# stopped UI discrete time martingale is UI

If $$\{X_n\}$$ is a uniformly integrable discrete time Martingale, and if $$\tau$$ is a (possibly $$\infty$$-valued) stopping time, then $$\{X_{\tau \wedge n}\}$$ is uniformly integrable (note I don't care that $$\{X_{\tau \wedge n}\}$$ is a martingale). The only proof I've seen of this uses stuff about test functions. Is it possible to give an elementary proof without this test function definition of uniform integrability?

Since $$(X_n)$$ is UI, we know there exists $$X_\infty \in L^1(\mathcal F_\infty)$$ such that $$(X_n) \rightarrow X_\infty$$ a.s. and in $$L^1$$. Furthermore, the uniform integrability and optional stopping theorem guarantee that $$X_{n \wedge \tau} = \mathbb{E}[X_\infty | \mathcal F_{n \wedge \tau}]$$. For the definition of uniform integrability, we want to show that $$\sup_{n} \mathbb{E}[|X_{n \wedge \tau}| 1_{|X_{n \wedge \tau}| > R}]$$ tends to $$0$$ as $$R$$ tends to $$\infty$$. We compute
\begin{align*} \mathbb{E}[|X_{n \wedge \tau}| 1_{|X_{n \wedge \tau}| > R}] &= \mathbb{E}[|\mathbb{E}[X_\infty|\mathcal F_{n \wedge \tau}]| 1_{|X_{n \wedge \tau}| > R}] \\ &\le \mathbb{E}[\mathbb{E}[|X_\infty||\mathcal F_{n \wedge \tau}] 1_{|X_{n \wedge \tau}| > R}] \\ &= \mathbb{E}[\mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}|\mathcal F_{n \wedge \tau}]] \\ &= \mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}] \end{align*}
and by Markov's inequality $$P(|X_{n \wedge \tau}| > R) \le \frac 1R \mathbb{E}[|X_{n \wedge \tau}|] \le \frac 1R \mathbb{E}[|X_{\infty}|]$$, where the last inequality comes from the same argument as above. Therefore we can choose $$R$$ such that $$P(|X_{n \wedge \tau}| > R)$$ is arbitrarily small for all $$n$$, and therefore $$\mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}]$$ can be made arbitrarily small as well. Since $$\mathbb{E}[|X_{n \wedge \tau}| 1_{|X_{n \wedge \tau}| > R}] \le \mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}]$$ this proves the uniform integrability of $$(X_{n \wedge \tau})$$.
• Thank you! I was actually trying to prove OST with my question but I already have the bounded version and thankfully that bounded version suffices to prove $X_{n \wedge \tau} = \mathbb{E}[X_\infty | \mathcal F_{n \wedge \tau}]$ since we have $$X_{n \wedge \tau} = \mathbb{E}[X_n | \mathcal F_{n \wedge \tau}]= \mathbb{E}[ \mathbb{E}[X_\infty| \mathcal F_{n}]| \mathcal F_{n \wedge \tau}]= \mathbb{E}[X_\infty | \mathcal F_{n \wedge \tau}]$$