Bounded predictable transforms preserve martingales

Put $D_k=M_k-M_{k-1}$. It is integrable since $E\mathrm{\mid }{D}_k\mathrm{\mid }\le E\mathrm{\mid }{M}_k\mathrm{\mid }+E\mathrm{\mid }{M}_{k-1}\mathrm{\mid }$. In the bounded case [F1] gives $E\mathrm{\mid }{H}_k{D}_k\mathrm{\mid }\le C_k(E\mathrm{\mid }{M}_k\mathrm{\mid }+E\mathrm{\mid }{M}_{k-1}\mathrm{\mid })<\mathrm{\infty }$; in the general case this is assumed. Consequently $Z=H\cdot M$ is adapted, integrable, and $Z_0=0$. This is verified before conditioning any product.

For each $n\ge 0$, $H_{n+1}$ is finite ${\mathcal{F}}_n$-measurable and $D_{n+1},H_{n+1}{D}_{n+1}$ are integrable. The unbounded clause of [F2] therefore gives $E[H_{n+1}{D}_{n+1}\mid {\mathcal{F}}_n]=H_{n+1}E[D_{n+1}\mid {\mathcal{F}}_n]=H_{n+1}(M_n-M_n)=0$. Linearity and known-variable conditioning now give $E[Z_{n+1}\mid {\mathcal{F}}_n]=Z_n$. This proves the martingale assertion Martingale submartingale and supermartingale even for signed $H$. AC is inherited from the conditional classes in this calculation.

If a uniform bound $C$ is supplied, choose $C_k=C$ in step 1.1. Another sufficient domain condition at a fixed $k$ is $H_k\in L^p$ and $D_k\in L^q$ with conjugate $p,q$ under the clauses of [F6]: then $E\mathrm{\mid }{H}_k{D}_k\mathrm{\mid }\le \mathrm{\parallel }{H}_k{\mathrm{\parallel }}_p\mathrm{\parallel }{D}_k{\mathrm{\parallel }}_q<\mathrm{\infty }$. The proof of step 2.1 only needs the resulting product integrability.