Is $M^2-[M]$ a local martingale when $M$ is a local martingale? I've learned that for each continuous local martingale $M$, there's a unique continuous adapted non-decreasing process $[M]$ such that $M^2-[M]$ is a continuous local martingale.
For a local martingale $M$, is there a adapted non-decreasing process $[M]$ such that $M^2-[M]$ is a local martingale? (i.e. Do we have an analogous result for discontinuous local martingales?)
Thank you.
(The notes I have only consider the continuous case. I tried to adapt the argument, but ran into various problems...)

The answer is yes. For a good exposition of the semimartingale theory (includes local martingales, not necessarily continuous), I recommend Peter Medvegyev's "Stochastic Integration Theory". 
And the general discontinuous (but still cadlag) theory is harder than continuous case, but also fun to learn!
