A non-principal [probability] measure on a set X is a function $\mu$ defined on all subsets of $X$, with values in $[0,1]$, which is finitely additive, satisfies $\mu(X)=1$, and vanishes on singletons.

Can one prove in ZF + DC that the existence of such a measure on $\bf N$ (or on $\omega$), the set of natural integers, implies that of a non-Lebesgue measurable subset of $\bf R$?

Comments: 1) Sierpinski proved this in 1938 in the special case that $\mu$ takes values in $\{0,1\}$ (non-principal, or non-free ultrafilter). It suffices to take $X=\{A\subset{\bf N}\mid\mu(A)=0\}\subset\{0,1\}^{\bf N}$, equipped with the standard Bernoulli measure $\lambda$ (so that $\{0,1\}^{\bf N},\lambda)\approx([0,1],{\rm Leb}$). If $X$ were $\lambda$-measurable, one would have $\mu(X)={1\over2}$ since $X^c=\{A^c\mid A\in X\}$. But $X$ is a queue event, thus by Kolmogorov, $\mu(X)=0$ or $1$, contradiction.

2) In the 1998 book by Howard and Rubin, Consequences of the axiom of choice, it is stated that the existence of a non principal measure on $\bf N$ implies that of a non-Lebesgue measurable set [Form 222 implies Form 93], and that it can be found in articles by Pincus in 1972 and by Foreman-Wehrung in 1991. However, I could not find this implication in these articles.

3) The most famous non-principal measures on countable sets are invariant means on amenable groups, starting with $\bf Z$ [mean is another name for a finitely additive probability measure defined on all subsets]. Question: does the existence of a non-principal measure on $\bf N$ (or $\bf Z$) imply (in ZF + DC) that of an invariant mean on $\bf Z$?

  • $\begingroup$ The Hahn-Banach theorem is equivalent to the existence of a finitely additive probability measure (FAPM) on every Boolean algebra. The existence of non-measurable sets follows (by Foreman-Wehrung) already from the existence of a FAPM on a specific Boolean algebra, namely the free sum of a large collection of algebras $P(A_i)$, each $A_i$ countable. It is not clear (to me) if FAPM for the single algebra $P(\omega)$ is sufficient. $\endgroup$
    – Goldstern
    Dec 1, 2016 at 22:40
  • 1
    $\begingroup$ The existence of a FAPM for the single algebra $P(\omega)$ is trivial (just let $P(\{0\})=1$). I assume @Goldstern meant the existence of an atomless FAPM. As I remember, Foreman-Wehrung use the existence of a FAPM on the free sum of the $P(A_i)$ in order to get a choice function for the FAPMs (i.e., a function $\mu$ from the index set $I$ such that $\mu_i$ is a FAPM on $P(A_i)$), a very nice trick. But nothing like this trick seems to be available when the assumption is merely that $P(\omega)$ has an atomless FAPM. $\endgroup$ Sep 14, 2017 at 14:04
  • $\begingroup$ @AlexanderPruss Thank you, that is what I meant. $\endgroup$
    – Goldstern
    Jul 15, 2018 at 13:20

1 Answer 1


This is stated as an open conjecture in Pincus' 1974 paper The Strength of the Hahn-Banach Theorem, which is a pretty good sign that he didn't prove it in a paper published in 1972. That same paper contains the first proof of the analogous question for the Baire Property.

As far as I am aware, this problem is still open, even assuming the stronger assumption of a Banach limit (such a measure that is translation-invariant and whose integral gives the limit for convergent sequences).

  • 1
    $\begingroup$ So, maybe this should be an erratum to Howard and Rubin? Their website has a link for "Changes and additions to the database" but unfortunately it is broken. $\endgroup$ Sep 21, 2020 at 0:18
  • $\begingroup$ Minor point, but I thought that there was a standard method of constructing a Banach limit from an arbitrary non-principal measure (by taking the integral of the moving average). Is having a Banach limit actually a stronger assumption? $\endgroup$ Aug 14, 2023 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.