In a larger proof I'm stuck on finding the quadratic variation of the martingale given by $$M_n = X_n-Y_n = \sum_{k=1}^n1_{A_k} - P(A_k\vert \mathcal F_{k-1}),$$ where $A_k\in\mathcal F_k$. I already have that $M_n$ is a martingale. Further, we have $$\begin{aligned} E(M_n^2)&=E\left(\left(\sum_{k=1}^n1_{A_k}-P(A_k\vert\mathcal F_{k-1})\right)^2\right) \\ &\le E\left(\left(\sum_{k=1}^n1_{A_k}\right)^2\right)\\ &\le E\left((n1_{\cup A_k})^2\right)=n^2E(1_{\cup_{k=1}^nA_k})\\ &= n^2P(\cup_{k=1}^n A_k)<\infty, \end{aligned}$$ which (unless I made an error) shows that the quadratic variation in fact exists. However when I plug the martingale in to the formula for the quadratic variation, I'm stuck on $$\begin{aligned} \langle M_n\rangle &= \sum_{j=1}^n E((M_j-M_{j-1})^2\vert \mathcal F_{j-1})\\ &= \sum_{j=1}^n E\left(\left(1_{A_j}-P(A_j\vert\mathcal F_{j-1})\right)^2\vert \mathcal F_{j-1}\right)\\ &= \sum_{j=1}^n E\left(1_{A_j}-2P(A_j\vert\mathcal F_{j-1})1_{A_j}+P^2(A_j\vert\mathcal F_{j-1})\vert\mathcal F_{j-1}\right)\\ \end{aligned}.$$ I'm sure I'm just overlooking something, but I can't simplify this thing further. But perhaps there is a better way of finding the quadratic variation that I don't know of?
EDIT: I noticed $P(A_j\vert\mathcal F_{j-1})$ is a constant, but I'm not exactly sure (how) I'm allowed to pull it out.