Construct a merely finitely additive measure on a $\sigma$-algebra 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 $\Omega = \mathbb{N}$.
 A: 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".)
A: The question is: 

"Is it possible to give an explicit construction of a set function, defined on a σ-algebra, with all the properties of a measure except that it is merely finitely additive and not countably additive?"

If you do not require the "measure" to be a "probability" (or to have only finite values), then the answer is YES. Here are two examples: 
Example 1:
Consider $\mathbb{N}$ and $\Sigma=P(\mathbb{N})$. Clearly $\Sigma$ is a   $\sigma$-algebra.  Define $\mu : \Sigma \to [0,+\infty]$ by 
$\mu(A)=0$ if $A$ is finite and $\mu(A)=+\infty$ if $A$ is infinite.
It is easy to see that $\mu$ is a finitely additive measure, but not a (countable additive) measure. 
Note that $\mu$ is not a finitely additive probability. In fact, it is not even a finite finitely additive measure. 
Example 2:.
Consider $[0,1]$ and $\Sigma$ be the set of countable or co-countable subsets of $[0,1]$. Clearly  $\Sigma$ is a   $\sigma$-algebra on $[0,1]$  Define $\mu : \Sigma \to [0,+\infty]$ by 
$\mu(A)=0$ if $A$ is finite and $\mu(A)=+\infty$ if $A$ is infinite.
It is easy to see that $\mu$ is a finitely additive measure, but not a (countable additive) measure. 
Note again that $\mu$ is not a finitely additive probability. In fact, it is not even a finite finitely additive measure. 
Remark 1: 
The examples above may be included in the following general case. 
Let $\Omega$ be any infinite set and let $\Sigma$ be a $\sigma$-algebra on $\Omega$, such that there are infinitely many finite sets in $\Sigma$, in other words,  the set $\{A \in \Sigma : A \textrm{ is finite }\}$ is infinite. Then, define $\mu : \Sigma \to [0,+\infty]$ by 
$\mu(A)=0$ if $A$ is finite and $\mu(A)=+\infty$ if $A$ is infinite.
It is easy to see that $\mu$ is a finitely additive measure, but not a (countable additive) measure. 
Remark 2: 
If the OP meant to ask:  

Given ANY σ-algebra, is it possible to give an explicit construction of a set function, defined on that σ-algebra, with all the properties of a measure except that it is merely finitely additive and not countably additive?

Then, the answer is NO, because any finitely additive measure on a finite σ-algebra is automatically a (countable additive) measure. 
A: 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.
