I want to find some condition to construct a continuous finitely additive measure on the natural numbers, i.e. $f:P(\mathbb{N})\rightarrow [0,1]$ such that $f(\{n\})=0$, and $f$ is an additive measure.

I know in ZFC we can use an ultrafilter $U$ and define $f$ by $f(A)=1\Leftrightarrow A\in U$, but this is too trival.

How about ZF? or some other condition? like large cardinal.

  • $\begingroup$ How do you define whether or not $f$ is continuous? $\:$ $\endgroup$
    – user5810
    May 4, 2012 at 5:03
  • $\begingroup$ What do you mean by 'nature'? $\endgroup$
    – David Roberts
    May 4, 2012 at 5:47
  • $\begingroup$ @David: I think this means the set of natural numbers $\endgroup$
    – Yemon Choi
    May 4, 2012 at 6:15
  • 1
    $\begingroup$ Conversely for any such $f$, $f^{-1}(\{1\})$ is a non-principal ultrafilter. So it's just the same as asking about the "construction" of a non-principal ultrafilter on $\mathbf{N}$. $\endgroup$
    – YCor
    Feb 19, 2020 at 16:13
  • $\begingroup$ The range of $f$ is $[0,1]$, not $\{0,1\}$. Thus any set $A\subset\Bbb N$ such that $0<f(A)<1$ intersects any set of $f^{-1}(\{1\})$, but $A\not\in f^{-1}(\{1\})$. $\endgroup$ Dec 1, 2020 at 5:26

2 Answers 2


In ZF you can't prove that there is a finitely additive measure on $\mathcal P(\mathbb N)$ (giving all singletons measure 0 and $\mathbb N$ measure 1). The existence of such a measure gives you a subset of $\mathbb R$ without the Baire property, and such a set cannot be constructed without some help of the axiom of choice. In other words, there is no explicit "construction" of a non-trivial finitely additive measure.

Since you don't like the ultrafilter measure, here is one that is less trivial:

Instead of $\mathbb N$, we use the integers $\mathbb Z$ as the underlying set. For $n\in\mathbb N$ let $d_n(A)=\frac{|A\cap\{-n,\dots,n\}|}{2n+1}$. The sequence $(d_n(A))_{n\in\mathbb N}$ typically does not converge. We fix this using an ultrafilter $U$ on $\mathbb N$ as follows:

For each $A\subseteq\mathbb Z$ let $\mu(A)$ be the unique element of the set $\bigcap_{S\in U}\mbox{cl}(\{d_n(A):n\in S\})$. Here the closure is taken in $\mathbb R$. Now $\mu$ is a finitely additive measure on $\mathcal P(\mathbb Z)$ that is even translation invariant and hence very different from the ultrafilter measure that you find too trivial.

Notice that I have just proved the well known fact that the integers are an amenable group.

  • $\begingroup$ In ZF,The existence of such a measure gives you a subset of R without the Baire property,is this set ${X\subset N:f(X)=1}$? Can you tell me whether the set is nonmeasure? $\endgroup$ May 4, 2012 at 13:20
  • 3
    $\begingroup$ Jialiang, the collection of sets of full measure might be meager, but if you look instead at the collection of sets with measure at least $1/2$ you get a non-BP set (since complementation is a homeomorphism, it would have to be nonmeager, and it's not hard to show that any nonmeager BP subset of the power set of $\mathbb{N}$ contains three almost disjoint sets). You cannot rule out a measurable fapm on $\mathbb{N}$; in fact ZFC+MA proves that there are universally measurable fapms. $\endgroup$ May 4, 2012 at 14:08
  • $\begingroup$ Can you tell me why the full measure might be meager ? note,the measure is finitely additive measure ,giving all singletons measure 0 and N measure 1. I think I have a proof,fisrt the full measure set $F$ is a filter,by a theorem of Talagrand(Set theory on the Structure of the real line P206),it engough to prove:for every partition of $\omega$into finite sets$I_n$,there exists $X\in F$such that$\exists^\infty n X\cap I_n=\emptyset$.use a MAD fmmily and ccc of the measure we can get a measure zreo set have infinite interval. $\endgroup$ May 5, 2012 at 2:06

Eric van Douwen constructed a whole slew of finitely additive measures on the set of natural numbers, with various shifting and scaling properties in this paper. He discussed how much `choice' was needed for the constructions; often some version of the Hahn-Banach theorem would suffice.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .