(Example) Can a probability distribution on $\mathbb{R}$ be finitely additive but not countably additive? A probability distribution $P$ on $\mathbb{R}$ is a function from the Borel $\sigma$-algebra on $\mathbb{R}$ to [0,1] which satisfies $P(\emptyset)=0, P(\mathbb{R})=1$.
Can you give me an example for a finitely additive but not countably additive probability distribution in this setup? (If there is one)
I've browsed other questions on this topic but none gives a clear answer to my question.
 A: I don't think one can be constructed explicitly. Assuming the axiom of choice you can prove they exist as follows. First pick a non-principal ultrafilter $U$ on $\mathbb{R}$, that is a collection of subsets of $\mathbb{R}$ such that


*

*no finite set is in $U$,

*$U$ is closed under finite intersections,

*if $A\subset B$ and $A\in U$ then $B\in U$, and

*for every $ A\subset \mathbb{R}$ either $A$ or $\mathbb{R}\setminus A$ is in $U$.


These can be shown to exist using Zorn's lemma (a maximal collection satisfying the first three conditions van be shown to satisfy the fourth). The characteristic function of U is now a finitely additive measure on $\mathbb{R}$ (that only takes the values $0$ and $1$). This follows straightforwardly from the definition of ultrafilter.
Now we have to ensure that this measure $\mu$ is not also countably additive. Here's two ways to do that:


*

*Don't try to show it for arbitrary non-principal ultrafilters $U$, just produce some $U$ that have this property. If $U$ contains a countable set $A = \{a_n : n \in \mathbb{N} \}$, then $\mu(A) = 1 \neq 0 = \sum \mu(\{a_n\})$. It is easy to ensure that $U$ contains a given countable set $A$: apply Zorn's lemma not to all collections satisfying the first three conditions, but only to those that contain $A$.

*Show that this is true for arbitrary non-principal ultrafilters $U$! Assume $\mu$ is countably additive. It follows then that $U$ is closed under countable intersections: indeed, if $A_n \in U$, $$\mu\left(\mathbb{R} \setminus \bigcap_n A_n\right) = \mu\left(\bigcup_n (\mathbb{R} \setminus A_n)\right) \le \sum_n \mu(\mathbb{R}\setminus A_n) = 0.$$
We'll show this leads to a contradiction, but let's work with $\{0,1\}^{\mathbb{N}}$ instead of $\mathbb{R}$, for convenience. Let $S_n(t) = \{ x : \mathbb{N} \to \{0,1\} \mid x(n)=t \}$. For each $n$ exactly one of $S_n(0)$ and $S_n(1)$ is in $U$, let it be $S_n(t_n)$. Then we have that the singleton $\{n \mapsto t_n\} = \bigcap_n S_n(t_n)$ is in $U$, which is a contradiction.
