I denote $X^*=\sup_n |X_n|$ and suppose that $(X_n)$ is a positive submartingale indexed by a finite set $(0,...,N)$. I know that $$\lambda \mathbb P(X^*\geq \lambda )\leq \mathbb E[|X_N|\boldsymbol 1_{X^*\geq \lambda }].$$ Doob inequality says that $$\mathbb E[\sup_n|X_n|^p]\leq \left(\frac{p}{p-1}\right)\mathbb E[|X_N|^p].$$
The proof goes as follow : $$\mathbb E[(X^*\wedge k)^p]=\int_0^kp\lambda ^{p-1}\mathbb P(X^*\geq \lambda )\leq \frac{p}{p-1}\mathbb E[|X_N|(X^*\wedge k)^{p-1}],$$ and thus by Holder $$\mathbb E[(X^*\wedge k)^p]\leq \frac{p}{p-1}\mathbb E[(X^*\wedge k)^p]^{(p-1)/p}\mathbb E[|X_N|^p]^{1/p},$$ i.e. $$\mathbb E[(X^*\wedge k)^p]\leq \left(\frac{p}{p-1}\right)^p\mathbb E[|X_N|^p].$$ Taking $p\to \infty $ gives the result.
Question : Why do we use $\mathbb E[(X^*\wedge k)^p]$ and not directly $\mathbb E[(X^*)^p]$. If we replace all $X^*\wedge k$ by $X^*$, I don't see where the proof fails.
Truncation ensures that $\mathsf{E}[(X^*\wedge k)^p]^{(p-1)/p}<\infty$ so that you can divide both sides of the 4th inequality by this term. The result then follows by the monotone convergence theorem.