Let $\tau$ be a stopping time, and let $X=M-A$ be the doob-decomposition of the super martingale $X$. Proof that the Doob decomposition of $X^{\tau}$ is $X^{\tau}=M^{\tau}-A^{\tau}$.
So I know that a) the Doob decomposition is unqiue b) the stopped process $X^{\tau}$ is again a supermartingale (Doob optional sampling).
can I use these information to prove that the decomposition is indeed the one above?