Let $(X_n)_{n \in \mathbb N}$ a sequence of independent random variables, where $X_n$ is uniformly continuous distributed across $(0,n)$. Show that $\liminf_{n\to \infty} X_n = 0$ almost surely. Hint: Use sets of the form $\{ X_n \leq c_n \}$ for a suitable $c_n \to 0$.
I've tried this so far:
(Probably wrong because first line takes a too large subset)
$$ \liminf \{X_n \leq c_n\} \subseteq \{\liminf X_n \leq c_n\}$$
$$ (\liminf \{X_n \leq c_n\})^c=\limsup \{X_n > c_n\}$$
Therefore:
$$P(\liminf X_n =0) = 1-P(\limsup \{X_n > c_n\})$$
Using Borel Cantelli Lemma for $P(\limsup \{X_n > c_n\}$ for independent $\{X_n> c_n\}$:
$$\sum_{n=0} ^\infty {P(\{X_n > c_n \})} = \sum_{n=0} ^\infty {\frac{n -c_n}{n}} = \sum_{n=0} ^\infty {1 -\frac{c_n}{n}} \to \infty$$
if $c_n \to 0$, therefore that would contradict the exercise by:
$$P(\liminf X_n =0) = 1-1=0$$
2nd Try:
$$ P(\liminf \{X_n \leq c_n\}) = P(\bigcap \bigcup \{X_n \leq c_n\}) \stackrel{decreasing}{=}\lim P(\bigcup_{m \geq n} \{X_m \leq c_m\})
$$
$$=\lim 1 – \prod_{m=n}^\infty {(1 – 1/m)} \to 1$$
Best Answer
Let $\varepsilon>0$ and set $$A_n:=A_n(\varepsilon)=\{X_n<\varepsilon\}.$$ Since $X_n$ are independent, then so is $A_n$'s. Since $\mathbb P(A_n)=\frac{\varepsilon}{n}$, we have that $$\sum_{n=1}^\infty \mathbb P(A_n)=\infty .$$
Using second Borel-Cantelli lemma, we get $$\mathbb P\{A_n\ \ i.o\}=\mathbb P\{X_n<\varepsilon\ \ i.o.\}=1.$$ Therefore, $$\mathbb P\left\{\liminf_{n\to \infty }X_n\leq \varepsilon\right\}=1.$$ Since $\varepsilon>0$ is unspecified, the claim follow.
Small exercise :-)
If you want, you can prove that if $X_n$ are i.i.d. uniform $[0,n^\alpha ]$ for a $\alpha >1$, then $$\liminf_{n\to \infty }X_n=+\infty \ \ \ a.s.$$