If we have the martingale $\{X_n\}_{n \in \mathbb{N}}$ where $X_n = \mathbb{E}[X|\mathcal{F}_n]$ for some $X \in L^1$ and filtration $\{\mathcal{F}_j\}$
We have the martingale maximal function defined as: $$(MX)(x) = \sup_{n \geq 0} |X_n(x)|$$
If we define $$Z_n = \sup_{0 \leq j \leq n} |X_j(x)|$$
How does one prove the following inequality? $$\mu(\{x: |Z_n(x)| > \alpha\}) \leq \frac{C}{\alpha}||X_n||_{L^1}$$
If $\alpha > 0$, then let $$ \tau := \inf \{ k\ge 0 : |X_k| > \alpha \}, $$ so that $\tau \wedge n \le n$ is a bounded stopping time. Notice that $$ \{ x : Z_n(x) > \alpha \} = \{ x : \tau(x) \le n \}. $$ Since $x\mapsto |x|$ is convex, we know that $|X_k|$ is a submartingale. The optional stopping theorem implies that $$ \mathbb{E} |X_{\tau \wedge n}| \le \mathbb{E} |X_n|. $$ However, $$ \mathbb{E} |X_{\tau\wedge n}| \ge \alpha \cdot \mu(\{ x: \tau(x) \le n \}) $$ and so, $$ \mu(\{x : Z_n(x) > \alpha\}) \le \frac{1}{\alpha}\mathbb{E}|X_n| = \frac{1}{\alpha}\|X_n\|_{L^1}. $$