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Put $Z_n=X_{n\wedge T}$. The stopped martingale in discrete time identityshows that $Z$ is a martingale: the indicator function is ${\mathcal{F}}_n$-measurable and the increment has zero conditional expectation. Integrability follows because $Z_n$ is selected from the finitely many integrable values $X_0,\dots ,X_n$.

For the bounded stopping time $\tau =n\wedge T$, the optional stopping theorem in its conditional form givesHere ${\mathcal{F}}_{\tau }$ is the stopping-time sigma-algebra. Indeed, for $A\in {\mathcal{F}}_{\tau }$, partition $A$ into $A\cap \{\tau =k\}\in {\mathcal{F}}_k$ and apply the martingale identity $\mathbb{E}[X_n\mid {\mathcal{F}}_k]=X_k$ on each piece. This proves the displayed conditional expectation identity directly.

Let $C={\sup }_n\mathbb{E}|X_n|<\infty$ and $A_n=\{|Z_n|>K\}\in {\mathcal{F}}_{n\wedge T}$. Conditional absolute-value domination and the Markov inequality give $\mathbb{P}(A_n)\le C/K$ and, for any $R>0$,First choose $R$ using the uniform integrability of $X$, and then choose $K$. The estimate is uniform in $n$, proving uniform integrability of a stopped uniformly integrable martingale. No finiteness assumption on $T$ is needed.

Recall the stopping-time sigma-algebra:For the proposed random time,The first set belongs to ${\mathcal{F}}_t$. Since $S\le T$, the second can be writtenwhich also belongs to ${\mathcal{F}}_t$. Therefore $U$ is a stopping time, as in pasting ordered stopping times. Moreover $S\le U\le T$, so $U$ is a bounded stopping time.