In our class we have proven Doob's inequality for discrete martingale, i.e. Let $(M_n)_{n \in \mathbb{N}}$ a martingale, then
$ \| \max_{0\le k\le n} M_k\|_p \le C_p \|X_n\|_p$
for $p\in (0,\infty)$. How can I use this to prove the same result for a continuous-time martingale which is right continuous?
EDIT:
After the hint of GEdgar, I did the following: Let $A=[0,T] \cap \mathbb{Q}$ a dense subset of $[0,T]$. For each $n$ let $A_n$ be a finite set, such that $A_n \subset A_{n+1}$ and $A=\cup_{n\in \mathbb{N}} A_n$. I'm quite sure, that I can choose such sets, but could someone give a concrete example here? Now we have: $\sup_{t\in A_n} M_t\le \sup_{0\le t \le T} M_t\, \forall n,$ hence $\sup_n\sup_{t\in A_n} M_t\le \sup_{0\le t \le T} M_t.$ On the other hand, let $(q_m)$ be a decreasing sequence of rationals in $[0,T]$ such that $M_t=\lim_m M_{q_m}$. Let $B_m$ be the set in $(A_n)$ with $q_m\in A_n$ then for all $m\ge1$ $M_{q_m}\le \sup_{t\in B_m}\le\sup_n\sup_{t\in A_n}M_t.$ Hence $M_t=\lim_m M_{q_m}\le\sup_n\sup_{t\in A_n}M_t$. Therefore we have shown: $\sup_{0\le t\le T}M_t=\sup_n\sup_{t\in A_n}M_t.$
Clearly $N_n:=\sup_{t\in A_n}M_t$ is increasing in $n$, measurable and converges to $\sup_{0\le t\le T}M_t$. Applying discrete Doob to $A_n$ leads to $E[N_n^p]\le c(p)E[M_{\max A_n}^+]$. The left hand side converges to $E[\sup_{0\le t\le T}]$ by Monotone Convergence. Why does the RHS converge to the right thing? Beside of my two questions, is my proof correct?