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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 $\bf Z$?

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  • $\begingroup$ 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
    Commented Dec 1, 2016 at 22:40
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    $\begingroup$ 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. $\endgroup$
    – Alexander Pruss
    Commented Sep 14, 2017 at 14:04
  • $\begingroup$ @AlexanderPruss Thank you, that is what I meant. $\endgroup$
    – Goldstern
    Commented Jul 15, 2018 at 13:20

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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).

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    $\begingroup$ 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. $\endgroup$
    – Nate Eldredge
    Commented 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? $\endgroup$
    – James E Hanson
    Commented Aug 14, 2023 at 23:42

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