Q: How can I show that $|X|^a$ is Lebesgue integrable if for $a \in [1,\infty)$ $\iff \sum_{k=1}^\infty k^{a−1}P(|X| \ge k) < \infty$?
I think that $X$ is only to be integrable if $E[X^+]$ and $E[X^−]$ are both finite. So is it OK to show that $E[(|X|^a)^+]$ and $E[(|X|^a)^−]$ are finite? If so then I guess It makes sense that $E[(|X|^a)^+]$ is finite as this is the positive part but for negative values I am not so sure how to show? Is this right? I don't really understand how to link this with the right hand side of equation.
First remember the well-known formula $$\mathbb{E}[|X|^a] = a \int_0^\infty x^{a-1} P( |X| \geq x) \, \mathrm{d} x,$$ which can be proven by an application of Fubini's theorem (write $|X|^a = \int_0^{|X|} a \, x^{a-1} \, \mathrm{d} x$). By the usual monotonicity argument we see as well that$$\sum_{n=0}^\infty n^{a-1} P(|X| \geq n+1) \le\int_0^\infty x^{a-1} P( |X| \geq x) \, \mathrm{d} x \le \sum_{n=1}^\infty n^{a-1} P(X \geq n-1).$$ To prove the equivalence stated in the question, we need to add some technical arguments only.
$\Longrightarrow$: If $|X|^a$ is integrable, then we have $$\sum_{n=0}^\infty n^{a-1} P(|X| \geq n+1) < \infty.$$ Since $a>1$, one has $$\sum_{n=0}^\infty n^{a-1} P(|X| \in [n,n+1)) =\sum_{n=0}^\infty n^{a-1} \int 1_{[n,n+1)}(|X|) \, \mathrm{d}P \leq \int |X|^{a-1} \, \mathrm{d} x \leq \int 1+ |X|^{a} \, \mathrm{d} P<\infty$$ and thus we may conclude finally that $\sum_{n=0}^\infty n^{a-1} P(|X| \geq n) < \infty$.
$\Longleftarrow$: On the other hand, if $\sum_{n=0}^\infty n^{a-1} P(|X| \geq n) < \infty$, then we may use that $$n^{a-1} \leq 1 + 2^{a-1} (n-1)^{a-1}$$ to see that \begin{align} \sum_{n=1}^\infty n^{a-1} P(X \geq n-1) &\leq \sum_{n=0}^\infty P(X \geq n) + 2^{a-1} \sum_{n=0}^\infty n^{a-1} P(|X| \ge n) \\ &\leq 1 + (1+2^{a-1}) \sum_{n=0}^\infty n^{a-1} P(|X| \ge n) < \infty \end{align} and thus $|X|^a$ is integrable.