Let $W$ be a standard Brownian motion under given probability space. 
For a given constant $a$, $W^a$ is a truncated Brownian motion by stopping time 
$T^a = \inf(t>0:W(t) = a)$. That is, $W^a(t) = W(t \wedge T^a)$.  

We want to consider the following question: Is the process $W^1$ of [the class DL](http://almostsure.wordpress.com/2009/12/22/class-d-processes/)?


(Solution1): Yes. Indeed, for any fixed $t>0$, we can prove the collection of random variables 
$( W(s), 0< s< t)$ 
is uniformly integrable by definition, since $E [|W^1(t)|] < \infty$.


We provide completely different answer using the following proposition from the Problem 1.5.19 (i) of Book [Karazas and Shereve 98].


[Proposition] A local martingale of class DL is martingale.


(Solution2): No. $W^1$ is strict local martingale, since $E [W^1(T^1)] = 1> E [W(0)]$. By [Proposition], $W^1$ is not of class DL.


In the above, we obtained completely two different solutions. Where is wrong?



Is the truncated Brownian motion of the class DL?
$W^a$ is in fact a martingale. To see this, write $W^a(t) = W(t \land T\_a)$. See also Theorem 3.39 [here](http://books.google.com/books?id=JYzW0uqQxB0C&pg=PA85).


When you write an expression like $\mathbb{E}(W^a(T^a))$ you are implicitly assuming that $W^a(T^a)$ is measurable. This requires $t \ge T\_a$ (and trivializes the expectation).


