Given a uniformly integrable discrete martingale $M_n$ on prob. space $(\Omega, \mathcal{F}, \mathbb{P})$, and a.s. finite stopping times $T$ and $S$ with $T\geq S$. Show that $E[M_T|\mathcal{F}_S] = M_S$.
I show that $M_n \rightarrow M$ in $\mathcal{L}^1$ and a.s. by uniform integrability. Therefore $M_T = E[M|\mathcal{F}_T]$ and $E[M_T|\mathcal{F}_S] = E[E[M|\mathcal{F}_T]|\mathcal{F}_S] = E[M | \mathcal{F}_S] = M_S$, however it seems there is something missing here.
Ok, so to show that $E[M |F_T] = M_T$ we have that letting $B \in \mathcal{F}_T$ and $A_i = B \cap \{T = i\}$ that $E[M_T 1_B] = \sum_i E[M_i 1_{A_i}] = \sum_i E[\, E[M | \mathcal{F}_i] 1_{A_i}] = \sum_i E[M 1_{A_i}]= E[M1_B]$. Now, it is clear that $E[M|F_T] = M_T$ and the above works fine.