Is the truncated Brownian motion of the class DL? 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?
(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?
 A: $W^a$ is in fact a martingale. To see this, write $W^a(t) = W(t \land T_a)$.  See also Theorem 3.39 here.
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).
A: Hi kenneth
Have a look at the following document http://www.ma.utexas.edu/users/gordanz/teaching/10_Spring_M385D/lecture16.pdf
(In particular to Propositions 16.24,16.25, 16.26, and 16.30)
First $W^1$ will be of class DL as soon as it is a martingale by proposition 16.25. So showing that $W^1$ is a martingale is sufficient to prove your claim (which follows from Proposition 16.30 for example if you know that a Brownian Motion is a martingale)
Second here is why Solution 2 doesn't work
By proposition 16.26 if $M_\tau$ is in L^1 for every bounded stopping times (which is the case here).
We have $X_t$ is a martingale if $E[M_\tau]=E[M_0]$ for every bounded stopping time $\tau$.
This is the criteria you are trying to apply to get your contradiction.
The problem with this, is that $T_1$ is not bounded almost surely so you cannot apply the preceding criteria to show that $W^1$ is not of class DL.
I hope I didn't make any mistake 
Regards
