Isometry for elementary predictable counting-process integrals

On each $(t_j,t_{j+1}]$, write $\mathrm{\Delta }{M}_j=M_{t_{j+1}}-M_{t_j}$ and $\mathrm{\Delta }{A}_j=A_{t_{j+1}}-A_{t_j}$. The martingale $M^2-A$ gives $\mathbb{E}[M_{t_{j+1}}^2-M_{t_j}^2\mid {\mathcal{F}}_{t_j}]=\mathbb{E}[\mathrm{\Delta }{A}_j\mid {\mathcal{F}}_{t_j}]$; expand the left square as $(\mathrm{\Delta }{M}_j{)}^2+2M_{t_j}\mathrm{\Delta }{M}_j$. Since $M_{t_j}$ and $\mathrm{\Delta }{M}_j$ are in $L^2$, their product is integrable by [F5], so [F3] applies with the possibly unbounded factor $M_{t_j}$. Use $\mathbb{E}[\mathrm{\Delta }{M}_j\mid {\mathcal{F}}_{t_j}]=0$ to get $\mathbb{E}[(\mathrm{\Delta }{M}_j{)}^2\mid {\mathcal{F}}_{t_j}]=\mathbb{E}[\mathrm{\Delta }{A}_j\mid {\mathcal{F}}_{t_j}]$. By [F3], $\mathbb{E}[H_j^2(\mathrm{\Delta }{M}_j{)}^2]=\mathbb{E}[H_j^2\mathrm{\Delta }{A}_j]$.

If $i<j$, then $H_i\mathrm{\Delta }{M}_i{H}_j$ is ${\mathcal{F}}_{t_j}$-measurable and its product with $\mathrm{\Delta }{M}_j$ is integrable by boundedness of $H$ and [F5]; [F3] and $\mathbb{E}[\mathrm{\Delta }{M}_j\mid {\mathcal{F}}_{t_j}]=0$ give $\mathbb{E}[H_i\mathrm{\Delta }{M}_i{H}_j\mathrm{\Delta }{M}_j]=0$. Expanding the finite-sum square [F2], all off-diagonal terms vanish, and step 1.1 gives $\mathbb{E}[I_T(H{)}^2]={\sum }_j\mathbb{E}[H_j^2(\mathrm{\Delta }{M}_j{)}^2]={\sum }_j\mathbb{E}[H_j^2\mathrm{\Delta }{A}_j]=\mathbb{E}{\int }_0^T{H}_s^{2}d{A}_s$; the martingale assertion is [F4].

If two step representations agree in the compensator $L^2$ seminorm, their difference has seminorm zero; applying the isometry in step 2.1 to the refined difference representation makes the squared $L^2$ norm of their terminal-integral difference zero, so they define the same terminal class.