TITLE: Construct a merely finitely additive measure on a $\sigma$-algebra
QUESTION [7 upvotes]: 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 $\Omega = \mathbb{N}$.

REPLY [2 votes]: The way I understand the question is this:
Can you construct

A measurable space $(\Omega,\Sigma)$ where $\Sigma$ is a $\sigma$-algebra
A finitely additive finite measure $\mu$ on $\Sigma$ that is not countably additive?

Previous answers showed that

The answer is yes if we allow $\mu$ to be infinite
The answer is no if $\mu$ is required to be finite and $\Omega$ is countable or finite.

But if I understand the owners intentions correctly from comments the exact question, where $\Omega$ may be uncountable and $\mu$ must be finite, has not been answered.
As I was interested in the question specified above I gave it some thought and found the following result, which means that the answer is essentially no:
Either it is not possible to construct such an example or if it is, then we cannot constructively show that the example is not countably additive.
Proof by contradiction:
Assume we have $(\Omega,\Sigma,\mu)$ as required. Assume also that we can construct a sequence $A_n$ of disjoint sets in $\Sigma$ which shows that $\mu$ is not countably additive, ie $\mu(\bigcup A_n)\ne\sum\mu(A_n)$.
Then we could construct another example where the underlying set is countable

$\Omega^\star=\{A_1,A_2,A_3,\dots\}$
$\Sigma^\star=2^{\Omega^\star}$
$\mu^\star(B)=\mu(\bigcup_{i:A_i \in B} A_i)$

Since $\mu^\star$ is essentially a restriction of $\mu$ it is also a finite, finitely additive measure and it is not countably additive by construction.
We already know from other answers that constructing an example with a countable $\Omega$ is not possible so we have a contradiction.
I would appreciate if anyone could tell which of the two possibilities is true.