Elementary predictable counting-process integrals are martingales

Each summand in [F2] is ${\mathcal{F}}_t$-measurable at its current endpoint. It is square-integrable because $H_j$ is bounded and each martingale increment is in $L^2$. A finite sum is therefore adapted and square-integrable at every $t\le T$.

Fix $0\le s\le t\le T$ and $B\in {\mathcal{F}}_s$. Refine the original deterministic partition by inserting $s$ and $t$. After subtracting $I_s(H)$ from $I_t(H)$, each remaining piece has the form $H_j(M_v-M_u)$ on an interval $(u,v]$ with $u\ge s$ and $t_j\le u$. Thus ${\mathbf{1}}_B{H}_j$ is bounded and ${\mathcal{F}}_u$-measurable. Since $\mathbb{E}[M_v-M_u\mid {\mathcal{F}}_u]=0$, [F3] gives $\mathbb{E}[{\mathbf{1}}_B{H}_j(M_v-M_u)]=0$. Summing the finitely many pieces yields $\mathbb{E}[{\mathbf{1}}_B(I_t(H)-I_s(H))]=0$. The event-integral characterization of conditional expectation proves $\mathbb{E}[I_t(H)\mid {\mathcal{F}}_s]=I_s(H)$. Thus the elementary integral is a square-integrable martingale.