continuous martingale are always bounded?

Question

Consider a continous martingale $M_t$ for $t \in [0,T]$.

Since each path $M(w)$ is bounded, I can take the maximum of the bounds for each path to deduce that M is bounded uniformly.

Why do I need the concept of local boundedness then? My textbook suggests you should consider the stopping time $\tau_k =\inf\{t \in [0,T] : M_t\geq k\}$ to make an continuous martinagle at least locally bounded. But why do I need that since I can get a upper bound for all paths?

Is there anything wrong with my first argumentation?

Answer 1

Your argument tells you that for each $\omega$ there exists $C(\omega) < \infty$ such that for $t \in [0,T]$, $M_t(\omega) \leq C(\omega)$. Taking suprema as you suggest would yield $\sup_\omega \sup_{t \in [0,T]} M_t(\omega) \leq \sup_\omega C(\omega)$. You do not know a priori that the right hand side is finite.

Answer 2

Brownian motion is a continuous martingale which is obviously not bounded since at every $t$ it is normally distributed and a normally distributed variable is unbounded.