Doob Decomposition is $L^1$ bounded

Question

Suppose $X_n$ is a martingale that is $L^p$ bounded for some $p > 1$. Then the problem asks to show that the Doob Decomposition of the submartingale $|X_n|^p = M_n + A_n$ where $M_n$ is a martingale and $A_n$ is predictable admits $M_n$ and $A_n$ that are both $L^1$ bounded. This is from a past final exam. Thank you!

Answer 1

Hints: Without loss of generality, we may assume $X_0=0$ (otherwise consider $X_n-X_0$).

1. Since $(X_n)_{n \in \mathbb{N}}$ is a martingale, $(|X_n|^p)_{n \geq 0}$ is a submartingale.
2. By the Doob decomposition, $A_n \geq 0$ is increasing. $$\mathbb{E}(|A_n|) = \mathbb{E}(A_n) \leq \sup_{k \in \mathbb{N}} \mathbb{E}(|X_k|^p) + \mathbb{E}(|M_0|).$$ Conclude that $A_n$ is $L^1$-bounded.
3. Deduce from Step 2 that $(M_n)_{n \in \mathbb{N}}$ is $L^1$-bounded.