It follows from the optional stopping theorem that $(X_{n \wedge \tau})_{n \in \mathbb{N}}$ is a martingale; in particular, $\mathbb{E}(X_{n \wedge \tau}) = \mathbb{E}(X_0)=0$. Since \begin{align*} 0 = \mathbb{E}(X_{\tau \wedge n}) &= \mathbb{E}(X_{\tau \wedge n} 1_{\{\tau \leq n\}}) + \mathbb{E}(X_{\tau \wedge n} 1_{\{\tau>n\}}) \\ & = 1 \cdot \mathbb{P}(\tau \leq n) + \mathbb{E}(X_{\tau \wedge n} 1_{\{\tau>n\}}) \end{align*} we find that $$\mathbb{E}(X_{\tau \wedge n} 1_{\{\tau>n\}}) = - \mathbb{P}(\tau \leq n). \tag{1}$$
On the other hand, we have for any fixed $R>1$ that
\begin{align*} \mathbb{E}(X_{\tau \wedge n} 1_{\{\tau>n\}}) &= \mathbb{E}(X_{\tau \wedge n} 1_{\{X_{\tau \wedge n} \leq -1\}}) \\ &= \mathbb{E}(X_{\tau \wedge n} 1_{\{-R < X_{\tau \wedge n} \leq -1\}}) + \mathbb{E}(X_{\tau \wedge n} 1_{\{X_{\tau \wedge n} \leq -R\}}) \\
&\geq -R \mathbb{P}(\tau>n) + \mathbb{E}(X_{\tau \wedge n} 1_{\{X_{\tau \wedge n} \leq -R\}}). \tag{2} \end{align*}
Hence,
$$\mathbb{E}(X_{\tau \wedge n} 1_{\{X_{\tau \wedge n} \leq -R\}}) \leq R \mathbb{P}(\tau>n) - \mathbb{P}(\tau \leq n) \xrightarrow[]{n \to \infty} 0-1 = -1.$$
In particular, there exists for any $R>0$ some $n \gg 1$ sufficiently large such that
$$\mathbb{E}(X_{\tau \wedge n} 1_{\{X_{\tau \wedge n} \leq -R\}}) \leq - \frac{1}{2},$$
i.e.
$$\mathbb{E}(|X_{\tau \wedge n}| 1_{\{X_{\tau \wedge n} \leq -R\}}) \geq \frac{1}{2}$$
which clearly shows that $(X_{n \wedge \tau})_{n \in \mathbb{N}}$ fails to be uniformly integrable.
Remark: A useful criterion to prove that a sequence $(Y_n)_n$ fails to be uniformly integrable is to show that $(Y_n)_{n \in \mathbb{N}}$ is unbounded in $L^1$, i.e. $\sup_{n \in \mathbb{N}} \mathbb{E}(|Y_n|)<\infty$. However, we cannot use this criterion here since $(X_{n \wedge \tau})_{n \in \mathbb{N}}$ is bounded in $L^1$.
Apply the definition of uniform intergablity. Suppose $\tau \leq N$. Then $E|X_{\tau \wedge n}|I_{\{|X_{\tau \wedge n}| >\Delta }\}$ $\leq E|X_n|I_{\{ \tau \geq nX_{\tau \wedge n >\Delta }\}}$ $+E(|X_1|+|X_2|+...+|X_N|) I_{\{{\tau <n,Y>\Delta }\}}$ where $Y=|X_1|+|X_2|+...+|X_N|$ The first term does not exceed $E|X_n|I_{\{|X_n| >\Delta\}}$. In the second term note that $P(Y >\Delta ) \to 0$ as $\Delta \to \infty$.
Best Answer
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 $(X_{n \wedge \tau})$.