= Solution
The original PDF has $\mathbb E|X_T|<\infty$ and $\mathbb E[|X_n|\mathbf1_{\{T>n\}}]\to0$. The TeX transcription drops the <expected value> and absolute value in the first condition, and the absolute value in the second. The proof uses the PDF's conditions.
Because the <stopping time> $T$ is finite <almost surely>, $X_{n\wedge T}\to X_T$ <almost surely>. More strongly,
$$
\mathbb E|X_{n\wedge T}-X_T|
=\mathbb E\left[|X_n-X_T|\mathbf1_{\{T>n\}}\right]
\leq\mathbb E\left[|X_n|\mathbf1_{\{T>n\}}\right]
+\mathbb E\left[|X_T|\mathbf1_{\{T>n\}}\right]\longrightarrow0.
$$
The first term tends to zero by hypothesis; the second does so by the <dominated convergence theorem>. Thus there is <convergence in L1>.
Here is the needed <L1 convergence implies uniform integrability> argument. For any <integrable random variables> $Z,Y$ and $K>0$, splitting according to $|Y|>K/2$ gives
$$
\mathbb E\left[|Z|\mathbf1_{\{|Z|>K\}}\right]
\leq2\mathbb E|Z-Y|+\mathbb E\left[|Y|\mathbf1_{\{|Y|>K/2\}}\right].
$$
Take $Z=X_{n\wedge T}$ and $Y=X_T$. For large $n$, the first term is uniformly small by <convergence in L1>, and the second is small for large $K$ by <integrable random variable>[integrability]. The finitely many remaining $n$ are handled individually by <integrable random variable>[integrability]. Hence \b[the stopped process is <uniformly integrable>]. This proves the <stopped-martingale uniform integrability criterion>; the <martingale> assumption is not needed for this particular implication.
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