Let $(X_n)_{n \in \Bbb N}$ be i.i.d.
I now have to show $\forall \epsilon \gt 0$ $$P(\mid X_n \mid \ge \epsilon \cdot n \:\:\mathrm{infinitely}\:\:\mathrm{often}) = 0 \Leftrightarrow \Bbb E[\mid X_1 \mid] \lt \infty$$
$$P(\mid X_n \mid \ge \epsilon \cdot n \:\:\mathrm{infinitely}\:\:\mathrm{often}) = 1 \Leftrightarrow \Bbb E[\mid X_1 \mid] = \infty$$
intuitivly the statements are somewhat clear but I don't know how to connect the expectation value and the infinitely often statement. I'm sure one must use Borel-Cantelli somewhere but I'm stuck.
Any ideas? Thanks in advance!
Since $\{X_n\}$'s are all independent, the probability of the following event $$ \{\omega\in\Omega:|X_n(\omega)|\geq n\epsilon \text{ infinitely often}\} $$ is directly dictated by the sum $$ \sum_{n=1}^{\infty}\mathbb{P}(|X_n|\geq n\epsilon). $$ Indeed, if this sum is finite, then by the first Borel-Cantelli lemma that the aforementioned event has probability $0$. On the other hand, if this sum is $\infty$, then second Borel-Cantelli lemma tells us that the aforementioned probability is $1$.
Next, using the identicality of distibutions, we have $$ \sum_{n=1}^{\infty}\mathbb{P}(|X_n|\geq n\epsilon)=\sum_{n=1}^{\infty}\mathbb{P}(|X_1|\geq n\epsilon). $$ Now we make the following claim.
Claim $$ \sum_{n=1}^{\infty}\mathbb{P}(|X_1|\geq n\epsilon) \text{ converges (diverges) }\iff \mathbb{E}[|X|]\text{ converges (diverges) }. $$ To see why this holds, recall that $\mathbb{E}[|X|]=\int_0^\infty \mathbb{P}(|X|\geq x)dx$, and consider a partition $\{n\epsilon:n\in\mathbb{N}\}$ of real numbers. We can bound the integral via upper and lower Riemann sums corresponding to this partition via $$ \sum_{n=1}^\infty \mathbb{P}(|X|\geq n\epsilon)\leq \int_0^\infty \mathbb{P}(|X|>x)dx \leq \sum_{n=0}^\infty \mathbb{P}(|X|\geq n\epsilon), $$ hence the sum and integral converge (diverge) together.