Continuous local martingale of finite variation is constant Is a continuous local martingale $M$ of finite variation constant?
We know that there exists a sequence of stopping times $T_n\nearrow \infty$ a.s. as $n\to\infty$ such that the stopped process $M^{T_n}$ is a continuous bounded martingale. And $M^{T_n}$ has finite variation, because $M$ has finite variation.
Since if $X$ is a continuous bounded martingale with finite variation, then $X=X_0$ a.s. So $M_t^{T_n}=M_0^{T_n}$ a.s. $\forall t\ge 0$ and by letting $n\to\infty$ we obtain that $M\equiv M_0$.


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*Is my reasoning correct?

*The statement "continuous bounded martingale with finite variation implies constant" is a lemma from my notes, but we did not prove it. How would one prove it?
 A: Your intuition is good, but there are some technicalities you should be careful about. WLOG $M_0 = 0$. Start by assuming $M$ is a continuous martingale of bounded variation. Then if $B$ is a bound on the variation of $M$, and $(t_i)$ is a partition of $[0,t]$,
$$
E[M_t^2] = E\sum (M_{t_{i+1}}-M_{t_i})^2 \leq B E[\sup_i|M_{t_{i+1}}-M_{t_i}|].
$$
Since $M$ is continuous the supremeum tends to $0$ as the partition size goes to $0$. Moreover, the supremum is dominated by $B$, and hence dominated convergence tells us that $E[M_t^2] = 0$, in particular $M_t=0$ a.s. Let $t$ run over rationals and use the continuity of $M$ to conclude that $M=0$.
Next, if $M$ is a continuous local martingale, take a localizing sequence of stopping times $(\tau_n)$. We then find that $M^{\tau_n}_t = 0$ for all $n$, and so taking the limit in $n$ gives $M_t = 0$ a.s. Again let $t$ range over rationals. 
A: I am not sure if it is that simple. I detest posting a link only answer but I feel I am not adding any value by posting anything in addition to this excellent link.
see theorem 3 and lemma 4 of
http://almostsure.wordpress.com/2010/04/01/continuous-local-martingales/
