SLLN for iid $X_n$

Question

For i.i.d $X_n$'s with $E[(X_1)_-]<+\infty$ and $E[(X_1)_+]=+\infty$, I want to prove that $$\frac{1}{n}\Sigma_{i=1}^nX_i\xrightarrow{\text{a.e}} +\infty$$ as $n\xrightarrow{}+\infty$.

I know that I have to use the Strong Law of Large numbers but I am confused on how to implement it. I think I have to find out that the mean of $\Sigma X_i$ is infinity.

I also found in my notes that $n^{-1}S_n\xrightarrow{\text{a.e}}\mu$ as $n\xrightarrow{}\infty$ whenever $E[X_1]<\infty$ and $S_n=\Sigma_{i=1}^nX_i$.

I would appreciate any answers to this.

Answer 1

Let $X_i^M(\omega) = \min\{X_i(\omega), M\}$, and $S_n^M=\sum_{i=1}^nX_i^M$. Then almost surely: $$\liminf_{n\rightarrow\infty}\frac{S_n}n\ge\lim_{n\rightarrow\infty}\frac1nS_n^M = E(X_1^M)$$ By monotone convergence, $E(X_{1,+}^M)\rightarrow\infty$ as $M\rightarrow\infty$, and since $E(X_{1,-}^M)=E(X_{1,-})<\infty$, the claim follows.