^{2}ds$, finite because there are finitely many bounded coefficients.

Steps 2.1, 3.1 and [F3] show the martingale and isometry assertions on $[0,T]$. The constant extension from the statement is adapted and continuous; if $s<T<t$, the already proved identity gives $E[I_t(H)\mid {\mathcal{F}}_s]=E[I_T(H)\mid {\mathcal{F}}_s]=I_s(H)$, while for $T\le s\le t$ both sides equal $I_T(H)$. Thus [F5] makes the extended process a continuous square-integrable martingale on $[0,\mathrm{\infty })$. Independence of the representation is the content of item 6. AC is used only through the conditional-expectation facts [F2], [F4] and the Brownian interface (H); the partition, the blocks and the sums are fixed by the representation.