Finite-variation integration by parts for counting-process martingales

For finitely many distinct jump times $0<s_1<\cdots <s_q\le T$, choose disjoint intervals $(u_i,s_i]$ with $u_i\uparrow s_i$. A partition containing all their endpoints has variation at least ${\sum }_i\mathrm{\mid }{x}_{s_i}-x_{u_i}\mathrm{\mid }$, so [F2] and the left limits give ${\sum }_i\mathrm{\mid }\mathrm{\Delta }{x}_{s_i}\mathrm{\mid }\le {\mathrm{Var}}_{[0,T]}(x)$. Hence each set of jumps of size at least $1\mathrm{/}m$ is finite, in its canonical time order, and the nonzero jumps form a countable set with ${\sum }_{0<s\le T}\mathrm{\mid }\mathrm{\Delta }{x}_s\mathrm{\mid }\le {\mathrm{Var}}_{[0,T]}(x)$. Let $J_t={\sum }_{0<s\le t}\mathrm{\Delta }{x}_s$. Absolute convergence makes omitted jump tails uniformly small, so finite jump partial sums converge uniformly to a path $J$; right continuity passes by an $\epsilon \mathrm{/}3$ estimate, and their left limits are Cauchy under the uniform tail bound, so completeness of $\mathbb{R}$ gives left limits for $J$. Absolute convergence gives $\mathrm{\Delta }{J}_s=\mathrm{\Delta }{x}_s$. For any partition, each increment of $J$ is a subseries of its jumps, so ${\mathrm{Var}}_{[0,T]}(J)\le {\sum }_{0<s\le T}\mathrm{\mid }\mathrm{\Delta }{x}_s\mathrm{\mid }$; the same argument bounds the variation of any omitted jump tail by its absolute sum. Therefore $x^c:=x-J$ is continuous and of finite variation. Choose a finite retained jump sum $z$ with omitted tail $r:=J-z$ satisfying ${\mathrm{Var}}_{[0,T]}(r)<\epsilon$, and set $y=x-r=x^c+z$. By [F3] and [F4], the continuous part has squared-increment sum at most ${\omega }_{x^c}(\mathrm{\parallel }{\pi }_n\mathrm{\parallel }){\mathrm{Var}}_{[0,T]}(x^c)\to 0$, where ${\omega }_{x^c}(\delta ):={\sup }_{\mathrm{\mid }u-v\mathrm{\mid }\le \delta }\mathrm{\mid }{x}_u^c-x_v^c\mathrm{\mid }$; sufficiently fine partitions separate the finitely many retained jump times, so ${\sum }_j({\mathrm{\Delta }}_j^{n}y(t){)}^2$ tends to the sum of their squared jumps, with cross terms between each retained jump and the continuous increment in its partition cell tending to zero by the same modulus.

For every partition, expansion of squares and [F5] give $\mathrm{\mid }{\sum }_j({\mathrm{\Delta }}_j^{n}x(t){)}^2-{\sum }_j({\mathrm{\Delta }}_j^{n}y(t){)}^2\mathrm{\mid }\le 2{\mathrm{Var}}_{[0,T]}(y){\mathrm{Var}}_{[0,T]}(r)+{\mathrm{Var}}_{[0,T]}(r{)}^2\le 2{\mathrm{Var}}_{[0,T]}(x)\epsilon +3{\epsilon }^2$, since ${\mathrm{Var}}_{[0,T]}(y)\le {\mathrm{Var}}_{[0,T]}(x)+\epsilon$. Also $x=y+r$ gives $\mathrm{\Delta }{x}_s=\mathrm{\Delta }{y}_s+\mathrm{\Delta }{r}_s$; applying the same finite-list expansion and [F5] to the jump sums, then using their absolute convergence [F1], yields $\mathrm{\mid }[x{]}_t-[y{]}_t\mathrm{\mid }\le 2{\mathrm{Var}}_{[0,T]}(y){\mathrm{Var}}_{[0,T]}(r)+{\mathrm{Var}}_{[0,T]}(r{)}^2\le 2{\mathrm{Var}}_{[0,T]}(x)\epsilon +3{\epsilon }^2$. Since the squared-increment sums for $y$ converge to $[y{]}_t$ by step 1.1, these two error bounds show that the limsup distance between the squared-increment sums for $x$ and $[x{]}_t$ is at most $4{\mathrm{Var}}_{[0,T]}(x)\epsilon +6{\epsilon }^2$. Letting $\epsilon \to 0$ proves ${\sum }_j({\mathrm{\Delta }}_j^{n}x(t){)}^2\to {\sum }_{0<s\le t}(\mathrm{\Delta }{x}_s{)}^2$.

For every finite partition, telescoping differences of squares gives $x_t^2-x_0^2=2{\sum }_j{x}_{t\wedge t_j^n}{\mathrm{\Delta }}_j^{n}x(t)+{\sum }_j({\mathrm{\Delta }}_j^{n}x(t){)}^2$. Step 2.1 gives the limit of the second sum, and rearrangement gives the stated limit of the left-endpoint sums.