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Let $U_n\sim \mathcal U[0,1]$ iid. Is it possible that $$\liminf_{n\to \infty } nU_n>0\ \ ?$$

I said no. Indeed, $\mathbb P\{U_n\leq u\}= u$. Therefore, $$\sum_{n=1}^\infty \mathbb P\{nU_n\leq \varepsilon\}=\sum_{n=1}^\infty \mathbb P\{U_n\leq \frac{\varepsilon}{n}\}=\varepsilon\sum_{n=1}^\infty \frac{1}{n}=+\infty .$$ Therefore, by Borel-Cantelli 2nd Lemma, $$\mathbb P\{\limsup_{n\to \infty }nU_n\leq \varepsilon\}=1,$$ and thus, $$\limsup_{n\to \infty }nU_n=0\ \ a.s.$$ Since $nU_n\geq 0$ a.s. we have that $\liminf_{n\to \infty }nU_n=0$ a.s. what conclude the claim. Is it correct ?

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    You are incorrectly reaching the correct conclusion "$\liminf_{n\rightarrow\infty} nU_n=0$ almost surely." However, it is easy to show that $\limsup_{n\rightarrow\infty} nU_n=\infty$ almost surely, which is different from what you state. Your mistake is in the application of the Borel-Cantelli Lemma. The fact $\sum_{n=1}^{\infty} P[n U_n \leq \epsilon] = \infty$ means that the events $\{nU_n \leq \epsilon\}$ almost surely occur infinitely often (i.e., for infinitely many indices $n$). From this you can conclude $\liminf_{n\rightarrow\infty} nU_n \leq \epsilon$ almost surely. – 2017-01-01
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    @Michael You are right. – 2017-01-02
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    @Michael: but $\{nU_n\leq\varepsilon\}$almost surely occur infinitely often doesn't mean that $\mathbb P\{\limsup_{n\to \infty }nU_n\leq \varepsilon\}=1$ ? I always thought that "infinitely often" and limsup was the same... – 2017-01-02

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Fix $\epsilon>0$. Notice that, with prob 1, the events $\{n U_n \leq \epsilon\}$ occur infinitely often (i.e., for infinitely many indices $n \in \{1, 2, 3, ...\}$), but the events $\{n U_n > \epsilon\}$ also occur infinitely often.

Your confusion is because there is a concept of “$\limsup$ of a sequence of events” that is different from the concept of “$\limsup$ of a sequence of random variables.” In this problem, the following two equalities can be proven: \begin{align} P\left[\limsup_{n\rightarrow\infty}\{n U_n \leq \epsilon\}\right] &=1 \\ P\left[\left(\limsup_{n\rightarrow\infty} nU_n\right) \leq \epsilon\right] &=0 \end{align} where the first is the probability that the events $\{n U_n \leq \epsilon\}$ occur infinitely often, and the second is the probability that the random variable sequence $n U_n$ has a $\limsup$ that is less than or equal to $\epsilon$. These two equations are different and you cannot prove one from the other. Your derivation using Borel-Cantelli proves the first. The second is proven by observing that $\{U_n>1/2\}$ occurs infinitely often (with prob 1) and so we can find arbitrarily large indices $n$ such that $n U_n \geq \epsilon + 1$.

Here are the basic definitions of $\limsup$ and $\liminf$ of a sequence of events, using language of "infinitely often" (i.o.) and "finitely often" (f.o). Suppose we have a sequence of events $\{A_n\}_{n=1}^{\infty}$ in some sample space $S$. So $A_n \subseteq S$ for all $n \in \{1, 2, 3, …\}$. We define:

\begin{align} \limsup_{n\rightarrow\infty} A_n &\equiv \cap_{n=1}^{\infty}\left[\cup_{m\geq n} A_m\right] \equiv \{\{A_n\} \quad i.o\} \equiv \{\mbox{An infinite number of events $A_n$ occur}\} \\ \liminf_{n\rightarrow\infty} A_n &\equiv \cup_{n=1}^{\infty} \left[\cap_{m \geq n} A_m\right] \equiv \{\{A_n^c\} \quad f.o.\} \equiv \{\mbox{All but finitely many events $A_n$ occur}\} \end{align} where $A_n^c$ is the complement of the event $A_n$. It can be shown that: $$ \liminf_{n\rightarrow\infty} A_n \subseteq \limsup_{n\rightarrow\infty} A_n $$ which means that if the events occur for all but finitely many indices $n$, then they must occur for infinitely many indices $n$. In general, the reverse is not true: If the events occur for infinitely many indices, they do not necessarily occur for all but finitely many indices.