Angle-brackets process, 12.13 Probability with Martingales williams, p122

Question

Maybe this is obvious, but how did we deduce $\langle M^{S(k)} \rangle = A^{S(k)}$? Note: for a martingale $M$, $\langle M \rangle$ here refers to the doob decomposition $A_n$. $A_\infty:= \lim A_n$.

Answer 1

$\langle M^{S_k} \rangle$ is the unique non-decreasing previsible process such that $$N:=(M^{S_k})^2- \langle M^{S_k} \rangle, \qquad N_0 = 0$$ is a martingale. The author has shown that $$(M^{S_k})^2- A^{S_k}$$ is a martingale, and since $A^{S_k}$ is non-decreasing and previsible, it follows from the uniqueness of the Doob decomposition that $\langle M^{S_k} \rangle = A^{S_k}$.