Let $ \{b_n\} $ be a sequence of non-zero complex numbers. We have $ N(t)=\#\{n \geq 1:|b_{n}|\leq t\} $, $ \displaystyle\limsup_{t\rightarrow\infty} N(t)/t^p<\infty (1\leq p <2)$, for $ X_1\in L_p(P) $ , why do we have: $ E[N(|X_1|)]<\infty $?
http://www.ee.bgu.ac.il/~guycohen/clsurvey.pdf see also in this link, page 5, theorem 2.2
We have $E(|X_1|^p)<\infty$ because $X_1\in L^p$ and that $N(|X_1|)=\#\{n\geq1 : |b_n|\leq|X_1|\}$. $E(N(|X_1|))=\displaystyle\sum_{n=1}^\infty P(N(|X_1|)\geq n)$. There must exist a constant $C$ such that $N(t)\leq Ct^p$ for all $t$ sufficiently large.
Thus we get $N(|X_1|)\leq C |X_1|^p$ when $|X_1|$ takes sufficiently large values, say when $|X_1|>K$.
Let $M=\sup_{t\leq K} N(t)$. We have $N(|X_1|)\leq M$ when $|X_1|\leq K$ and that $N(|X_1|)\leq C |X_1|^p$ when $|X_1|> K$.
$$ \begin{aligned} E(N(|X_1|)) &\leq M P(|X_1| \leq K)+C\sum_{n=K+1}^\infty n^pP(|X_1|= n)\\ &\leq M+C\cdot E(|X_1|^p) < \infty. \end{aligned}$$