Strong Law of Large Numbers for a i.i.d. sequence whose integral does not exist

Question

Prove : let $X_1 ,X_2 , ... $ be i.i.d. random variables with $\mathbb{E}[X_1^+]=\mathbb{E}[X_1^-]=+\infty$, and $S_n=\sum_{i=1}^{n}{X_i}$. Then $$\limsup_{n\rightarrow\infty}{\frac{S_n}{n}=+\infty}\text{ a.s., }\liminf_{n\rightarrow\infty}{\frac{S_n}{n}=-\infty}\text{ a.s.}$$

I have proven that $$\mathbb{P}\left(\left\{\omega :\limsup_{n\rightarrow\infty}{\left|\frac{S_n}{n}(\omega)\right|=+\infty}\right\}\right)=1,$$

Since the events $$\left\{\omega:\limsup_{n\rightarrow\infty}{\frac{S_n}{n}(\omega)=+\infty}\right\} \text{ and } \left\{\omega:\liminf_{n\rightarrow\infty}{\frac{S_n}{n}(\omega)=-\infty}\right\}$$ are asymptotic with regard to the sequence $X_1,X_2,...$ their probability is $0$ or $1$, and at least one has probability one, but I don't know how to prove both of them.

Update: According to the paper The strong law of large numbers when the mean is undefined (Erickson K B, 1973), this proposition is wrong.

Answer 1

Corollary 3 (p. 1195) in [Kesten (1970)][1] states:

If $$\mathbb E[X_1^+]=\mathbb E[X_1^-]=\infty,$$ then one of the following three cases must prevail

(i) $\displaystyle\lim_{n\to\infty}\frac{S_n}n = \infty$ w.p. 1

(ii) $\displaystyle\lim_{n\to\infty}\frac{S_n}n = -\infty$ w.p. 1

(iii) $\displaystyle\liminf_{n\to\infty}\frac{S_n}n = -\infty$ and $\displaystyle\limsup_{n\to\infty}\frac{S_n}n = +\infty$ w.p. 1

From the equivalence of (b) and (c) in Theorem 6:

If $\mathbb E[X_1^+]=\infty$ then the following statements are equivalent

(b) $\displaystyle\mathbb P\left(\limsup_{n\to\infty}\frac{S_n}n>-\infty\right)=1$

(c) $\displaystyle\mathbb P\left(\limsup_{n\to\infty}\frac{S_n}n=+\infty\right)=1$

and the Hewitt-Savage zero-one law, if (ii) holds then $$\mathbb P\left(\limsup_{n\to\infty}\frac{S_n}n=\infty\right)=1. $$ If neither (i) nor (ii) hold, then similarly $$\mathbb P\left(\liminf_{n\to\infty}\frac{S_n}n=-\infty\right)=1,$$ from which the result follows.