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 $M\equiv M_0$.
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?
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/