TITLE: Doob's supermartingale inequality
QUESTION [8 upvotes]: I'm trying to prove that For a non-negative supermartingale $M$ it holds that for all $\lambda>0$ we have
$$\lambda P\{\sup_{n}M_{n}\geq\lambda\}\leq E(M_{0})$$
My idea was to use Markov's inequality which states that
$$\lambda P(M_{n}\geq\lambda)\leq E(M_{n})$$
As it holds for all $M_{n}$ it must also hold for the supremum of $M_{n}$ and using the supermartingale property $E(M_{n})\leq E(M_{0})$ one finds the desired result.
However I'm not sure if I can just say that it also holds for the supremum, could anyone help me out with this?
Help is much appreciated.

REPLY [11 votes]: Hint:
Define a stopping time $T=\inf\{k: X_k\geq \lambda \}$
Write $\mathbb{E}X_{n\wedge T}=\mathbb{E}X_T1_{\{T\leq n\}} + \mathbb{E}X_n1_{\{T> n\}} $ 
Use optional sampling theorm: $\mathbb{E}X_{n\wedge T}\leq\mathbb{E}X_0$ drop something you don't want because it is positive.
fill in the details yourself :)