Stopped version of UI martingale is UI Suppose $(X_n)_{n\geq0}$ is a uniformly integrable martingale. Let $T$ be any stopping time. Show that the stopped process $(X_n^T)_{n\geq0}=(X_{n\wedge T})_{n\geq0}$ is uniformly integrable.
My only idea was to work on the two events $\{T<\infty\}$ and $\{T=\infty
\}$. The result is obvious on the second event by assumption, but I'm not sure how to formally prove that the limit must go to zero (as in the definition of uniform integrability). Any advice would be greatly appreciated!

Any uniformly integrable martingale can be written in the form $(E(X|\mathcal F_n))$ with $E|X| <\infty$. Optional Sampling Theorem (SeeTheorem 9.3.3, p.324 of  K L Chung's "A Course in Probability Theory")  shows that $X_T=E(X|\mathcal F_T)$ holds for any stopping time $T$. Since $\{E(X|\mathcal G): \mathcal G \subset \mathcal F\}$ is itself uniformly integrable, we are done.
