Is there a random variable where E[X] exists but the expectation of the negative and positive part do not? Is there a random variable where $E[X]$ exists but the expectation $E[X^{+}]$ and $E[X^{-}]$ do not? If not, how can I show this?
I know that it is usually written that
$$ X = X^{+} - X^{-}$$ 
and that the expectation operator is linear, giving
$$ E[X] = E[ X^{+} - X^{-}] = E[X^{+}] - E[X^{-}].$$ 
But I'm not convinced. One might just argue that 
$$E[ X^{+} - X^{-}] = E[X^{+}] - E[X^{-}]$$
holds only when $E[X^{+}]$ and $E[X^{-}]$ are finite. 
Thus I ask, are there cases when $E[X^{+}]$ and $E[X^{-}]$ are not finite but $E[ X^{+} - X^{-}]$ is. 
If no, how can I show this more convincingly?
 A: What is the basic definition of $E(X)$? It involves summing or integrating over possible $X$-values. Consider restricting the summation/integration over only positive, or only negative $X$-values. If both of these things are infinite in opposite direction, then summing/integrating over all $X$ values introduces a convergence issue. The order that you sum or integrate can influence the value/existence of the result. So unless you provide a canonical order to sum over, there is no meaning to the sum/integral.
For example, suppose $P(X=n)=\frac{1}{n^2}$, for $n\in\mathbb{Z}\setminus{0}$. Of course, this would need to be normalized to make the total probability $1$. 
Now the expected value of the positives brings you to the Harmonic series, and the expected value of the negatives brings you to the negative Harmonic series, both of which diverge.
To "find" $E(X)$, one person might sum $$\frac{1}{1}-\frac{1}{1}+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\cdots$$ which converges to $0$.
But someone else might sum $$\frac{1}{1}+\frac{1}{2}-\frac{1}{1}+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+\frac{1}{5}+\frac{1}{6}-\frac{1}{3}+\cdots$$ with a "next two positives, next one negative" pattern. Here, the sum clearly either converges to a postive number or diverges, since each triple is a positive amount. So there's inconsistency with the first ordering.
The first approach may seem more natural, and that might be OK. But you would have to specify that this is your chosen order for summing.
A: So, I think I've got an answer.
If X is a real-valued random variable and $X \in \mathcal{L}^1(P)$, then X is called integrable and the expectation is defined
$$ E[X] := \int X dP. $$
We can then see the result by looking at how the definitions here.
For f measurable, 
$$ \|f\|_p \equiv \left({\int_S |f|^p\;\mathrm{d}\mu}\right)^{\frac{1}{p}} $$
if $p \in [1,\infty]$,
and 
$$\mathcal{L}^p(\mu) := \{f : \Omega \rightarrow \bar \Re \; \text{is measurable and} \|f\|_p < \infty \}. $$
So by definition $E[X^k] < \infty$ implies that $E[|X|^k] < \infty$.
("Probability Theory" by Klenke, p.101)
We can then use the equality $X = X^{+} - X^{-}$ and $|X| = X^{+} + X^{-}$ to answer the desired question.  
