Optional stopping with a nonnegative supermartingale I encountered the following problem:
Let $ X_n $ be a nonnegative supermartingale. For two stopping times $ M \le N \le \infty $, prove that $$ \mathbb{E}[ X_{N} | \mathcal{F}_{M}] \le X_{M} . $$
At first I thought this is not true, because to have the optional stopping theorem we need something as powerful as the uniform integrability. Since a nonnegative supermartingale is not necessarily uniformly integrable, I started there and tried to construct a counterexample where $ \mathbb{E} X_M < \mathbb{E} X_N $. However it seems it's not that simple.
So my question is

*

*Is this true?

*If it is true, how can I prove it? If not, how can I get to a counterexample? Even a small hint would be very much appreciated.

 A: This is Theorem 28 in Chapter V of Meyer's classical "Probability and potentials", but, candidly, I don't like the proof very much, because the pertinent ideas are obscured by technicalities.
First of all: the condition $M \le N \color{red}{\le} \infty$ does make sense, because non-negative supermartingales converge so some integrable $X_\infty$, a.s. Still better: we can extend our supermartingale $(X_n)_{n\in\mathbb{N}}$ to a supermartingale $(X_n)_{n\in\mathbb{N}\cup\{\infty\}}$. That's because the supermartingale property
$$X_m\ge\mathbb{E}(X_n|\mathcal{F}_m)\quad\mbox{for}\quad m<n\quad\mbox{a.s.}$$ is equivalent to $$\int_{A}X_m\,dP\ge\int_{A}X_n\,dP\quad\mbox{for}\quad m<n\quad\mbox{and all}\quad A\in\mathcal{F}_m.\tag{1}$$ This implies
$$\int_{A}X_m\,dP\ge\liminf_{n\to\infty}\int_{A}X_n\,dP\ge\int_{A}X_\infty\,dP\quad\mbox{for all}\quad A\in\mathcal{F}_m$$ by Fatou's lemma, and that's equivalent to
$$X_m\ge\mathbb{E}(X_\infty|\mathcal{F}_m)\quad\mbox{a.s.}\tag{2}$$
So we proved the proposition for $N=\infty$ and a deterministic $M=m$. Let's extend that to a non-deterministic stopping time $M$: we have to prove
$$X_M\ge\mathbb{E}(X_\infty|\mathcal{F}_M)\quad\mbox{a.s.}\tag{3}$$ and that's equivalent to
$$\int_{A}X_M\,dP\ge\int_{A}X_\infty\,dP\quad\mbox{for all}\quad A\in\mathcal{F}_M.$$
Here, we have to remember the definition of $\mathcal{F}_M$:
$$\mathcal{F}_M=\{A\subset\Omega: A\cap\{M=m\}\in\mathcal{F}_m\quad\mbox{for all}\quad m\in\mathbb{N}\}$$ (if I were a textbook author, I'd add "the proof that this is indeed a $\sigma$-algebra is left as an exercise to the reader"). In fact, you may more often find $\{M\le m\}$ instead of $\{M=m\}$ in definitions, but for discrete time, both definitions are equvalent.
Now we can split $A$ like $\displaystyle A=\bigcup_{m\in\mathbb{N}\cup\{\infty\}} A\cap\{M=m\}$, and $A\in\mathcal{F}_M$ means $A\cap\{M=m\}\in\mathcal{F}_m$, and $X_M=X_m$ on $A\cap\{M=m\}$, so
$$\int_{A}X_M\,dP=\sum_{m\in\mathbb{N}\cup\{\infty\}}\int_{A\cap\{M=m\}}X_m\,dP\ge\sum_{m\in\mathbb{N}\cup\{\infty\}}\int_{A\cap\{M=m\}}X_\infty\,dP=\int_{A}X_\infty\,dP,$$ indeed, due to (2).
The rest is easy: if $X_n$ is a non-negative supermartingale and $N\le\infty$ a stopping time, $Y_n=X_{N\wedge n}$ is a non-negative supermartingale as well, $Y_\infty=X_N$, $Y_M=X_M$ for a stopping time $M\le N$, so (3) becomes
$$X_M\ge\mathbb{E}(X_N|\mathcal{F}_M)\quad\mbox{a.s.}$$
