Example for finitely additive but not countably additive probability measure A probability measure defined on a sample space $\Omega$ has the following properties:


*

*For each $E \subset \Omega$, $0 \le P(E) \le 1$

*$P(\Omega) = 1$

*If $E_1$ and $E_2$ are disjoint subsets $P(E_1 \cup E_2) = P(E_1) + P(E_2)$


The above definition defines a measure that is finitely additive (by induction) but not necessarily countably additive.
What is a probability measure that would be finitely additive but not countably additive (for a countable sample space  $\Omega$)?
The example that I have seen most commonly on forums (this and elsewhere) is to set $P(E) = 0$ if $E$ is finite and $P(E) = 1$ if $E$ is co-finite. But that is not a probability measure as defined above since it is not defined on every subset of $\Omega$. 
So an example of such a probability measure, or what is the reasoning that a finitely additive probability measure is not always countably additive?
 A: This question was from years ago, but I was just about to ask a similar question (I found this page from the stackexchange list of similar questions).  My own question is whether it is possible to have an explicit example. The answers above are all non-explicit.  Here is another (non-explicit) answer in a different form that I found to be helpful.  It uses a Banach limit from functional analysis. 
Define the natural numbers $\mathbb{N} = \{1, 2, 3, \ldots\}$ and define $2^{\mathbb{N}}$ as the set of all subsets of $\mathbb{N}$.  Define $P:2^{\mathbb{N}}\rightarrow\mathbb{R}$ as follows: For each set $A \subseteq \mathbb{N}$, define $P(A)$ as a Banach limit of the sequence $\left\{\frac{|A \cap \{1, 2, ..., k\}|}{k}\right\}_{k=1}^{\infty}$. 
Banach limit properties:
A Banach limit can be proven to exist and to have the following properties: 
1) It is defined for all bounded real-valued sequences $\{x_k\}_{k=1}^{\infty}$, regardless of whether or not $x_k$ has a limit.  In fact, the Banach limit is always a real number between $\liminf_{k\rightarrow\infty} x_k$ and $\limsup_{k\rightarrow\infty} x_k$.
2) The Banach limit is the same as the regular limit whenever the regular limit exists.
3) The Banach limit of a sum of two bounded sequences is the sum of the Banach limits of the individual sequences. 
4) The Banach limit is nonnegative whenever $x_k \geq 0$ for all $k$. 
There are many ways to define a real-valued function on bounded sequences that satisfies the above properties,  so it is implicitly assumed that we consistently use one such function.  The value of that function on a given bounded sequence is what we shall call the "Banach limit" of that sequence.   Proofs of existence of such functions use nonexplicit things like axiom of choice or ultrafilters. 
Using these properties:
Now if $A$ is a finite subset of $\mathbb{N}$ then $\frac{|A \cap \{1, ..., k\}|}{k} \leq \frac{|A|}{k}\rightarrow 0$, so the limit exists and $P(A)=0$.  In particular, $P(\{n\})=0$ for all $n \in \mathbb{N}$. So:
$$ 1=P[\mathbb{N}] = P[\cup_{n=1}^{\infty} \{n\}] \neq \sum_{n=1}^{\infty}P[\{n\}]=0$$
Furthermore, $P(A)$ is nonnegative for all $A\subseteq \mathbb{N}$ (by the 4th property of Banach limits above). It also satisfies $P(A \cup B)=P(A)+P(B)$ whenever $A$ and $B$ are disjoint (which can be shown by the 3rd property above). So this $P(A)$ function is indeed a finitely-additive measure on all subsets of $\mathbb{N}$, but not a countably-additive one. 
Remaining question:
The above pushes a bit more towards an explicit answer, but still uses Banach limits and hence is not explicit.  Can a more explicit answer can be given?  Now I'm not sure if I should formally ask this question on stackexchange or not, I suspect I would just get pointers back to your question.
A: In the following note the author shows that a finitely additive diffused measure on $\mathcal{P}(\omega)$ can be used to define a non Ramsey family. Combining this with a result of Mathias, it follows that it is consistent with $ZFC$ that there is no (ordinal) definable finitely additive diffused total measure on $\mathcal{P}(\omega)$.
A: In this interesting note, the author proves the following: It is consistent to have a finitely additive total extension of Lebesgue measure on $[0, 1]$ such that, although, the measure zero sets form a sigma ideal, there is no real valued measurable cardinal below continnum.
A: Let $\mathcal{U}$ be a free ultrafilter on $\mathbb{N}$. Let $P(A)=1$ if $A\in\mathcal{U}$ and $P(A)=0$ if $A\notin\mathcal{U}$. I think it is impossible to give an explicit example of a finitely additive measure on a $\sigma$-algebra that is not countably additive, but our resident set theorists might be able to tell you more about that.
