Let $M_0,M_1,\dots$ be a martingale with respect to $X_0,X_1,\dots$ and $T$ be a stopping time with respect to $X_0,X_1,\dots$ Define $T_n=\min\{n,T\}$ and let $M_{T_n}$ be the stopped martingale.
By Optional Stopping Theorem, I know that $EM_{T_n}=EM_0=EM_n$. However, my question is: how might I show $E|M_{T_n}|\leq E|M_n|$?
Hint: Use the optional sampling theorem for the submartingale $|M_n|$.
Direct proof: First, note that it suffices to show $\mathbb{E}(|M_S|) \leq \mathbb{E}(|M_n|)$ for any stopping time $S$ such that $S \leq n$.
Since $(M_n)_{n \in \mathbb{N}}$ is a martingale, we know from Jensen's inequaliy that $(|M_n|)_{n \in \mathbb{N}}$ is a submartingale, i.e.
$$|M_k| \leq \mathbb{E}(|M_n| \mid \mathcal{F}_k) \qquad \text{for all $k \leq n$}. \tag{1}$$
Using $\{S=k\} \in \mathcal{F}_k$ and the tower property, we get
$$\begin{align*} \mathbb{E}(|M_k| \cdot 1_{\{S=k\}}) &\stackrel{(1)}{\leq} \mathbb{E}(1_{\{S=k\}} \mathbb{E}(|M_n| \mid \mathcal{F}_k)) \\ &= \mathbb{E}[ \mathbb{E}(1_{\{S=k\}} |M_n| \mid \mathcal{F}_k)] \\ &= \mathbb{E}(1_{\{S=k\}} |M_n|) \tag{2}\end{align*}$$
for all $k \leq n$. Since $S \leq n$, we have
$$\Omega = \bigcup_{k=0}^n \{S=k\}$$
and therefore
$$\begin{align*} \mathbb{E}(|M_S|) &= \sum_{k=0}^n \mathbb{E}(|M_S| 1_{\{S=k\}}) \\ &= \sum_{k=0}^n \mathbb{E}(|M_k| 1_{\{S=k\}}) \\ &\stackrel{(2)}{\leq} \sum_{k=0}^n \mathbb{E}(|M_n| 1_{\{S=k\}}) \\ &= \mathbb{E}(|M_n|). \end{align*}$$