TITLE: Uniform integrability of Doob's process
QUESTION [5 upvotes]: Let $Z,Y_0,Y_1,\cdots$ be joint random variables with $\Bbb E[|Z|]<\infty$. Doob's martingale process is defined by $X_n=\Bbb E[Z\mid Y_0,\cdots,Y_n]$ for $n\geq 0$.
In Karlin and Taylor's A First Course in Stochastic Processes (p. 296), the authors prove that the martingale is uniformly integrable as follows:
\begin{align*} |\Bbb E[X_nI\{X_n>c\}]|&\leq \Bbb E[I\{X_n>c\}\Bbb E[|Z|\mid Y_0,\cdots,Y_n]]\\
& \leq \Bbb E[|Z|I\{|X_n|>c\}]\\
& \leq \Bbb E[|Z|I\{U>c\}]\rightarrow 0
\end{align*}
as $c\rightarrow \infty$. (Here, $U=\sup_{k\geq 0} |X_k|$ and $I\{\cdot\}$ is the indicator function.)
I understand all the steps except the second inequality; why does it hold?

REPLY [4 votes]: \begin{align}
&\mathsf{E}[I\{|X_n|>c\}\mathsf{E}[|Z|\mid Y_0,\cdots,Y_n]] \\
&\qquad=\mathsf{E}[\mathsf{E}[|Z|I\{|X_n|>c\}\mid Y_0,\cdots,Y_n]] \\
&\qquad=\mathsf{E}[|Z|I\{|X_n|>c\}] \le \mathsf{E}[|Z|I\{U>c\}]
\end{align}