Let $X$ be a non-negative random variable then $\mathbb{E}(|X|) <\infty$ if and only if $\sum_{n}^{\infty}\mathbb{P}(|X| > \epsilon n) < \infty $ for every $\epsilon > 0$.
I tried in the following way,
let $\epsilon = 1$
Now $$ \mathbb{E}(|X|) = \int_0^{\infty}\mathbb{P}(|X|>t)dt\\
= \sum_{n=1}^{\infty} \int_{n-1}^{n}\mathbb{P}(|X|>t)dt\\
$$
Since $\mathbb{P}(|X|>t)$ is non-increasing function then
$$
\sum_{n=1}^{\infty}\mathbb{P}(|X|>n)\leq\mathbb{E}(|X|)\leq \sum_{n=1}^{\infty}\mathbb{P}(|X|>n-1)\\
\Rightarrow \sum_{n=1}^{\infty}\mathbb{P}(|X|>n)\leq\mathbb{E}(|X|)\leq 1 +\sum_{n=1}^{\infty}\mathbb{P}(|X|>n)
$$
Hence the result follow by comparison test.
How can I prove this for every $\epsilon >0$.
Thanks in advance
Best Answer
What you have done is correct if you repalce $i$ by $n$ throughout.
Note that $E|X|<\infty$ iff $E|\frac X {\epsilon} | <\infty$ for any $\epsilon >0$. Apply what you have proved to the random variable $\frac X {\epsilon}$.