$E[X]$ finite iff $\sum\limits_{n} P(X>an)$ converges Show that: 
$$\sum\limits_{n \in N } P(X>an) < \infty\ \text{for some}\ a > 0 \Rightarrow E[X] < \infty \Rightarrow \sum\limits_{n \in N } P(X>an) < \infty\ \text{for every}\ a > 0.$$
We have an unknown probability space ($\Omega,\mathbb{F},P)$, where $\mathbb F$ is a sigma-algebra, and $P:\mathbb F \rightarrow [0,1]$ is a probability measure, $X$ is a $\mathbb{F}/\mathbb{B}(\mathbb{R})$-measurable random variable and $E[X]$ is the expectation of the variable $X$. 
I've been working through past exam papers, and am stuck on this. Would appreciate a solution! 
 A: Here's a hint: for $X>0$,
$$E[X] = \int_0^\infty x f(x) \mathrm d x = \sum_{n=0}^\infty \int_{na}^{(n+1)a} x f(x) \mathrm d x$$
Can you bound these integrals in terms of integrals of $f$? What can you do with the factors?
A: Try to integrate the pointwise inequality $X\leqslant a\sum\limits_{n=0}^{+\infty}\mathbf 1_{X\gt an}$ to get one implication. For the other implication, try to find a similar pointwise inequality, only reversed.
Edit: Consider $a\sum\limits_{n=1}^{+\infty}\mathbf 1_{X\gt an}$ (note that the sum is starting at $n=1$, not $n=0$), can you compare it to $X$?
A: We prove the first implication.
Let's take $a>0$ such that $\sum_{n \in N } P(X>an) < \infty$.
First of all, $P(X>an)=\sum_{k=n}^{\infty}P(X\in(an,a(n+1)])$. Hence, we write
$\infty>\sum_{n \in N } P(X>an) = \sum_{n \in N } \sum_{k=n}^{\infty} P(X\in(ak,a(k+1)])$. By Tonelli's theorem we can change the order of summation. We arrive at
$\sum_{k \in N } \sum_{n=0}^{k} P(X\in(ak,a(k+1)])=\sum_{k \in N }  (k+1) P(X\in(ak,a(k+1)])$.
If $d\mu(x)$ represents our probability measure (i.e. $P(X\in A)=\int_{x\in A} d\mu(x)$), then we can continue 
$ \sum_{k \in N }  (k+1) P(X\in(ak,a(k+1)]) = \frac {1}{a}\sum_{k \in N }  a(k+1) \int_{(ak,a(k+1)]}d\mu$
$\ge \frac {1}{a}\sum_{k \in N }   \int_{x\in (ak,a(k+1)]}xd\mu(x)=\frac {1}{a}    \int_{x>0}xd\mu(x)\ge \frac {1}{a}E[X]$ by definition. In other words, $E[X]<\infty$. Note that there can still be $E[X]=-\infty$.
In order to prove the second implication, a little modofication of this trick is enough.
