Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 30 / 1 / a

Let $N$ be the zero-starting continuous local martingale, and let $V_t$ be the total variation of its path on $[0,t]$. Stop when $|N|$ or $V$ reaches $n$, and also at time $n$. The stopped process is bounded, hence a true square-integrable martingale by the bounded local martingale criterion, and its total variation is bounded by $n$.

For a deterministic partition $0=t_0<\cdots <t_m=t$, martingale increments are orthogonal in $L^2$, soThe sum inside the expectation is at most the largest absolute increment times the stopped total variation. As the mesh tends to zero, path continuity makes the largest increment tend to zero. The product is bounded by $2n^2$, so the dominated convergence theorem makes the displayed expectation zero. Thus the stopped process vanishes at every fixed time. Use rational times, continuity and ${\tau }_n\uparrow \infty$ to concludeThis proves that a continuous finite-variation local martingale is constant.

Write a normalized semimartingale decomposition as $X=X_0+N+A$, with both $N_0=A_0=0$, $N$ a continuous local martingale and $A$ of finite variation. The difference of two such martingale parts is also the difference of two finite-variation parts, and hence vanishes by the result just proved. The other parts coincide too. The normalized continuous semimartingale decomposition is unique. Fixing the initial constants is necessary: without normalization a constant can be transferred between the two parts.