Consider an arbitrary space $X$ equipped with the discrete topology $2^X$, and then consider the induced Borel s-algebra, which is also $2^X$.
I've always assumed that the Borel probability measures on $(X,2^X)$ had a simple characterization: namely, they are those supported on some finite or countably infinite subset. Is this correct without any further hypotheses?
When $X$ has cardinality $\aleph_0$ or smaller, this seems trivial.
If $X$ has cardinality greater than $\aleph_0$, then my intuition is that the only Borel probability measures are those that concentrate on a finite or countably infinite set. Well, at least, I don't see any issue with constructing a Borel probability measure $\mu$ from a finite or countably infinite sequence $x_1,x_2,\dots$ and sequence $p_1,p_2,\dots \ge 0$ with $\sum_i p_i = 1$ by taking $\mu$ on $A \subseteq X$ to be $\mu(A) = \sum_{i : x_i \in A} p_i$.
So are these all the Borel probability measures? I suppose this has to do with whether there can be a Borel probability measure on $(X,2^X)$ that has no atoms.
I'm going to post this, while I continue to think about it.
UPDATE: So I've considered the question of what atoms must look like in a Borel probability measure $(X,2^X,\mu)$. Recall that an atom in this context is a subset $B \subseteq X$ with $\mu(B) > 0$, such that, for all subsets $A \subseteq B$, either $\mu(A) = 0$ or $\mu(B) = \mu(A)$.
Claim Assume that there exists a sequence $A : \mathbb N \to 2^X$ of sets such that every point is identified by the subset of sets containing it. Then every atom contains a singleton atom.
Proof sketch. If $B$ is an atom, then one of $B \cap A_1$ or $B \setminus A_1$ has full mass. Let $B_1$ be the set with full mass. We repeat this process for all $A_i$. At the end of the process, we have taken the intersection with every $A_i$ or its complement $X \setminus A_i$, and so we must be down to one point. Hence every atom is a singleton.
Remark. The hypothesis rules out some big sets. Perhaps it can be generalized using some transfinite argument?