Is Discrete expectance defined for all reals? This may be a bit of a dumb question, but Let $(S,P)$ be a discrete probability space, $S$ a countable set and $P$ a probability distribution, is the expectance defined for all $\alpha \in \mathbb{R}$, provided $|\mathbb{E}(X)| < \infty$? That is, do we have
$$\mathbb{E}(X) = \sum_{\alpha \in \mathbb{R}} X(\alpha)P(\alpha)?$$
I think the answer is yes, since we would just set $P(\alpha) = 0$ for all $\alpha \notin S$.
 A: In short the answer is yes, but there are some technicalities. These basically boil down to the fact that $\sum_{\alpha \in \mathbb{R}} F(\alpha)$ is not so simple to define. One way to define it if $F \geq 0$ is as
$$\sup_{S \subset \mathbb{R},S \text{ is countable }} \sum_{\alpha \in S} F(\alpha).$$
If $F \leq 0$ then you can look at:
$$\inf_{S \subset \mathbb{R},S \text{ is countable }} \sum_{\alpha \in S} F(\alpha)$$
instead. 
If $F$ takes on both signs then you can look at $F^+=\max \{ F,0 \}$ and $F^-=\max \{ -F,0 \}$ so that $F=F^+-F^-$. Then the sum would be:
$$\sup_{S \subset \mathbb{R},S \text{ is countable }} \sum_{\alpha \in S} F^+(\alpha) - \sup_{S \subset \mathbb{R},S \text{ is countable}} \sum_{\alpha \in S} F^-(\alpha)$$
provided that this is not $\infty - \infty$.
All of these things will work properly and will agree with the usual definition of the expectation if the absolute expectation exists.
Note that you can replace "countable" with "finite", the result will turn out to be the same.
A: Yes, this definition can be made to work. We have a random variable $X$ taking values in a probability space $(S,P)$ with $S$ discrete and the $\sigma$-algebra is the power set of $S$. We identify the measure $P$ with the values of its probability mass function $\{P(\alpha)\}_{\alpha\in S}$. Since $S$ is countable, there is a bijection $f\colon S\to \mathbb N$. Define a new probability measure $Q$ on $\mathbb R$ as a sum of the following dirac measures on $\mathbb R$:
$$
Q=\sum_{\alpha\in S}P(\alpha)\delta_{f(\alpha)}.
$$
Then $f(X)$ is a random variable taking values in the probability space $(\mathbb R,Q)$, and
$$
\mathbb E\ f(X)=\sum_{\alpha\in\mathbb R}X(\alpha)\ P(\alpha),
$$
which is what you wanted ($f(X)$ is just a way of relabeling $X$ so that it becomes a random variable taking values in $\mathbb R$).
