Given $\left\{A_n\right\}$ are i.i.d. Show that $E(|A_1|)< \infty$ $\iff \ P\left\{|A_n| > n \ \ \text{i.o}\right\} = 0$.
My attempt: ($\Rightarrow$) Assume $E(|A_1|)< \infty$. Since $\left\{A_n\right\}$ are i.i.d, by the Strong Law of Large Number, $\frac{1}{n}\sum_{i=1}^{n} |A_i|\rightarrow E(|A_1|)$ almost surely. This is equivalent to $\forall\ \epsilon > 0$, $\lim_{n\rightarrow \infty} P(\cup_{k\geq n}|(\frac{1}{k}\sum_{i=1}^{k} |A_i|)-E(|A_1|)|\geq \epsilon)= 0$. This implies for sufficiently large $k$, $\frac{1}{k}\sum_{i=1}^{k} |A_i| - E(|A_1|)\ <\ \epsilon$. Since we don't know whether $E(|A_1|) > 1$ or not, we cannot choose $\epsilon = 1-E(|A_1|)$ to get $|A_k|$ bounded above by $k$. Could someone please help with this part?
($\Leftarrow$) If $P\left\{|A_n| > n \ \ \text{i.o}\right\} = 0$, this implies for sufficiently large $k$, $k$ is fixed, $|A_k|< k$. This implies $E(|A_k|) = E(|A_1|) < E(k) = k < \infty$ (the first equality is due to $\left\{A_n\right\}$ are iid).
My question: Could someone help complete my "solution" above for the forward direction? If I'm on the wrong track, please help point that out to me.
The forward part is an easy consequence of Borel-Cantelli'e lemma. First of all see that: $$ \mathbb E(|A_1|)\geq \sum_{n=1}^\infty \mathbb P(|A_1|>n). $$ Using i.i.d. assumption: $$ \sum_{n=1}^\infty \mathbb P(|A_1|>n)= \sum_{n=1}^\infty \mathbb P(|A_n|>n). $$ So $\sum_{n=1}^\infty \mathbb P(|A_n|>n)<\infty$ and therefore from Borel-Cantelli's lemma: $$ \mathbb P(|A_n|>n , \rm{ i.o. })=0. $$