Let $T > 0$ and consider a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ on which there is a filtration
$$\{\mathcal{F}_t: 0 \leq t \leq T\}$$
Let $M:=\{M_t: 0 \leq t \le T\}$ be a martingale w.r.t. this filtration such that $M$ is right continuous.
If $$\mathbb{E}\left(\sup_{0 \leq s \leq T} M_s^2\right)< \infty$$
then a proof I'm reading claims the following:
(1) $\sup_{0 \leq s \leq T} M_s = 0 $ almost surely
(2) Consequently, because the martingale is right continuous, $M_s = 0$ for every $0 \leq s \leq T$ with probability one.
Why are $(1)$ and $(2)$ true?
Thanks in advance!
Best Answer
This is not true: take $(B_s)_{0 \leq s \leq T}$ a Brownian motion. By Doob's maximal inequality, $$\mathbb{E} \left[\sup_{0\leq s\leq T}B_s^2\right] \leq 4 \mathbb{E}\left[B_T^2\right]<\infty.$$ But $B$ is certainly not the zero function with probability $1$.