This is a simple exercise needed to prove the Optional Stopping Theorem that I'm working on.
Suppose $(X_n)$ is a supermartingale and we have stopping times $T, S$. Then we already know in general that $\mathbb{E}(X_T|F_s)\leq X_{\min(S,T)}$ by splitting $X_T=X_{\min(S,T)} +\sum_{k=0}^n (X_{i+1}-X_i)1_{S\leq k\leq T}$, however I want to show directly with $T\leq S$ that $$\mathbb{E}[X_T|F_S]\leq X_T$$ This seems trivial if the expectation was conditioned upon $F_n$ where $n\leq T$ almost surely, but I'm unsure of the proof for a stopped $\sigma$-algebra. I can do this:
$$\mathbb{E}(\mathbb{E}[X_T|F_S]1_{A})\leq \mathbb{E}(X_T 1_A)$$ for any $A \in F_S$, i.e. $A\cap\{S\leq n\}\in F_n$ for any $n\geq 0$. This yields, by earlier assumption, that $A\cap \{S\leq n\}\cap \{T\leq S\}\in F_n$, but can we then assume from there that $A\cap \{T\leq n\}\in F_n$?
I guess the above is the same as trying to show that $X_T$ is $(F_S)$-measurable if $T\leq S$ a.s. I would appreciate any brief pointers on how I can go on about this, as well as anything I should add to better clarify my current thinking.
For real $x$ you have $$ \{X_T\le x\}\cap\{S\le n\} = \cup_{k=0}^n\{X_k\le x, T=k, S\le n\}. $$ which is clearly $\mathcal F_n$-measurable, for each $n$. It follows that $X_T$ is $\mathcal F_S$-measurable when $T\le S$. In this case, $\Bbb E[X_T|\mathcal F_S] = X_T$ (ignoring questions of integrability of $X_T$).