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This is probably a stupid question, but am I mislead if I think that as soon as a stochastic process indexed by $t$ (continuous time) is not uniformly integrable (UI) for a certain range of $t$, say on $[0,1]$, it is also not UI on $[0,\infty]$? And if that is wrong, then why so?

Thank you in advance!

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    Oh, I made a mistake, very sorry, I wanted to know if it holds for a process, which is NOT UI, changed my question.2012-04-25

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For any collections $C$ and $D$ of random variables, if $D$ is uniformly integrable and $C\subseteq D$, then $C$ is uniformly integrable.

To see this, recall that $D$ is uniformly integrable if and only if $S(x,D)\to0$ when $x\to+\infty$, where $ S(x,D)=\sup\{\mathrm E(|X| ; |X|\geqslant x)\mid X\in D\}. $ Now, if $C\subseteq D$, then $S(x,C)\leqslant S(x,D)$ for every $x$, hence the conclusion follows.

Thus, the contraposition holds: if $C$ is not uniformly integrable and $C\subseteq D$, then $D$ is not uniformly integrable.

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I guess the definition of a stochastic process on an interval $I$ you are using is the following:

$\{X_t,t\in I\}$ is uniformly integrable if and only if $\lim_{R\to\infty}\sup_{t\in I}\int_{\{|X_t|\geq R\}}X_t(\omega)\mathrm d\mathcal P(\omega)=0$.

With this definition, we can have that $\{X_t,0\leq t\leq 1\}$ is uniformly integrable but \{X_t,0\leq t<\infty\} is not. We can take a deterministic example, like X_t(\omega)=\begin{cases}0&\mbox{ if }0\leq t\leq 1\\\ t&\mbox{ if }1