I have the following problem on my problem set:
I approached the problem this following way, which I think is the right path but I'm still struggling:
We have $A_{n} = \{n-1\leq |X| < n\}$ and we know that $X \in \mathcal{L}^p$ if $\mathbb{E}(|X|^p)$ is finite. For every $\omega \in A_{n}$, it holds that $(n-1)^p\leq |X|^p < n^p$. For any $\omega \in \Omega$, since $(A_{n})$ defines a partition: \begin{gather*} \sum\limits_{n=1}^{\infty}(n-1)^p 1_{A_{n}} \leq |X|^p < \sum\limits_{n=1}^{\infty}n^p 1_{A_{n}}\\ \sum\limits_{n=1}^{\infty}(n-1)^p \mathbb{P}(A_{n}) \leq \mathbb{E}(|X|^p) < \sum\limits_{n=1}^{\infty}n^p \mathbb{P}(A_{n}) \end{gather*} Now, if $\sum\limits_{n=1}^{\infty}n^p \mathbb{P}(A_{n})$ is finite, it implies that $\mathbb{E}(|X|^p)$ is also finite, which means that $||X||_{p} = [\mathbb{E}(|X|^p)]^{1/p} < +\infty$ and $X \in \mathcal{L}^p$.
This is only halfway. Now I need to prove that if $\sum\limits_{n=1}^{\infty}n^p \mathbb{P}(A_{n})$ if finite, then $\mathbb{E}(|X|^p)$ is also finite. I procede the following way:
Using the above inequalities, if $\mathbb{E}(|X|^p)$ finite, then $\sum\limits_{n=1}^{\infty}(n-1)^p \mathbb{P}(A_{n})$ is finite and also is $\sum\limits_{n=1}^{\infty}(n)^p \mathbb{P}(A_{n+1})$ since we also have that $\sum\limits_{n=1}^{\infty}(n)^p \mathbb{P}(A_{n+1}) \leq \mathbb{E}(|X|^p) < \sum\limits_{n=1}^{\infty}(n+1)^p \mathbb{P}(A_{n+1})$. The same logic tells us that $\sum\limits_{n=1}^{\infty}(n)^p \mathbb{P}(A_{n+1}) < \sum\limits_{n=1}^{\infty}(n)^p \mathbb{P}(A_{n})$.
I couldn't compare anything else in order to achieve the result. Any ideas? Am I on the right path? Thanks a lot in advance!!