A local martingale is a martingale iff it is in the class DL.


The condition: for every $t\in[0,\infty)$


$$E[\sup\limits\_{0\leq s\leq t} |M\_s|]<\infty\tag1$$


guarantees a local martingale $M$ is a martingale by ensuring it satisfies the condition for being in the class DL. Moreover, by Burkholder-Davis-Gundy, this means: for every $t\in[0,\infty)$,


$$E[\langle M\rangle^{1/2}\_t]<\infty$$


My question is: do there exist (simple?) examples of continuous **martingales** where these two conditions are violated? 


What I am really asking is that these are known to be sufficient conditions to guarantee local martingales are martingales, why are they not necessary?



Examples of a continuous martingale with $E[\sup\limits_{0\leq s\leq t} |M_s|]=\infty$?
There exist indeed a uniformly integrable martingale $X$ whose max $\sup\_{k\in \mathbb{N}} |X\_k|$ is not integrable. 
A simple example in discrete time can be found here <http://www.math.fsu.edu/~nichols/martingalezoo.pdf>
(see the last 2 lines of point 5). 


A related construction is found in example 4.1 of the paper by A.S. Cherny "Some particular problems of martingale theory", which produces a uniformly integrable martingale $(X\_n)\_{n\in \mathbb{N}}$ and a bounded process $(H\_n)\_{n\in \mathbb{N}}$ such that the stochastic integral $M\_n:=\sum\_{k=0}^n H\_k (X\_{k+1}-X\_k)$ is a martingale which is not uniformly integrable (by the Burkholder-Davis-Gundy inequality this implies that $\sup\_{k\in \mathbb{N}} |X\_k|$ is not integrable) 


A second (completely different) example in continuous time is sketched in Exercise 3.15 of the 2nd Chapter of Revuz-Yor's Book "Continuous Martingales and Brownian Motion, 3rd ed"


