Suppose $(\Omega ,\mathscr{F}, P)$ be probability space. Let $X_n$ be independent identically distribution random variables, and $S_n:=X_1+\cdots +X_n$. If $p>0$, $\lim _{n \to \infty} \frac{S_n}{n^{1/p}}=0$ a.s., then $E|X_1|^p<\infty$
I know by Fubini's theorem, $E|X_1|^p=\int P(|X_1|^p>x)dx =\int P(|X_1|>x^{1/p})dx$
But, I can't prove it is finite.
Assume that $E|X_1|^p=\infty$,
As you noted \begin{align*} E|X_1|^p&=\int_0^{\infty} P(|X_1|^p>x)dx \\ &\leq \sum_{n=0}^{\infty} P(|X_1|^p>n)\\ &=\sum_{n=0}^{\infty}P(|X_n|^p>n) \end{align*} Hence, by Borel-Cantelli $\{|X_n|> n^{1/p} \,\,\text{i.o.}\}$ holds almost surely.
So, $$\frac{S_{n}}{n^{1/p}} = \frac{S_{n-1}}{n^{1/p}} +\frac{X_n}{n^{1/p}}\Longrightarrow \frac{S_{n}}{n^{1/p}} -\frac{S_{n-1}}{(n-1)^{1/p}} \frac{(n-1)^{1/p}}{n^{1/p}}=\frac{X_n}{n^{1/p}},$$ but the LHS goes to zero almost surely, whereas the RHS is not. Which gives the desired contradiction.