Brackets of a compensated simple counting process

Since $A$ is continuous, $\mathrm{\Delta }{M}_s=\mathrm{\Delta }{N}_s$; each jump of $N$ is zero or one, so the jump-square definition gives $[M{]}_t={\sum }_{0<s\le t}(\mathrm{\Delta }{N}_s{)}^2=N_t$, using $N_0=0$.

Let $S=N+A$ and ${\tau }_n=\inf \{t:S_t\ge n\}\wedge T$ for integers $n\ge 1$. The adapted nondecreasing càdlàg process $S$ satisfies $\{{\tau }_n\le t\}=\{S_t\ge n\}$ for $t<T$, so ${\tau }_n$ is bounded stopping; at the first crossing, continuity of $A$ and unit jumps of $N$ give $S_{{\tau }_n}\le n+1$, hence $\mathrm{\mid }{M}_{t\wedge {\tau }_n}\mathrm{\mid }\le n+1$. By [F5], $M^{{\tau }_n}$ is a square-integrable martingale.

The variables $M_T$ and $A_T$ are integrable, so $N_T=M_T+A_T$ is integrable; since $M$ is a martingale starting at zero, $\mathbb{E}{N}_T=\mathbb{E}{A}_T<\mathrm{\infty }$. The pathwise total variation of $M$ is at most $N_T+A_T$, hence is integrable.

Choose nested deterministic partitions ${\pi }_m=\{0=t_0^m<\cdots <t_{q_m}^m=T\}$ with mesh tending to zero. The coefficients $H_j^{n,m}=M_{t_j^m\wedge {\tau }_n}$ are ${\mathcal{F}}_{t_j^m}$-measurable and bounded by $n+1$, so [F3] makes the step integral against $M^{{\tau }_n}$ a martingale. The partition identity [F4] gives $(M_{t\wedge {\tau }_n}{)}^2-{\sum }_j(M_{t\wedge t_{j+1}^m\wedge {\tau }_n}-M_{t\wedge t_j^m\wedge {\tau }_n}{)}^2=2I_t(H^{n,m})$. The stopped process has the same unit jumps as $N^{{\tau }_n}$, so the squared-increment sum converges to $[M^{{\tau }_n}{]}_t=N_{t\wedge {\tau }_n}$ by [F4] and step 1.1. Also $\mathrm{Var}(M^{{\tau }_n}{)}_T\le N_{T\wedge {\tau }_n}+A_{T\wedge {\tau }_n}\le n+1$, so the terms are bounded by a deterministic constant depending only on $n$; dominated convergence in the martingale event identities proves $(M_{t\wedge {\tau }_n}{)}^2-N_{t\wedge {\tau }_n}$ is a martingale.

Since $S_T=N_T+A_T<\mathrm{\infty }$ a.s., ${\tau }_n=T$ eventually on each path. On nested deterministic grids, [F6] gives $\mathbb{E}{\max }_{u\in {\pi }_m}\mathrm{\mid }{M}_u{\mathrm{\mid }}^2\le 4\mathbb{E}\mathrm{\mid }{M}_T{\mathrm{\mid }}^2$; right continuity identifies the increasing grid limit with ${\sup }_{u\le T}\mathrm{\mid }{M}_u\mathrm{\mid }$, and [F8] gives $\mathbb{E}{\sup }_{u\le T}\mathrm{\mid }{M}_u{\mathrm{\mid }}^2<\mathrm{\infty }$. Thus $R={\sup }_{u\le T}\mathrm{\mid }{M}_u{\mathrm{\mid }}^2+N_T$ is integrable and dominates every stopped value $\mathrm{\mid }(M_{t\wedge {\tau }_n}{)}^2-N_{t\wedge {\tau }_n}\mathrm{\mid }$. For $B\in {\mathcal{F}}_s$, [F7] and the event identities from step 2.2 let $n\to \mathrm{\infty }$, proving $M^2-N$ is a martingale.

Since $N-A=M$ is a martingale, $M^2-A=(M^2-N)+M$ is a martingale by step 3.1 and linearity, so [F2] and [F1] make $A$ a scalar bracket candidate; uniqueness gives $⟨M⟩=A$. Also $[M,M]=[M]=N$, so $[M,M]-A=N-A=M$ is a martingale, making $A$ a cross-bracket candidate too; uniqueness gives $⟨M,M⟩=A$.