In an interview (at http://www.alainconnes.org/docs/Inteng.pdf) Connes remarks that

I had been working on non-standard analysis, but after a while I had found a catch in the theory.... The point is that as soon as you have a non-standard number, you get a non-measurable set. And in Choquet’s circle, having well studied the Polish school, we knew that every set you can name is measurable; so it seemed utterly doomed to failure to try to use non-standard analysis to do physics.

What does he mean; what is he referring to?

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    $\begingroup$ "What does he mean" - do you mean in reference to 'every set you can name is measurable'? Or in reference to 'the Polish school'? Or the relation to physics? $\endgroup$
    – David Roberts
    Mar 2, 2011 at 4:14
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    $\begingroup$ Did you ask Connes? He may know the answer. $\endgroup$
    – user6976
    Mar 2, 2011 at 4:16
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    $\begingroup$ @Yemon: The question is perfectly clear -- AmbroseH wants to understand the entire passage. The fact that there are several statements in the short passage that need clarification doesn't make this a bad question at all. Breaking the question up (as you seem to suggest) would miss the point of understanding the passage as a whole. $\endgroup$ Mar 2, 2011 at 14:27
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    $\begingroup$ Ah, c'est vraiment une remarque conne ! (I'm sorryyyyy! It had to be done.) $\endgroup$ Nov 15, 2012 at 16:04
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    $\begingroup$ @gunnar: or: caisse-toi, pauv' Connes? $\endgroup$
    – Yemon Choi
    Nov 15, 2012 at 16:45

6 Answers 6


...as soon as you have a non-standard number, you get a non-measurable set.

Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$, by saying that a set $X\subset\mathbb{N}$ is in $U$ if and only if $N\in X^*$, the nonstandard analogue of $X$. In other words, the ultrafilter says $X$ is large if it expresses a property that the nonstandard number $N$ has. We may regard $U$ as a subset of $2^\mathbb{N}$, which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the full bit-flipping operation, which is measure-preserving, carries $U$ exactly to its complement, so $U$ would have to have measure $\frac12$, but $U$ is invariant by the operation of flipping any finite number of bits, and so must have measure $0$ or $1$ by Kolmogorov's zero-one law. (See also this article by Blackwell and Diaconis proving the same fact.)

...every set you can name is measurable.

Another way of saying that a set is easily described is to say that it lies low in the descriptive set-theoretic hierarchy, and the lowest such sets are necessarily measurable. For example, every set in the Borel hierarchy is measurable, and the Borel context is often described as the domain of explicit mathematics. Under stronger set-theoretic axioms, such as large cardinals or PD, the phenomenon rises to higher levels of complexity, for under these hypotheses it follows even that all sets in the projective hierarchy are Lebesgue measurable. This would include any set that you can define by quantifying over the reals and the integers and using any of the basic mathematical operations.

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    $\begingroup$ Note that from any nonstandard real number $x$, one gets a nonstandard natural number $N$ either by using $\lfloor x\rfloor$, if $X$ is infinite, or by using the first $N$ with $\frac 1N\lt |x-st(x)^*|$, if $x$ is bounded. So from any nonstandard $x$ we get the ultrafilter $U$ and thus the non-measurable set, confirming Rachel's remarks $\endgroup$ Mar 2, 2011 at 14:40
  • $\begingroup$ What is the "non-standard analog of X"? $\endgroup$ Mar 2, 2011 at 17:16
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    $\begingroup$ The Transfer Principle (see en.wikipedia.org/wiki/Transfer_principle) of nonstandard analysis asserts that every mathematical object $X$ has a nonstandard analogue $X^*$ with all the same expressible properties in the nonstandard realm as $X$ had in the standard realm. If you build your nonstandard universe by an ultrapower, as is often done, then $$X^*$ is simply the image of $X$ under the canonical map, that is, represented by the constant $X$ sequence. $\endgroup$ Mar 2, 2011 at 17:31
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    $\begingroup$ The second part of your answer also explains the mention of "Polish school". First, descriptive set theory studies subsets of Polish spaces. Second, the axiom of determinacy, which allows for consideration of some generalized hierarchies and which implies that every subset of $\mathbb{R}$ is Lebesgue measurable, was introduced by Polish mathematicians, Hugo Steinhaus and Jan Mycielski: MR0140430 Mycielski, Jan; Steinhaus, H. A mathematical axiom contradicting the axiom of choice. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 10 1962 1–3 $\endgroup$ Nov 15, 2012 at 19:11

Connes' critique was recently analyzed by Kanovei, Katz, and Mormann in this article in Foundations of Science (see also arXiv 1211.0244). Here is the abstract:

We examine some of Connes' criticisms of Robinson's infinitesimals starting in 1995. Connes sought to exploit the Solovay model S as ammunition against non-standard analysis, but the model tends to boomerang, undercutting Connes' own earlier work in functional analysis. Connes described the hyperreals as both a 'virtual theory' and a 'chimera', yet acknowledged that his argument relies on the transfer principle. We analyze Connes 'dart-throwing' thought experiment, but reach an opposite conclusion. In S, all definable sets of reals are Lebesgue measurable, suggesting that Connes views a theory as being 'virtual' if it is not definable in a suitable model of ZFC. If so, Connes' claim that a theory of the hyperreals is 'virtual' is refuted by the existence of a definable model of the hyperreal field due to Kanovei and Shelah. Free ultrafilters aren't definable, yet Connes exploited such ultrafilters both in his own earlier work on the classification of factors in the 1970s and 80s, and in Noncommutative Geometry, raising the question whether the latter may not be vulnerable to Connes' criticism of virtuality.

We analyze the philosophical underpinnings of Connes' argument based on Goedel's incompleteness theorem, and detect an apparent circularity in Connes' logic. We document the reliance on non-constructive foundational material, and specifically on the Dixmier trace (featured on the front cover of Connes' magnum opus) and the Hahn-Banach theorem, in Connes' own framework. We also note an inaccuracy in Machover's critique of infinitesimal-based pedagogy.

A brief review of Kanovei-Shelah is here.

Connes wrote in 2001 as follows:

What conclusion can one draw about nonstandard analysis? This means that, since noone will ever be able to name a nonstandard number, the theory remains virtual (Connes et al. 2001, p. 16)

The exact meaning of the verb "to name" used by Connes here is not entirely clear. Connes provided a hint as to its meaning in 2000, in the following terms:

if you are given a non standard number you can canonically produce a subset of the interval which is not Lebesgue measurable. Now we know from logic (from results of Paul Cohen and Solovay) that it will forever be impossible to produce explicitely [sic] a subset of the real numbers, of the interval [0, 1], say, that is not Lebesgue measurable (Connes 2000a, p. 21, 2004, p. 14).

The reference to Solovay indicates that Connes is relying on the result, which may be found in Solovay (1970, p. 3, Theorem 2) on the existence of a model S of set theory ZFC, in which (it is true that) every set of reals definable from a countable sequence of ordinals is Lebesgue measurable. Thus when Connes claims that the theory remains virtual he apparently means that the theory remains undefinable.

This however is contradicted by the existence of a hyperreal field definable in ZF, with transfer probable using only the countable axiom of choice, and a further mild well-orderability assumption (weaker than well-orderability of the reals taken for granted by Connes) ensuring properness; see this answer for details.

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    $\begingroup$ Some of us think of the Annals paper on injective factors as the "magnum opus" ... $\endgroup$
    – Yemon Choi
    Nov 15, 2012 at 16:44
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    $\begingroup$ Also, would it be too much trouble for you to turn the abstract of your paper into an answer that is less compressed? $\endgroup$
    – Yemon Choi
    Nov 15, 2012 at 16:48
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    $\begingroup$ Have just thought of other magnum opus: Connes's IHES publication on NCDG... $\endgroup$
    – Yemon Choi
    Nov 15, 2012 at 16:49
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    $\begingroup$ See my comment below (it was too long to place here). $\endgroup$ Nov 20, 2012 at 16:48

A Google search turns up a page on nonstandard analysis on WorldLingo that cites a quote from Connes in the section on criticisms. This may give you more understanding of what Connes is referring to, given that this seems to be a re-statement of what you have quoted above:

The answer given by nonstandard analysis, a so-called nonstandard real, is equally deceiving. From every nonstandard real number one can construct canonically a subset of the interval [0, 1], which is not Lebesgue measurable. No such set can be exhibited (Stern, 1985). This implies that not a single nonstandard real number can actually be exhibited.

The next remark is:

A. Connes Noncommutative Geometry and Space-Time, Page 55 in The Geometric Universe, Huggett et al. The point of Connes' criticism is that nonstandard hyperreals are as fictitious as non-measurable sets. These sets can be shown to exist, assuming the axiom of choice of set theory, but are not constructible. Non-measurable sets are usually considered pathological, a sort of irritant that must be tolerated in order to have the axiom of choice available.

As there are sources listed, you may be able to get some additional insight in reading some of the references.



A recent article by Leichtnam and myself (arxiv) in the American Mathematical Monthly contains a "theorem" to the effect that, in the presence of a construction of the hyperreals, the following is true: as soon as you have a Connes infinitesimal, you get a non-measurable set.


I don't think this answer is fundamentally different from Joel's, but perhaps the differing exposition may help.

Every irrational real number in $[0,1]$ has a unique binary expansion, and so every irrational real number in ${}^\ast[0,1]$ has a unique $\ast$-binary expansion. The set of irrational real numbers whose $N$-th binary digit is 0 (where $N$ is an infinite nonstandard natural number) is not Lebesgue measurable. This follows from the Lebesgue Differentiation Theorem.

I'm slowly beginning to grasp that adjectives like "countable" and "measurable", although applied to individual sets, actually describe the ambient set theory and not the set itself.

Edit: The weakest form of the Lebesgue Differentiation Theorem is this: Let $A\subseteq[0,1]$ be measurable. For almost all $x\in A$, $$ \lim_{r\to0^+} \frac{\lambda(A \cap (x-r,x+r))}{2r} = 1.$$ Such $x$ are often called "points of density". More involved statements allow one to consider more general measure spaces than $[0,1]$, to integrate functions (instead of taking the measure of a set), and most interestingly, to consider more general types of neighborhoods than balls of radius $r$. For "nice" neighborhoods this generalizes, and there is still some issue as to whether particular neighborhoods are too ugly to be nice, or not.

The take-away is this: a measurable set has its positive measure in clumps. The set described above is too uniform to be measurable.

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    $\begingroup$ Kevin, indeed, I think we have literally the same set, since a real is in your set if and only if $N$ is in the star of the set $X$ of places where the real has a $1$, if and only if this set is in my ultrafilter, if and only if the corresponding real is in my set. So your set is really a nonprincipal ultrafilter. But could you provide more details about your non-measurability argument using the Lebesgue Differntiation Theorem? $\endgroup$ Mar 2, 2011 at 21:28
  • $\begingroup$ Oh, I see you had digit $0$ instead of $1$, so our sets are complements. $\endgroup$ Mar 2, 2011 at 23:39

Specifically with regard to the issue of what Connes means when he says that you can't "name" an infinitesimal: this issue was discussed above on this page, and I think we provided an answer that's different from what was given above. Namely, Connes specifically refers to Solovay. Recall that in Solovay's model, all measurable sets are definable. This indicates that what Connes means by "naming" is that infinitesimals are not definable. From this Connes jumps to the conclusion that the hyperreal theory is "virtual". This, we argue, is a non-sequitur, or even an error. A hyperreal field is indeed definable, as was shown by Kanovei and Shelah to everybody's surprise. Therefore Connes' claim is erroneous. As far as the fact that infinitesimals are not definable, well Connes himself used ultrafilters in an essential manner in his earlier work in functional analysis (including the papers mentioned by Choi above). In this sense Connes is criticizing his own earlier work, in a way. Note that a generic real number is arithmetically not definable, so the whole thing is a non-starter.

An additional point is that Skolem's non-standard model of the integers imbeds in *R. A non-standard number in Skolem's model is represented by a definable function on N, and therefore represents a Robinson hypernatural. Skolem's nonstandard integers can be constructed without the axiom of choice. Yet they can also be viewed as Robinson non-standard integers. This makes it particularly clear that the fact that a non-standard integer produces a non-measurable set is due not the "chimerical" nature of the integer (as Connes repeatedly claimed) but rather to the power of the transfer principle (which is available in Robinson's framework but not in Skolem's).

  • $\begingroup$ Regarding the technical point: since you had already 50 points, I think, it is strange you cannot comment (wherever you like, you can always comment on your own things, in principle). Two things to consider: first, there is character limit of 600 for comments, different browsers handle this differently, it might be that (you'd need to break up the comment in pieces). Second, you get only a comment box when you are logged in (while the answer box is always visible). So if ever you did not login before starting to write this could be an explanantion. $\endgroup$
    – user9072
    Nov 20, 2012 at 16:41
  • $\begingroup$ Thanks. Do you have rights to format the above comment as a response to Choi? $\endgroup$ Nov 20, 2012 at 16:45
  • $\begingroup$ You are welcome! No, this is not possible for me. (I am not even sure moderators can do this.) If you like, and now could comment, you could however post the comments and then delete this answer (link below the text of the question). [I do not think it is necessary, just of you prefer.] $\endgroup$
    – user9072
    Nov 20, 2012 at 16:53
  • $\begingroup$ Hi @quid, these were very helpful remarks. I hope we can continue in this spirit from now on. $\endgroup$ May 13, 2014 at 8:45

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