Let $(X_t)_{t\in [0,T]}$ a stochastic process. If $$\mathbb E\left[\sup_{t\in [0,T]}|X_t|\right]<\infty \,,$$ is $(X_t)_{t\in [0,T]}$ uniformly integrable ?
For me it's uniformly integrable if and only if $$\lim_{M\to \infty }\sup_{t\in [0,T]}\int_{|X_t|\geq M}|X_t|d\mathbb P=0.$$
I'm rather sure that it's not true since if $(X_t)$ is a continuous submartingal, we have that $$\mathbb E[\sup_{t\in [0,T]} |X_t|^p]\leq C\sup_{t\in [0,T]}\mathbb E[|X_t|],$$ so uniformly integrable would implies $\mathbb E[\sup_t |X_t|]<\infty $. But I can't find an example where the converse doesn't hold. Any idea ?
The statement holds, consider the process $Y_n:= \sup_{t \in [0,T] } \mathbb{1}_{|X_t|\geq n}|X_t|$. This process is dominated by the integrable rv $\sup _{[0,T]}|X_t|$ and is monotone decreasing, therefore we have by monotone convergence $\lim _{n \to \infty}\mathbb{E}[Y_n] = \mathbb{E} [\lim _{n \to \infty} \sup _{t \in [0,T]}Y_n ]=\mathbb{E}[0]$ because $ \mathbb{P} (\exists_{t_m \in [0,T]} \lim _{m \to \infty}|X_{t_m}| = \infty) = 0 $ and this is the only set where $\lim _{n \to \infty} \sup _{t \in [0,T]}Y_n $ does not converge to zero. We now see that $\sup _{t \in [0,T]} \mathbb{E} [\mathbb{1}_{|X_t| \geq n} |X_t|] $ is dominated by $\mathbb{E} [Y_n]$ and therefore also converges to 0.