For a Brownian motion $B_t$ started from $0$ (it is also a martingale), denote by $T_a=\inf\{t\ge 0: B_t=a\}$ the stopping time. Consider the stopped process $B_{t\land T_a}$ (this is still a martingale), then $|B_{t\land T_a}|$ is bounded by $a$. (Recall that $T_a<\infty$ a.s.) Why can we say $B_{t\land T_a}$ is uniformly integrable?
At first I thought this is because that $E\left[\lvert B_{t\land T_a}\rvert^2\right]\le C$ for some constants $C$, then $B_{t\land T_a}$ is uniformly integrable.
However, I just saw this question:Almost sure bounded imply finite expectation?. I believe the answer means the finite bounded cannot imply the expectation. Thus, we cannot get $E[|B_{t\land T_a}|^2]\le C$ from $|B_{t\land T_a}|\le a$.
Uniform boundedness of $\{B_{t\land T_a}\}_{t \geq 0}$ implies that it is uniformly integrable. Indeed, by noting that for an $t\geq 0$, $$E(B_{t \land T_a } 1(|B_{t\land T_a}| \geq 2a )) = E(0) = 0$$ we immediately see that the family $\{B_{t\land T_a}\}_{t \geq 0}$ is uniformly integrable, by appealing to the definition of uniform integrability.