# Why is $E[\ \sup_{ 0\leq u \leq t } |X_u|\ ]<\infty$ where $X$ is a right continuous submartingale?

Suppose that $\{ X_t, \mathcal{F}_t \ : \ 0 \leq t < \infty \}$ is a right-continuous submartingale and $T$ is a stopping time of $\{ \mathcal{F}_t \}$. I have been trying to show unsuccessfully that

$E[\ \sup_{ 0\leq u \leq t } |X_u|\ ] < \infty$.

Is this even true? I get a feeling that s not true but i can't show it either. I was trying to understand the solution of Exercise 3.24 in Chapter 1 of Karatzas and Shreve from

No, in general the assertion does not hold true - not even for martingales. You can find several counterexamples here.

What is, however, true is the following statement:

Let $(M_t)_{t \geq 0}$ be a martingale (or a positive submartingale) with càdlàg sample paths. Then $$\mathbb{E} \left( \sup_{s \leq t} |M_s| \right) \leq \frac{e}{e-1} \mathbb{E}(|M_t| |\log M_t|).$$

You can find a (sketch of the) proof for discrete martingale for instance in Revuz & Yor, Exercise 2.1.16; using standard approximation it can be easily extended to the time-continuous setting.

• So it means that the part of the proof here math.stackexchange.com/questions/1500273/… where they prove the integrability of $X_t$ doesnt work? Commented Mar 4, 2018 at 21:54
• @user3503589 Yes, it seems so.
– saz
Commented Mar 5, 2018 at 6:39