Let $p\in[1,\infty)$, M is called a $L^p$ martingale if it's a martingale and $M_t\in L^p$ for all $t\in\mathbb R_+$, if $\sup_{t\in\mathbb R_+} \mathbb E(|M_t|^p)<\infty$ then $M$ is $L^p$-bounded.
I don't really see the difference between this two cases, doesn't the fact that $M_t\in L^p$ for all $t$ imply that $\mathbb E(|M_t|^p)<\infty$ for all $t$ and hence also for the $\sup$?
Thanks in advance.
More a comment than an answer (I haven't enough reputation).
As example for $p=1$ and working on the discrete index set $\mathbb{N_0}$, try to construct a martingale such that $E[|X_n|] = n$ for every $n \in \mathbb{N}$. Now if you take the supremum over $n$ this is clearly infinite. The fact is that before you take the supremum, $t$ ( or $n$ in my case) are fixed