# Finitely Additive not Countably Additive on $\Bbb N$

Does there exist a function defined on the power set of the natural numbers to the interval from $0$ to $1$, $p:2^{\Bbb N}\rightarrow [0,1]$, such that $p$ is finitely additive, i.e. $p(A_1\cup\ldots\cup A_n)=p(A_1)+\ldots +p(A_n)$ for all pairwise disjoint subsets $A_i$ of $\Bbb N$ and some $n\in\Bbb N$, but $p$ is not countably additive. $p$ also satisfies the condition that $p(\Bbb N)=1$, $p(\emptyset)=0$.

• There are many different ways of constructing such a $p$, using the axiom of choice in many forms and disguises. It might be amusing to collect several of them here. In addition, some of the set theorists among us could contribute a "No" answer :) – Nate Eldredge Sep 6 '13 at 4:44

You want to look at the version of this question on MO. In his answer, Stefan Geschke indicates there that the axiom of choice is needed to exhibit examples, since any such finitely additive measure gives us sets of reals without the Baire property. (In a comment, Clinton Conley provides details.) The point is that there are models of set theory without the full axiom of choice, but where the axiom of dependent choice holds (so basic classical analysis can be carried out), and where all sets of reals have the Baire property.

Stefan also shows how there are examples different from ultrafilter measures (the example given by Ross Millikan here).

As pointed out in the other answer to the MO question, by KP Hart, Eric van Douwen wrote a nice paper on this, indicating many examples, and the different properties they may have:

Eric K. van Douwen. Finitely additive measures on $\mathbb N$. Topology Appl., 47 (3), (1992), 223–268. MR1192311 (94c:28004).

The paper provides many additional references. For example, it indicates that Banach found a translation invariant example: For any $X$ and any $n$, $X$ and $X+n=\{m+n\mid m\in X\}$ have the same measure.

I do not know of a reasonable complete "classification", or whether this is possible (from a descriptive set theoretic point of view). Certainly similar problems have been studied. In particular, see

Piotr Borodulin-Nadzieja, and Mirna Džamonja. On the isomorphism problem for measures on Boolean algebras. J. Math. Anal. Appl., 405 (1), (2013), 37–51. MR3053484.

Let me add that William's answer here of the density measure can be extended to a finitely additive measure on all sets. For example, if $\mathcal U$ is a free ultrafilter, then rather than limits, we can compute "$\mathcal U$-limits", which always exist, and coincide with ordinary limits when these exist. The measures that extend the density measure can be nicely characterized, see

Martin Sleziak, and Miloš Ziman. Lévy group and density measures, J. Number Theory, 128 (12), (2008), 3005–3012. MR2464850 (2009j:11019).

Their result can be easily described: The Levy group $\mathcal G$ is the group of all permutations $\pi$ of $\mathbb N$ such that $$\lim_{n\to\infty}\frac{|\{k\mid k\le n<\pi(k)\}|}n=0.$$ It turns out that a finitely additive measure on $\mathcal P(\mathbb N)$ extends the density measure iff it is $\mathcal G$-invariant.

• What exactly does it mean to say that a (finitely additive) measure extends the density measure? Is it merely that the measure of $X$ is equal to the density of $X$ whenever the density exists, or is it the stronger condition, that the measure of $X$ is $\ge$ the lower density and $\le$ the upper density? – bof Jun 5 '14 at 6:14
• @bof I just meant the first interpretation. I agree the second version would be more interesting. – Andrés E. Caicedo Jun 5 '14 at 21:25

A free ultrafilter will do the job. I am not sure it is the only thing that will. Every subset or its complement is given measure $1$, and all finite subsets have measure $0$. It is not countably additive, as the countable union of all singletons adds to zero measure. If $A$ and $B$ are disjoint they cannot both have measure $1$ because if $A$ has measure $1$, $A \subset B^c$, so $B^c$ must have measure $1$ and so $B$ must have measure zero.

The natural density ($p(X)=\lim_{n\to\infty} \#\{k\in X\big|k\le n\}/n)$ satisfies most of the requirements of your $p$ (it's finitely additive, $p(\Bbb{N}) = 1$, and $p(\emptyset)=0$)

However, it's not defined on all subsets of $\Bbb{N}$. For example, for the set of integers with an odd number of binary digits, that limit does not exist.

• Could there be a way to make the natural density work through the Tauberian theorem? en.wikipedia.org/wiki/… – domoremath Sep 6 '13 at 5:01
• @domoremath Yes, and this approach yields the (sigma-additive) zeta probability measures, defined by $\mu_s(\{n\})=1/(n^s\zeta(s))$ for some real number $s\gt1$. – Did Sep 6 '13 at 6:04