# Question about a step in the proof of the bracket process

Let's first state the theorem

$\forall M$ continuous local martingale, there exists a unique increasing continuous process $\langle M\rangle$ zero at $t=0$ and such that $M^2-\langle M \rangle$ is again a continuous local martingale. Further for all stopping times $\tau$, we have $\langle M^\tau\rangle = \langle M \rangle^\tau$.

The last statement is not clear for me. I know that $L:=M^2-\langle M \rangle$ is a local martingale. Then we look at $L^\tau=(M^\tau)^2-\langle M \rangle^\tau$, and we apply the stopping theorem. Why do we can apply the stopping theorem? For that, $L$ should be uniformly integrable, or $\tau$ must be finite?

All you need to use is that the set of local martingales are stable under stopping, i.e. $M^\tau$ is local martingale if $M$ is a local martingale and $\tau$ is any stopping time. Then $$(M^{\tau})^2-\langle M\rangle^\tau=(M^2-\langle M\rangle)^{\tau}$$ yields the result.
But this is just, as far as I know, true if $\tau$ is bounded or the martingale is uniformly integrable? – user20869 Jul 22 '12 at 12:00
It is true for a general $\tau$, but it's true that it uses the optional sampling theorem on the bounded stopping time $\tau\wedge t$. – Stefan Hansen Jul 22 '12 at 16:58