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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.

Tim
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1 Answers1

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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:

  1. 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 $A$.

  2. 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.

Omar Antolín-Camarena
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  • Note there is a little more work to do to verify this is not countably additive. – Nate Eldredge Sep 24 '14 at 13:34
  • Well to be fair, @NateEldredge, I didn't show the verification of finite additivity either! – Omar Antolín-Camarena Sep 24 '14 at 13:53
  • Oh, I see what you meant now @NateEldredge (I mean that showing that this measure is not countably additive is not as easy as I thought it was). I'll update the answer with a proof of that. – Omar Antolín-Camarena Sep 24 '14 at 14:20
  • Thanks a lot! The above question originated from another, related but more confusing, question I have. Maybe you can take a look at this, as well: http://math.stackexchange.com/questions/944469/identifying-random-variables-with-their-generated-distribution-function-necess – Tim Sep 24 '14 at 15:47