A continuous finite-variation path has zero <quadratic variation>. Hence a continuous finite-variation martingale $M$ satisfies $\langle M\rangle=0$. After localizing to make it square-integrable,
$$
\mathbb E[(M_t-M_0)^2]=\mathbb E[\langle M\rangle_t]=0.
$$
Letting the localization level tend to infinity shows that \b[$M_t=M_0$ for every $t$ almost surely]; continuity makes the equality simultaneous in $t$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 6 / 1 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-202/6/1/solution