Is it possible to give an explicit construction of a set function, defined on a $\sigma$-algebra, with all the properties of a measure except that it is merely finitely additive and not countably additive?
Let me elaborate. By "explicit" I mean that the example should not appeal to non-constructive methods like the Hahn-Banach theorem or the existence of free ultrafilters. I'm aware that such examples exist, but I'm looking for something more concrete. If such constructions are not possible, I'm especially interested in understanding why that is so.
This question is similar, but, so far as I can tell, not identical to several other questions asked on this site and MO. For example, I've learned that proving the existence of the "integer lottery" on $P(\mathbb{N})$ requires the Axiom of Choice (https://mathoverflow.net/questions/95954/how-to-construct-a-continuous-finite-additive-measure-on-the-natural-numbers).
That's the sort of result I'm interested in, but it doesn't fully answer my question. My question doesn't require that the $\sigma$-algebra in question be $P(\Omega)$, and I'm interested in general $\Omega$, not just $\Omega = \mathbb{N}$.
Best Answer
The result that the Hahn-Banach is equivalent to the existence of a [nontrivial] finitely additive probability measure on an arbitrary Boolean algebra is due to Luxemburg. Note that we don't even require the Boolean algebra to be $\sigma$-complete. Going out on a limb, I'll guess that working with only $\sigma$-algebras you don't get the full strength of the HB theorem, but only something close enough.
You can find a reasonably detailed proof (along with many other equivalents of the Hahn-Banach theorem) in Eric Schechter's book "Handbook of Analysis and its Foundations" on page 620.
(When you look at the book, remember that Schechter calls a finitely additive measure a "charge".)