 Solution

Stop just before a large predictable coefficient would be used. Setwith the infimum of the empty set equal to infinity. Because $K_{t+1}$ is ${\mathcal{F}}_t$-measurable, ${\sigma }_j$ is a stopping time. The increment of the stopped martingale transform isIts coefficient is predictable and bounded by $j$: the first coefficient exceeding $j$ occurs at the step after stopping, and is never included. Part (c) makes $Y^{{\sigma }_j}$ a martingale. Since the finitely many $K_s$ on any fixed finite horizon are finite almost surely, ${\sigma }_j\uparrow \infty$ almost surely. ThusThis is predictable-coefficient localization of a martingale transform. Stopping after taking the large increment would not give the required bound.