Elementary Ito integrals do not depend on step representation

Let $0=w_0<w_1<\cdots <w_J=T$ be the common refinement of the two partitions, and write $H_s={\sum }_{j=0}^{J-1}{\eta }_j{1}_{(w_j,w_{j+1}]}(s)$ and $H_s^{\mathrm{\prime }}={\sum }_{j=0}^{J-1}{\eta }_j^{\mathrm{\prime }}{1}_{(w_j,w_{j+1}]}(s)$, where ${\eta }_j$ is the coefficient ${\xi }_k$ of the block containing $(w_j,w_{j+1}]$ and similarly for ${\eta }_j^{\mathrm{\prime }}$; then ${\eta }_j$ and ${\eta }_j^{\mathrm{\prime }}$ are bounded and ${\mathcal{F}}_{w_j}$-measurable, because ${\mathcal{F}}_{t_k}\subseteq {\mathcal{F}}_{w_j}$ whenever $t_k\le w_j$.

On each time interval the difference is the constant random variable ${\eta }_j-{\eta }_j^{\mathrm{\prime }}$: $H-H^{\mathrm{\prime }}={\sum }_j({\eta }_j-{\eta }_j^{\mathrm{\prime }})1_{(w_j,w_{j+1}]}$ identically on $[0,T]$, so Tonelli gives $0={\int }_0^T\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\int }_{\mathrm{\Omega }}\mathrm{\mid }H-H^{\mathrm{\prime }}\mathrm{\mid }dPdt={\sum }_{j=0}^{J-1}(w_{j+1}-w_j)E\mathrm{\mid }{\eta }_j-{\eta }_j^{\mathrm{\prime }}\mathrm{\mid }$. Every summand is nonnegative, $w_{j+1}-w_j>0$, and therefore $E\mathrm{\mid }{\eta }_j-{\eta }_j^{\mathrm{\prime }}\mathrm{\mid }=0$, that is, ${\eta }_j={\eta }_j^{\mathrm{\prime }}$ almost surely for each $j$.

On the common refinement the elementary sum of item 5 can be taken over the refined partition: within each original block the increments telescope, ${\sum }_{j:(w_j,w_{j+1}]\subseteq (t_k,t_{k+1}]}{\xi }_k(B_{w_{j+1}}-B_{w_j})={\xi }_k(B_{t_{k+1}}-B_{t_k})$, a finite rearrangement of the defining sum. Applying this to $H$ and to $H^{\mathrm{\prime }}$, the difference of the two terminal sums is $D:=I_T(H)-I_T(H^{\mathrm{\prime }})={\sum }_{j=0}^{J-1}({\eta }_j-{\eta }_j^{\mathrm{\prime }})(B_{w_{j+1}}-B_{w_j})$ identically.

Write ${\zeta }_j:={\eta }_j-{\eta }_j^{\mathrm{\prime }}$ and ${\mathrm{\Delta }}_j:=B_{w_{j+1}}-B_{w_j}$ for each $j$. By [F2] the increment ${\mathrm{\Delta }}_j$ is independent of ${\mathcal{F}}_{w_j}$ and has law $N(0,w_{j+1}-w_j)$. Hence [F3] gives $E[{\mathrm{\Delta }}_j\mid {\mathcal{F}}_{w_j}]=0$ and $E[{\mathrm{\Delta }}_j^2\mid {\mathcal{F}}_{w_j}]=w_{j+1}-w_j$ almost surely, because these conditional expectations of a variable independent of ${\mathcal{F}}_{w_j}$ equal its unconditional mean. Since ${\zeta }_j$ is bounded and ${\mathcal{F}}_{w_j}$-measurable, [F4] yields $E[{\zeta }_j{\mathrm{\Delta }}_j\mid {\mathcal{F}}_{w_j}]={\zeta }_{j}E[{\mathrm{\Delta }}_j\mid {\mathcal{F}}_{w_j}]=0$ and $E[{\zeta }_j^2{\mathrm{\Delta }}_j^2\mid {\mathcal{F}}_{w_j}]={\zeta }_j^2(w_{j+1}-w_j)$ almost surely.

Expanding the square, $E{D}^2={\sum }_{j,j^{\mathrm{\prime }}=0}^{J-1}E[{\zeta }_j{\zeta }_{j^{\mathrm{\prime }}}{\mathrm{\Delta }}_j{\mathrm{\Delta }}_{j^{\mathrm{\prime }}}]$. If $j<j^{\mathrm{\prime }}$, then ${\zeta }_j{\mathrm{\Delta }}_j{\zeta }_{j^{\mathrm{\prime }}}$ is ${\mathcal{F}}_{w_{j^{\mathrm{\prime }}}}$-measurable and integrable, so the tower property and step 3.1 give $E[{\zeta }_j{\zeta }_{j^{\mathrm{\prime }}}{\mathrm{\Delta }}_{j}E[{\mathrm{\Delta }}_{j^{\mathrm{\prime }}}\mid {\mathcal{F}}_{w_{j^{\mathrm{\prime }}}}]]=0$; by symmetry every off-diagonal term vanishes. The diagonal terms satisfy $E[{\zeta }_j^2{\mathrm{\Delta }}_j^2]=E[E[{\zeta }_j^2{\mathrm{\Delta }}_j^2\mid {\mathcal{F}}_{w_j}]]=E[{\zeta }_j^2](w_{j+1}-w_j)$ by step 3.1.

Combining steps 4.1 and 1.2, $E{D}^2={\sum }_{j=0}^{J-1}E[{\zeta }_j^2](w_{j+1}-w_j)\le {\max }_j\mathrm{\parallel }{\zeta }_j{\mathrm{\parallel }}_{\mathrm{\infty }}{\sum }_{j=0}^{J-1}(w_{j+1}-w_j)E\mathrm{\mid }{\zeta }_j\mathrm{\mid }=0$, where the last identity is the Tonelli computation of step 1.2. Since $D$ is square-integrable, $E{D}^2=0$ forces $D=0$ almost surely.

The same computation at a deterministic time $t\in [0,T]$ uses the truncated common partition $\{w_j\wedge t\}$: its nonempty blocks are $(w_j,w_{j+1}\wedge t]$ with $w_j<t$, whose left endpoints are $w_j$ and whose coefficients ${\zeta }_j$ are ${\mathcal{F}}_{w_j}$-measurable, so the truncated sum is elementary; the difference of the truncated sums is ${\sum }_j{\zeta }_j(B_{w_{j+1}\wedge t}-B_{w_j\wedge t})$, each increment is independent of ${\mathcal{F}}_{w_j}$ with second moment $(w_{j+1}\wedge t)-(w_j\wedge t)$, and the off-diagonal terms vanish by the same tower argument. The diagonal sum is ${\sum }_{j}E[{\zeta }_j^2]((w_{j+1}\wedge t)-(w_j\wedge t))\le {\max }_j\mathrm{\parallel }{\zeta }_j{\mathrm{\parallel }}_{\mathrm{\infty }}{\sum }_{j}E\mathrm{\mid }{\zeta }_j\mathrm{\mid }((w_{j+1}\wedge t)-(w_j\wedge t))$, and this is $0$ by the Tonelli identity of step 1.2 restricted to $[0,t]$. Hence $I_t(H)=I_t(H^{\mathrm{\prime }})$ almost surely for every $t$. A deterministic refinement of one partition is the special case in which the two representations are identically equal, and then the shared coefficients cancel in ${\zeta }_j$, recovering that refinement changes nothing. AC enters only through the conditional-expectation facts [F3] and [F4]; the grid, the coefficients and the limits in the argument are all determined by the given representations.