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I have a history question for which I've had trouble finding a good answer.

The common story about nonmeasurable sets is that Vitali showed that one existed using the Axiom of Choice, and Lebesgue et al. put the blame squarely on this axiom and its non-constructive character. It was noticed however that some amount of choice was required to get measure theory off the ground, namely Dependent Choice seemed to be the principle typically employed. But the full axiom of choice which allows uncountably many arbitrary choices to be made is of a different character, and is the culprit behind the pathological sets. This viewpoint was not really justified until Solovay showed in the 1960s that ZF+DC could not prove the existence of a nonmeasurable set, assuming the consistency of an inaccessible cardinal.

My question is, in the many years before Solovay's theorem, was there any effort aimed at showing the existence of a nonmeasurable set without the use of the full AC? Was something like the following question ever posed or worked on: "Can constructions similar to those of Vitali, Hausdorff, and Banach-Tarski be done without appeal to the Axiom of Choice?"

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I would assume that Solovay's theorem was so celebrated precisely because people had cared about this issue. So surely people thought about it? –  Joel David Hamkins Aug 28 at 23:44
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I wrote to Solovay about it, and we'll see if he has anything to say. –  Joel David Hamkins Aug 29 at 0:19
    
Thanks Joel! I will be very interested to hear what he says. –  Monroe Eskew Aug 29 at 0:32
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You may want to take a look at Lebesgue's writings. Bressoud's A radical approach to Lebesgue's theory of integration, claims (in p. 154) that "Vitali's nonmeasurable set, appearing less than a year later [than Zermelo's Well-ordering theorem], was greeted by Lebesgue and many others as an empty exercise. They wanted an example of a nonmeasurable set whose construction ould not depend on the axiom of choice." –  Andres Caicedo Aug 29 at 1:12

2 Answers 2

up vote 36 down vote accepted

Paul Cohen posed the question of getting a model of "All Sets Lebesgue Measurable" in his early talks on his own results. (He did not mention the principle of Dependent Choices. Adding that to the problem was my idea.) I know of no work trying to prove the Vitali result constructively. Certainly Cohen's conjecture (which I presume was widely shared) was that the use of choice was essential.

It is quite striking (if one works through Halmos) that all the positive results in measure theory can be carried out in ZF + DC. Only the counterexample section uses full choice.

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If my memory serves me right, first came the proof that the Lebesgue measure can be extended to all sets (without choice, and without large cardinals, of course), right? –  Asaf Karagila Aug 29 at 1:26
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Did Cohen not even mention countable choice of reals in this connection? That seems needed just to get Lebesgue measure theory started (e.g., proving countable additivity). As far as I can see, DC becomes important only in "higher" parts of the theory. (To formulate a specific conjecture: Countable choice suffices for all the measure theory that we require graduate students to know for their qualifying exams.) –  Andreas Blass Aug 29 at 14:59
    
As I recall, Cohen didn't mention "countable choice". But this was an off hand remark (posing the problem) at the end of his lecture. It seemed to me at the time I did this stuff that full DC was needed for the Radon-Nikodym theorem. But as I recall, I convinced myself a few years ago that a different proof of Radon-Nikodym than the one given in Halmos could be done with just AC_omega. –  Bob Solovay Sep 4 at 1:57
    
Re Asaf's remark. Yes, I had this result first. Something very like my proof of this was subsequently published by Sacks. –  Bob Solovay Sep 4 at 2:01

I would like to point out that the question on the existence of a non-measurable set without the use of the full AC was technically established in the literature before Solovay's result (which afaik goes back to March-July 1964).

In 1938 Sierpinski had established (cf. "Fonctions additives non complètement additives et fonctions non mesurables", Fund. Math.) that a non-measurable set could be constructed from the assumption (in modern terminology) that there is a prime ideal on the power set of the natural numbers extending the ideal of finite sets. He explains that the existence of such prime ideal was proven by Tarski (cf. "Une constribution à la théorie de la mesure", Fund. Math.) with the aid of the Axiom of Choice (he proved it by transfinite induction).

But it remained an open question, especially in the 50's after Henkin's results, whether the existence of such prime ideals in the power sets, or, more generally, in Boolean algebras, was or not weaker than the Axiom of Choice. This was eventually settled by Halpern, who proved that it was, and his results first appeared on his doctoral dissertation, submitted in the spring 1962.

Sierpinski construction is quite simple: define a function $f$ on a real number $x$ to be either $1$ or $0$, depending on whether the subset of ones in (the non integer part of) its dyadic expansion (choosing the finite development for the rationals) is or not in the prime ideal defined on the powerset of the natural numbers extending the ideal of finite sets. It follows that $f$ has arbitrarily small periods (all numbers of the form $2^{-n}$) and that $f(1-x)=1-f(x)$. From this and the fact that $f$ only takes the values $0$ and $1$ it is not hard to show that it cannot be measurable, and the preimage of $0$ provides a non-measurable set.

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This is a very good answer. It shows that there was some work on finding nonmeasurable sets without the full AC, perhaps not by outright trying to prove it with only DC, but by weakening the hypothesis. –  Monroe Eskew Aug 29 at 19:49
    
And Halpern did it without forcing! –  Monroe Eskew Aug 29 at 20:02
    
Now I am very curious whether anyone investigated measure theory in Fraenkel-Mostowski permutation models. –  Monroe Eskew Aug 29 at 20:23
    
@Monroe: Nothing to investigate. Since the reals are considered as pure sets. In permutation models pure sets are well-ordered. So everything works as in $\sf ZFC$. –  Asaf Karagila Aug 29 at 21:08
    
@Asaf: Maybe one could still investigate measurability in these models with respect to more general (outer-) measure spaces involving the urelements? –  Monroe Eskew Aug 29 at 21:45

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