While reading the proof of Theorem 3 from the following post : https://almostsuremath.com/2010/04/01/continuous-local-martingales/
I have come across a subtle point that is not clear to me.
So the general setting is that for fixed times $s<t$, $Y_u$ is a continuous local martingale that is $0$ up to time $s$, with the quadratic variation $[Y]_u$ equal to $0$ up to time $t$. Then we define $\tau = \inf \{u: [Y]_u>0\}$, which is a stopping time with respect to the right-continuous filtration $\mathscr{F}_{\cdot +}$. No assumptions on right continuity or completeness of the filtration are given here.
Then, it says by stopping, $Y^\tau$ is a local martingale.
Question I do not understand how we get this part. I believe the rest of the argument given here can follow (as it is proving a pathwise property) if $Y^\tau$ is a local martingale with respect to the right-continuous filtration. However, I don't know of any theorem that guarantees this. If we have a local martingale in a filtration $\mathscr{F}_t$ then is it still a local martingale in the right-continuous filtration $\mathscr{F}_{t+}$? If not, then perhaps, since we are really concerned with the paths here, is there a version of $Y$ in the right-continuous filtration that is a local martingale with respect to the right-continuous filtration?
