Does it hold for any random variable that $E[X]$ exists iff $\sum_{n\geq 1}P(X\geq n)<\infty$? 
If $X:\Omega \to \mathbb R$ is a random variable, show that $E[X]$ exists iff $\sum_{n\geq 1}P(X\geq n)<\infty$

I can prove it only under the additional assumption that $X\geq0$. So I was wondering how to prove it for $X:\Omega \to \mathbb R$ or is it just a typo in my textbook and there is a counterexample.
 A: You are right that it is a typo. Sometimes, instead of stipulating that $X$ be greater than zero, the summation is written by taking the absolute value of $X$, that is, $\mathbb P(|X| > n)$.
A: Note that when we say "$E(X)$ exists", we mean $X \in L^1(P)$ by definition. That is, the expectation of $X$ is only defined if $E(|X|) < \infty$. So what you are actually asking is how to show that $E(|X|)< \infty$ if and only if $\sum_{n=1}^{\infty}P(|X| \geq n) < \infty.$
To show this, I use the theorem which states that for any positive random variable $Y$, 
$E(Y)= \int_{0}^{\infty} P(Y \geq x) dx$. Applying this theorem, we see that
$$E(|X|) = \int_{0}^{\infty} P(|X| \geq x) = \sum_{n=0}^{\infty} \int_n^{n+1}P(|X| \geq x) dx.$$
Now for any $x \in [n,n+1],$ we have that $\{ |X| \geq n+1 \} \subset \{|X| \geq x \} \subset \{|X| \geq n\},$ and so $P(|X| \geq n+1) \leq P(|X| \geq x) \leq P(|X| \geq n).$ It follows that
$$\sum_{n=0}^{\infty}P(|X| \geq n+1) \leq E(|X|) \leq \sum_{n=0}^{\infty} P(|X| \geq n).$$
Therefore, $E(|X|)$ is finite if and only if these series converge. 
A: In some situations, we can still give a sense to $E(X)$, whereas $E|X| = + \infty$. Namely, we can do it when either $E(X^+)$ or $E(X^-)$ is finite, where $X^+$ and $X^-$ denote the positive and the negative parts of $X$ respectively.
In this case, we define $E(X) = E(X^+) - E(X^-)$, and this quantity is $+\infty$ or $-\infty$ when $E|X| = \infty$. In this case, $X$ is called quasi-integrable.
The only situation where we cannot define $E(X)$ is when both $E(X^+)$ and $E(X^-)$ are infinite.
To come back to the exercise, if we use the generalized definition of $E(X)$ I've mentionned, the quantity $E(X)$ is well defined if $\sum_n P(X \ge n) < \infty$, and can possibly be equal to $- \infty$.
Nevertheless, if $E(X)$, it doesn't imply that $\sum P(X \ge n) < \infty$, since $E(X)$ can be $+ \infty$.
All that being said, I think there is a typo in the exercise, and the right condition should be $\sum P(|X| \ge n)< \infty$, as it has been said, and that the usual definition of integrability is the one used there.
