 Solution

A continuous adapted process $M$ is a local martingale if there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $M_t^{{\tau }_n}=M_{t\wedge {\tau }_n}$ is a martingale. In this definition $M_0$ is integrable. A continuous adapted process $A$ has finite variation if, almost surely, its paths have finite total variation of a function on each compact time interval:The permitted fact that this total-variation process is continuous and adapted makes its level-hitting times stopping times.

Suppose $M=A$, and put $N=M-M_0$. DefineThese times increase to infinity almost surely. The optional stopping theorem applied to the localized martingale shows that $N^{{\sigma }_n}$ is a martingale; it is also bounded in absolute value by $n$, with total variation at most $n$. Fix $n,t$ and write $L=N^{{\sigma }_n}$. For a deterministic partition of an interval $0=t_0<\cdots <t_r=t$, orthogonality of square-integrable martingale increments givesPathwise the sum on the right is bounded by $n{\max }_j|L_{t_j}-L_{t_{j-1}}|$. For partitions with mesh tending to zero, continuity makes this converge to zero. It is bounded by $2n^2$, so dominated convergence gives $\mathbb{E}{L}_t^2=0$. First use rational $t$, then continuity, to obtain $L\equiv 0$ almost surely. Letting $n\to \infty$ provesThis proves continuous finite-variation local martingale is constant without assuming integrability of the unstopped total variation. Continuity matters: compensated jump martingales can have finite variation without being constant.