Is $M^2-[M]$ a local martingale when $M$ is a local martingale?

Question

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...)

Answer 1

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!

Answer 2

Yes this is a consequence of Doob-Meyer decomposition theorem I think.

For a reference you can also look at Philip Protter's book "Stochastic Integration and Differential Equations"

Another web-reference is George Lowther's blog but I think Doob-Meyer decomposition is not yet proved.

Regards