A problem about strong law of large numbers of Shiryaev's Probability This is a problem after the section "Strong Law of Large Numbers" of Shiryaev's Probability:

Let $\xi_1,\xi_2,...$ denote independent and identically distributed random variables such thatt $E|\xi_1|=\infty$. Show that
  $$\limsup_{n\to\infty}\left|\frac{S_n}{n}-a_n\right|=\infty\text{ (P-a.s.)}$$
  for every sequence of constants $\{a_n\}$.

I have no idea about it. Any hint please.
Thanks!
 A: Let $Y_i:=X_i-X'_i$, where $(X'_i)$ is an i.i.d. copy of $(X_i)_i$. Define 
$$A:=\left\{\limsup_{n\to \infty}\left|\frac 1n\sum_{i=1}^nX_i-a_n\right|<\infty\right\}$$
$$A':=\left\{\limsup_{n\to \infty}\left|\frac 1n\sum_{i=1}^nX'_i-a_n\right|<\infty\right\}.$$
The goal is to show that $\mathbb P(A)=0$. Since $A$ and $A'$ are independent and of equal probability, we are reduced to show that $\mathbb P(A\cap A')=0$. Notice that 
$$A\cap A'\subset \left\{\limsup_{n\to \infty}\left|\frac 1n\sum_{i=1}^nY_i\right|<\infty\right\},$$
and the sequence $(Y_i)_i$ is i.i.d., with $\mathbb E|Y_1|=\infty$. Indeed, by independence of $X_1$ and $X'_1$, 
$$\mathbb E|X_1-X'_1|=\int_\Omega\mathbb E|X_1-x'|\mathrm d\mathbb P_{X'_1}(x')=\infty.$$
Using the Borel-Cantelli lemma, we have that for each $R$, 
$\mathbb P(\limsup_i \{|Y_i|>iR\})=1$. 
A: I have a solution myself.
Let $X_i=\xi_i-\tilde{\xi}_i$, where $\tilde{\xi}_i$ is a independent copy of $\xi_i$. By the comment of mike, we have $E|X_i|=\infty$. And let $H_n=\sum_{i=1}^{n}X_i$. Since
$$\limsup_{n\to\infty}\frac{|H_n|}{n}\leq\limsup_{n\to\infty}\left|\frac{S_n}{n}-a_n\right|+\limsup_{n\to\infty}\left|\frac{\tilde{S}_n}{n}-a_n\right|$$
So by Kolmogorov $0$-$1$ law, it is sufficient to prove
$$\limsup_{n\to\infty}\frac{|H_n|}{n}\geq\infty\text{ (P-a.s.)}$$
Let $A>0$,
$$\infty=E\frac{|X_i|}{A}=\int_0^\infty P(|X_i|>\lambda A)d\lambda\leq\sum_{k=0}^\infty P(|X_i|>kA)$$
so
$$\sum_{i=1}^\infty P(|X_i|>iA)\geq\infty$$
Since $\{X_i\}$ are independent, by Borel-Catelli lemma
$$P(|X_i|>iA\text{ i.o.})=1$$
Since $H_{n+1}-H_n=X_{n+1}$, we have
$$P(|H_n|>\frac{nA}{2}\text{ i.o.})=1$$
i.e.
$$\limsup_{n\to\infty}\frac{|H_n|}{n}\geq\frac{A}{2}\text{ (P-a.s.)}$$
Take a sequence $0<A_m\uparrow\infty$, we get
$$\limsup_{n\to\infty}\frac{|H_n|}{n}\geq\infty\text{ (P-a.s.)}$$
So the problem has been proved.
