Let $(E,\mathcal{A},\mathbb{P})$ be a probability space $\{X\_n\}$ be a sequence of random variable, such that:


$$
(1)~.~~~\sup\_n\mathbb E (|X\_n|)<\infty\Rightarrow
$$
$$
(2)~.~~~\dfrac{M\_j}{2}<\int\_{j-1<|X\_n|\leq j}{|X\_{n}(t)|d\mathbb{P}(t)}\leq M\_j+\dfrac{1}{j^2} \qquad\forall n\geq 1 \text{ et }1\leq j\leq n^2
$$
with $M\_j>0$ such that $\sum\_{j=1}^{\infty}{M\_j}<\infty$,and
$$
(3)~.~~~\{|X\_n|\mathbb{1}\_{|X\_n|\leq n}\} \text{ is uniformly integrable},
$$  

furthermore 
$$ 
(4)~.~~~\sum\_{n=1}^{\infty}{\mathbb{P}(\{|X\_n|>n\})}<\infty.
$$


Show that :
$$
 \sum\_{n=1}^{\infty}{\frac{1}{n^{1+\epsilon}}\mathbb{E}\big((|X\_n|\mathbb{1}\_{|X\_n|\leq n})^{1+\epsilon}\big)}<\infty,~~\forall\epsilon>0
$$
An idea please.



$\sum_{n=1}^{\infty}{\frac{1}{n^{1+\epsilon}}\mathbb{E}\big((|X_n|\mathbb{1}_{|X_n|\leq n})^{1+\epsilon}\big)}<\infty,~~\forall\epsilon>0 $
Let $N\_j:=M\_j+1/j^2$, so that 
$$\sum\_1^\infty N\_j<\infty,$$
and for $j\le n$
$$E|X\_n|1\_{j-1<|X\_n|\le j}\le N\_j$$
and hence 
$$(j-1)P(j-1<|X\_n|\le j)\le E|X\_n|1\_{j-1<|X\_n|\le j}\le N\_j,$$
so that for $j=2,\dots,n$
$$P(j-1<|X\_n|\le j)\le \frac{N\_j}{j-1}\ll\frac{N\_j}j,$$
where $a\ll b$ means $a=O(b)$. Since $N\_1\ge1$, we have 
$$P(j-1<|X\_n|\le j)\ll\frac{N\_j}j$$
for all natural $j\le n$. 
So, for $p:=1+\epsilon>1$, 
$$E|X\_n|^p 1\_{|X\_n|\le n}\le\sum\_{j=1}^n j^p P(j-1<|X\_n|\le j)
\ll\sum\_{j=1}^n j^{p-1}N\_j$$
and hence 
$$\sum\_{n=1}^\infty\frac1{n^p}\,E|X\_n|^p 1\_{|X\_n|\le n}
\ll\sum\_{n=1}^\infty\frac1{n^p}\,\sum\_{j=1}^n j^{p-1}N\_j
=\sum\_{j=1}^\infty j^{p-1}N\_j\sum\_{n=j}^\infty\frac1{n^p}
\ll\sum\_{j=1}^\infty N\_j<\infty.$$
Thus,
$$\sum\_{n=1}^\infty\frac1{n^p}\,E|X\_n|^p 1\_{|X\_n|\le n}<\infty,$$
as desired.


