Question (informal)

Is there an empirically verifiable scientific experiment that can empirically confirm that the Lebesgue measure has physical meaning beyond what can be obtained using just the Jordan measure? Specifically, is there a Jordan non-measurable but Lebesgue-measurable subset of Euclidean space that has physical meaning? If not, then is there a Jordan measurable set that has no physical meaning?

If you understand my question as it is, great! If not, in the subsequent sections I will set up as clear definitions as I can so that this question is not opinion-based and has a correct answer that is one of the following:

  1. Yes, some Jordan non-measurable subset of Euclidean space has physical meaning.

  2. No, there is no physically meaningful interpretation of Jordan non-measurable sets (in Euclidean space), but at least Jordan measurable sets do have physical meaning.

  3. No, even the collection of Jordan measurable sets is not wholly physically meaningful.

In all cases, the answer must be justified. What counts as justification for (1) would be clear from the below definitions. As for (2), it is enough if the theorems in present scientific knowledge can be proven in some formal system in which every constructible set is Jordan measurable, or at least I would like citations of respected scientists who make this claim and have not been disproved. Similarly for (3), there must be some weaker formal system which does not even permit an embedding of Jordan sets but which suffices for the theorems in present scientific knowledge!


Now what do I mean by physical meaning? A statement about the world has physical meaning if and only if it is empirically verified, so it must be of the form:

For every object X in the collection C, X has property P.

For example:

For every particle X, its speed measured in any reference frame does not exceed the speed of light.

By empirical verification I mean that you can test the statement on a large number of instances (that cover the range of applicability well). This is slightly subjective but all scientific experiments follow it. Of course empirical verification does not imply truth, but it is not possible to empirically prove anything, which is why I'm happy with just empirical evidence, and I also require empirical verification only up to the precision of our instruments.

I then define that a mathematical structure $M$ has physical meaning if and only if $M$ has a physically meaningful interpretation, where an interpretation is defined to be an embedding (structure-preserving map) from $M$ into the world. Thus a physically meaningful interpretation would be an interpretation where all the statements that correspond to structure preservation have physical meaning (in the above sense).

Finally, I allow approximation in the embedding, so $M$ is still said to have (approximate) physical meaning if the embedding is approximately correct under some asymptotic condition.

For example:

The structure of $V = \mathbb{R}^3$ has an (approximate) physically meaningful interpretation as the points in space as measured simultaneously in some fixed reference frame centred on Earth.

One property of this vector-space is:

$\forall u,v \in V\ ( |u|+|v| \ge |u+v| )$.

Which is indeed empirically verified for $|u|,|v| \approx 1$, which essentially says that it is correct for all position vectors of everyday length (not too small and not too big). The approximation of this property can be written precisely as the following pair of sentences:

$\forall ε>0\ ( \exists δ>0\ ( \forall u,v \in V\ ( |u|-1 < δ \land |v|-1 < δ \rightarrow |u|+|v| \ge |u+v|-ε ) ) )$.

This notion allows us to classify scientific theories such as Newtonian mechanics or special relativity as approximately physically meaningful, even when they fail in the case of large velocities or large distances respectively.

Question (formal)

Does the structure of Jordan measurable subsets of $\mathbb{R}^3$ have (approximate) physical meaning? This is a 3-sorted first-order structure, with one sort for the points and one sort for the Jordan sets and one sort for $\mathbb{R}$, which function as both scalars and measure values.

If so, is there a proper extension of the Jordan measure on $\mathbb{R}^3$ that has physical meaning? More specifically, the domain for the sort of Jordan sets as defined above must be extended, and the other two sorts must be the same, and the original structure must embed into the new one, and the new one must satisfy non-negativity and finite additivity. Bonus points if the new structure is a substructure of the Lebesgue measure. Maximum points if the new structure is simply the Lebesgue measure!

If not, is there a proper substructure of the Jordan measure on $\mathbb{R}^3$ such that its theory contains all the theorems in present scientific knowledge (under suitable translation; see (*) below)? And what is an example of a Jordan set that is not an element in this structure?


A related question is what integrals have physical meaning. I believe many applied mathematicians consider Riemann integrals to be necessary, but I'm not sure what proportion consider extensions of that to be necessary for describing physical systems. I understand that the Lebesgue measure is an elegant extension and has nice properties such as the dominated convergence theorem, but my question focuses on whether 'pathological' sets that are not Jordan measurable actually 'occur' in the physical world. Therefore I'm not looking for the most elegant theory that proves everything we want, but for a (multi-sorted) structure whose domains actually have physical existence.

The fact that we do not know the true underlying structure of the world does not prevent us from postulating embeddings from a mathematical model into it. For a concrete example, the standard model of PA has physical meaning via the ubiquitous embedding as binary strings in some physical medium like computer storage, with arithmetic operations interpreted as the physical execution of the corresponding programs. I think most logicians would accept that this claim holds (at least for natural numbers below $2^{1024}$). Fermat's little theorem, which is a theorem of PA, and its consequences for RSA, has certainly been empirically verified by the entire internet's use of HTTPS, and of course there are many other theorems of PA underlying almost every algorithm used in software!

Clearly also, this notion of embedding is not purely mathematical but has to involve natural language, because that is what we currently use to describe the real world. But as can be seen from the above example, such translation does not obscure the obvious intended meaning, which is facilitated by the use of (multi-sorted) first-order logic, which I believe is sufficiently expressive to handle most aspects of the real world (see the below note).

(*) Since the 3-sorted structure of the Jordan measure essentially contains the second-order structure of the reals and much more, I think that all the theorems of real/complex analysis that have physical meaningfulness can be suitably translated and proven in the associated theory, but if anyone thinks that there are some empirical facts about the world that cannot be suitably translated, please state them explicitly, which would then make the answer to the last subquestion a "no".

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    $\begingroup$ Downvoters please make clear what is wrong with the question. I'm tired of not knowing what is wrong when I spent hours thinking carefully about the question and how to make it concretely answerable. There are so many questions about physical meaning that are well received even though they are so much more vague than mine. Perhaps it is because they are on fancy topics like Banach-Tarski or because they include the word "paradox"? $\endgroup$ – user21820 May 6 '16 at 12:17
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    $\begingroup$ I didn't downvote the question (although I probably should), but, to begin with, does $\mathbb{R}$ itself have physical meaning that can be empirically verified? $\endgroup$ – Alex Degtyarev May 6 '16 at 12:54
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    $\begingroup$ @AlexDegtyarev: That's a good question. There are two parts to this. (1) Let $R$ be the second-order structure of the reals. Embed it into the world as positions on the path of a photon (this is of course an approximate embedding), and interpreting arithmetic operations by physical implementation of geometric constructions. Take any true universal sentence over $R$. In many cases we can Skolemize it to obtain a $Π_1$-sentence that can be empirically verified in the sense of us substituting 'real' values obtained from measurements. [continued] $\endgroup$ – user21820 May 6 '16 at 13:22
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    $\begingroup$ [continued] You may object that measurements may not produce a representative set of the elements of $R$, but at least there is no empirical evidence against the interpretation, and in some sense it is empirically verified because every 'real'-valued physical measurement will seem to satisfy the sentence! [continued] $\endgroup$ – user21820 May 6 '16 at 13:22
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    $\begingroup$ @AlexDegtyarev : user21820 may have more luck embedding intervals of $\Bbb{R}$ into this world's dual space (quantum fields do appear to be continuous-valued) than in this world's primal space (for which there are discontinuous theories of spacetime). But I read the original Question as allowing for our incomplete present understanding of this universe, so model/object mismatch is expected and not an obstruction to answering. $\endgroup$ – Eric Towers May 6 '16 at 16:37

There are at least two different $\sigma$-algebras that Lebesgue measure can be defined on:

  1. The (concrete) $\sigma$-algebra ${\mathcal L}$ of Lebesgue-measurable subsets of ${\bf R}^d$.
  2. The (abstract) $\sigma$-algebra ${\mathcal L}/\sim$ of Lebesgue-measurable subsets of ${\bf R}^d$, up to almost everywhere equivalence.

(There is also the Borel $\sigma$-algebra ${\mathcal B}$, but I will not discuss this third $\sigma$-algebra here, as its construction involves the first uncountable ordinal, and one has to first decide whether that ordinal is physically "permissible" in one's concept of an approximation. But if one is only interested in describing sets up to almost everywhere equivalence, one can content oneself with the $F_\delta$ and $G_\sigma$ levels of the Borel hierarchy, which can be viewed as "sets approximable by sets approximable by" physically measurable sets, if one wishes; one can then decide whether this is enough to qualify such sets as "physical".)

The $\sigma$-algebra ${\mathcal L}$ is very large - it contains all the subsets of the Cantor set, and so must have cardinality $2^{\mathfrak c}$. In particular, one cannot hope to distinguish all of these sets from each other using at most countably many measurements, so I would argue that this $\sigma$-algebra does not have a meaningful interpretation in terms of idealised physical observables (limits of certain sequences of approximate physical observations).

However, the $\sigma$-algebra ${\mathcal L}/\sim$ is separable, and thus not subject to this obstruction. And indeed one has the following analogy: ${\mathcal L}/\sim$ is to the Boolean algebra ${\mathcal E}$ of rational elementary sets (finite Boolean combinations of boxes with rational coordinates) as the reals ${\bf R}$ are to the rationals ${\bf Q}$. Indeed, just as ${\bf R}$ can be viewed as the metric completion of ${\bf Q}$ (so that a real number can be viewed as a sequence of approximations by rationals), an element of ${\mathcal L}/\sim$ can be viewed (locally, at least) as the metric completion of ${\mathcal E}$ (with metric $d(E,F)$ between two rational elementary sets $E,F$ defined as the elementary measure (or Jordan measure, if one wishes) of the symmetric difference of $E$ and $F$). The Lebesgue measure of a set in ${\mathcal L}/\sim$ is then the limit of the elementary measures of the approximating elementary sets. If one grants rational elementary sets and their elementary measures as having a physical interpretation, then one can view an element of ${\mathcal L}/\sim$ and its Lebesgue measure as having an idealised physical interpretation as being approximable by rational elementary sets and their elementary measures, in much the same way that one can view a real number as having idealised physical significance.

Many of the applications of Lebesgue measure actually implicitly use ${\mathcal L}/\sim$ rather than ${\mathcal L}$; for instance, to make $L^2({\bf R}^d)$ a Hilbert space one needs to identify functions that agree almost everywhere, and so one is implicitly really using the $\sigma$-algebra ${\mathcal L}/\sim$ rather than ${\mathcal L}$. So I would argue that Lebesgue measure as it is actually used in practice has an idealised physical interpretation, although the full Lebesgue measure on ${\mathcal L}$ rather than ${\mathcal L}/\sim$ does not. Not coincidentally, it is in the full $\sigma$-algebra ${\mathcal L}$ that the truth value of various set theoretic axioms of little physical significance (e.g. the continuum hypothesis, or the axiom of choice) become relevant.

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    $\begingroup$ in the last parenthetical, it is a bit unclear which you meant by "the latter". Do you mean $\mathcal{L}$ or $\mathcal{L}/ \sim$? $\endgroup$ – Willie Wong May 6 '16 at 20:28
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    $\begingroup$ I don't know what "physically meaningful" means in this context, but from a mathematical point of view, most of the function spaces built upon Lebesgue measure foundations (e.g. $L^p({\bf R}^d)$, Sobolev spaces, spaces of distributions, etc.) can be constructed using ${\mathcal L}/\sim$ rather than ${\mathcal L}$. In particular the mathematical foundations of the quantum mechanics of nonrelativistic particles can be constructed in this fashion. (QFT is another story, as there are serious issues making it fully mathematical rigorous at present.) $\endgroup$ – Terry Tao May 9 '16 at 17:25
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    $\begingroup$ The one major thing one gives up when restricting to ${\mathcal L}/\sim$ instead of ${\mathcal L}$ is that one no longer has a set membership relation $x \in A$ between a real number $x$ and an element $A$ of ${\mathcal L}/\sim$; similarly, one cannot evaluate a measurable function $f$ at a specific real number $x$ unless $f$ has additional regularity beyond measurability (e.g. continuity) that would permit this. In particular it becomes difficult to establish results that hold at every real number $x$ using just ${\mathcal L}/\sim$, though almost everywhere results are OK. $\endgroup$ – Terry Tao May 9 '16 at 17:28
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    $\begingroup$ The extent to which set theoretic axioms impact "low-level" objects in mathematics, such as measurable functions on the reals up to almost everywhere equivalence, is still an area of ongoing research (see for instance the article of Harvey Friedman at u.osu.edu/friedman.8/files/2014/01/CCRTalk1.121905-xkz4qf.pdf ). For instance, there are some reasonably elementary (but by no means first-order) statements about the reals which are sensitive to set theory axioms, such as Borel's conjecture en.wikipedia.org/wiki/Strong_measure_zero_set . $\endgroup$ – Terry Tao May 9 '16 at 17:31
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    $\begingroup$ In any case, the main point of my post is that the $\sigma$-algebra ${\mathcal L}/\sim$ (together with its attendant Lebesgue measure) is very analogous to the reals ${\bf R}$ (with its attendant metric) in that both can be viewed as the metric completion of a countable collection of "rational" objects. As such, whatever ontological status you wish to confer on the reals ${\bf R}$ can be plausibly also conferred onto ${\mathcal L}/\sim$. $\endgroup$ – Terry Tao May 9 '16 at 17:35

Your question reminds me of Hamming's quote about why he believed there is no "physical significance" to the difference between Riemann and Lebesgue integration: if there were a plane whose ability to fly depends on the distinction between Riemann and Lebesgue integration then he would not care to fly in that plane.

The distinction is mathematical, not physical. A paper that explains this is http://www.mast.queensu.ca/~andrew/notes/pdf/2007c.pdf.

This kind of distinction arises much earlier in mathematics than measure theory. For example, in the study of infinite series, we have the nice theorem that the set of real numbers where a power series converges is an interval on the real line because of inequalities on power series and the completeness properties of the real numbers. Does this tidy result have a physical meaning when the most you can ever physically measure a length (using standard length units like meters) does not even extend out to something like the 25th digit after the decimal point? I do not think so, but even if you want to talk about the value of a Bessel function at a specific number, you need to have a general conception of a power series that is converging at all real numbers -- even numbers you do not care about for physical purposes -- to have a Bessel function in the first place. The set of numbers that have at most 25 digits after the decimal point has bad algebraic and analytic properties (not a field, not complete), so the set of "physically meaningful" real numbers is not something you can base a good mathematical theory on.

To put it more simply, is there a physical meaning in the trillionth digit of $\pi$? I would say no, and I've seen statements to the effect that you don't need more than 10 or 15 digits after the decimal point of $\pi$ to measure the radius of the universe down to the width of a proton, but that does not mean $\pi$ should be considered equal to 3.141592653589793.

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    $\begingroup$ That is exactly one of the quotes I had in mind, but didn't want to add unnecessary side remarks to my question. $\endgroup$ – user21820 May 6 '16 at 13:23
  • $\begingroup$ I must say I greatly appreciate your link, because I had never seen it before despite hearing the quote from Hamming. Unfortunately, your answer doesn't address my question because I already accept the mathematical elegance of the Lebesgue measure but am asking about the physical meaning. $\endgroup$ – user21820 May 6 '16 at 13:33
  • $\begingroup$ Wow so many points for me to address. =) (1) For concreteness, the claim in the linked article that the physical properties of an airplane can never be affected by functions that are not Riemann integrable but Lebesgue integrable is not justified. I would like to hear anyone who have justification for affirming or refuting this claim, and that is why I formulated my question to try to give a very concrete way to do just that. (2) Fourier transforms are always mentioned, but do the functions that require the Lebesgue integral have real-world significance? I don't think so. $\endgroup$ – user21820 May 6 '16 at 13:40
  • $\begingroup$ (3) Concerning many decimal places, I tackled that question by allowing approximate embeddings. Notice that my example about $\mathbb{R}^3$ embedding into spacetime (at a single point in time from a fixed reference frame) is only stated to be valid for vectors with length bounded away from $0$ (quantum effects become dominant) and away from $\infty$ (spacetime curvature becomes non-negligible). Furthermore I stated that we cannot expect to empirically verify anything beyond our instrument precision. $\endgroup$ – user21820 May 6 '16 at 13:44
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    $\begingroup$ In your first comment you say my answer doesn't address your question, but I think it does: the "problem" that seems to bother you about Lebesgue measure vs. Jordan measure is already at work at a more elementary level in calculus when you want to talk about an infinite series converging: essentially all the standard theorems about infinite series depend on completeness of $\mathbf R$, and the convenience of spaces like $L^1(\mathbf R)$ and the Hilbert space $L^2(\mathbf R)$ compared to their dense subspaces of Riemann integrable functions is their completeness. Do you care about completeness? $\endgroup$ – KConrad May 6 '16 at 19:21

Your question does not seem to be mathematically precise, but perhaps I can suggest some mathematically precise questions that approximate what you want.

Many people believe that the axiom of choice is not physically meaningful. One argument for this cites Shoenfield's absoluteness theorem, which implies in particular that any arithmetical (i.e., expressible in first-order Peano arithmetic) consequence of ZFC is already a consequence of ZF. For many people, "physically meaningful" mathematical statements are a subset of arithmetical statements.

The existence of non-Lebesgue measurable sets doesn't quite imply the axiom of choice, but is still widely regarded as not physically meaningful. In this regard, the Solovay model is frequently cited. The Solovay model shows that if there exists an inaccessible cardinal, then ZF is consistent with "all sets of reals are Lebesgue measurable."

If you find these arguments persuasive, perhaps because you believe something akin to "ZF suffices for physics," then maybe one way to phrase your question is whether ZF is consistent with "all subsets of $\mathbb{R}^2$ are Jordan-measurable." I don't know the answer to this question.

I'm less sure about how to interpret your question of whether there exists a physically unmeaningful Jordan-measurable subset of the plane. A trivial observation is that there are uncountably many Jordan-measurable subsets (in fact, there are uncountably many rectangles). Perhaps there can only be countably many physically meaningful subsets of the plane? But this is probably not what you are asking. In the comments you mention computable reals; this suggests that you may be interested not in ZF but in RCA0. But I am not sure what statement whose provability in RCA0 (or lack thereof) would capture your question.

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    $\begingroup$ Of course ZF is not consistent with "all sets are Jordan-measurable!" The set of rational points is not Jordan-measurable. $\endgroup$ – Monroe Eskew May 6 '16 at 17:48
  • $\begingroup$ @MonroeEskew: Indeed, and one side-question of mine is whether the set of rational points is physically meaningful or not! Anyway Timothy: thanks for your answer. I'm aware of absoluteness that does partially answer the question of Lebesgue non-measurable sets if indeed we only consider arithmetical sentences physically meaningful. However, I'm interested in the Jordan non-measurable sets that are Lebesgue measurable, and I doubt arithmetical sentences are the only physically meaningful ones. $\endgroup$ – user21820 May 7 '16 at 3:48
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    $\begingroup$ @user21820 : I don't have strong beliefs about what is physically meaningful; I'm just trying to probe what you mean by it. Are you saying that the existence of computable functions with no computable fixed points does not make you doubt that Brouwer's fixed point theorem is physically meaningful? $\endgroup$ – Timothy Chow May 8 '16 at 18:35
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    $\begingroup$ @user21820 : I would still like you to provide a direct answer to my question about Brouwer's fixed-point theorem specifically, because it affects whether one considers RCA0 or WKL0. $\endgroup$ – Timothy Chow May 9 '16 at 15:10
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    $\begingroup$ @user21820 : I'm not sure that your proposal works. Orevkov's example involves a function that is continuous on the set of computable reals in the unit square, but it is not uniformly continuous and therefore does not extend to a continuous function on the full unit square. So there is no natural way to define the value of your oracle. $\endgroup$ – Timothy Chow May 9 '16 at 20:33

Really comments, not an answer to the question.

(1) Mandelbrot has some comments on "physical meaning" in his book The Fractal Geometry of Nature. He relates that before 1970 or so, whenever he would try to put a Cantor set into a paper about physics, it would be rejected as "unphysical". But nowadays, physics journals are replete with papers on fractal this and fractal that.

(It sometimes becomes humorous (and painful) when an older physicist, who never learned about Lebesgue measure, attempts to write about fractals.)

(2) Quantum mechanics. It uses Hilbert space. Even for the simplest harmonic oscillator, you use a Hilbert space such as $L^2(\mathbb R)$. It doesn't work if you restrict only to Riemann integrable functions with $\int |f|^2 < \infty$. You have to use something more general, such as Lebesgue integrable functions.

Of course I suppose you can say quantum mechanics itself has no physical meaning. But I leave that to physicists to answer. I wonder what happens if you ask a physicist whether electron orbitals are Jordan measurable...


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    $\begingroup$ Why would you say that quantum mechanics has no physical meaning? $\endgroup$ – Nik Weaver May 6 '16 at 14:54
  • $\begingroup$ Is there a fundamental obstruction to developing quantum mechanics without Lebesgue integration? I know a lot of people say so, but don't see why. Specifically, can you prove that there is some theorem derived in the framework of quantum mechanics (with the full theory of Lebesgue integration) that concerns only functions that are physically realizable but that cannot be derived using only Jordan measurability? $\endgroup$ – user21820 May 6 '16 at 15:20
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    $\begingroup$ Questions on QM belong in physics.stackexchange.com not here. $\endgroup$ – Gerald Edgar May 6 '16 at 15:22
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    $\begingroup$ @user21820- certainly Lebesgue is practically (and I suspect strictly) essential. For QM of a free particle one wants delta functions, rigged Hilbert spaces, and the Fourier transform in sufficient generality to make the informal arguments translatable into math vs utter gibberish $\endgroup$ – Steve Huntsman May 6 '16 at 17:50
  • $\begingroup$ @SteveHuntsman: I agree that Lebesgue is practically very useful, but my question really is about whether or not it is "physically essential". $\endgroup$ – user21820 May 7 '16 at 3:38

In ergodic theory Lebesgue vs Jordan is a critical distinction. Whether or not the Stosszahlansatz etc. are physically meaningful as you mean here, such things are of fundamental import to the theory of statistical physics and chaos.

To give a particular toy example in this vein: the SRB measure of a hyperbolic toral automorphism is Lebesgue measure. Consider the pushforwards of a small ball: in the limit, that set will be Lebesgue but not Jordan measurable. But one wants (needs?) the Liouville theorem and ergodic hypothesis.

  • $\begingroup$ I'm sorry I can't see how your post answers my question, where I said: "I'm not looking for the most elegant theory that proves everything we want, but for a (multi-sorted) structure whose domains actually have physical existence." I'm not surprised that one wants a more powerful theory in mathematical work, but that isn't what my question is about. $\endgroup$ – user21820 May 7 '16 at 5:43
  • $\begingroup$ @user21820- does a chaotic attractor in classical mechanics physically exist? I'm pretty sure this breaks before getting down to the Planck scale $\endgroup$ – Steve Huntsman May 7 '16 at 14:44
  • $\begingroup$ Actually I believe chaos theory from classical mechanics does give good approximation for everyday time scales but breaks down at small or large time scales precisely due to quantum effects. But as I said, I would like actual justification for your views, which is the very reason I ask my question; before that I've looked everywhere I could but I've only seen opinions and nothing more. Terry Tao's post seems to imply that the correct answer to my question is my option (2), but he hasn't explicitly said it nor given full justification, but what he wrote comes the closest I think. $\endgroup$ – user21820 May 7 '16 at 15:14

I believe the crucial point behind the physical significance of Lebesgue measure as opposed to Jordan measure boils down to the issue of completeness, as Gerald Edgar remarked in part (2) of his answer. I'm answering here anyway because I want to elaborate a bit on this.

Since a bounded set is Jordan measurable if and only if its characteristic function is Riemann integrable, it is clear that the space of $p$-th power integrable functions with respect to Jordan measure cannot be complete, for the same reason that the space of $p$-th power (improper) Riemann-integrable functions is not complete (both are dense in the space of $p$-th power Lebesgue-integrable functions, of course). In other words, your question (as I see it) essentially amounts to asking for the physical meaning of completeness of the above function spaces. This also begs the question about the physical meaning of completeness of the real line itself, as pointed several times by several people here, but I won't touch it in what follows (apart from the last paragraph below)...

For $p=2$, this implies asking "Why do we need Hilbert spaces? Don't pre-Hilbert spaces suffice?" as asked here in math.SE in the context of Quantum Mechanics (pointed by Gerald Edgar's comment while I was writing this). The many mathematical uses of completeness in this case are generously illustrated in this MO question, including applications to signal processing (see e.g. Alain Valette's answer). Without completeness, many theorems of fundamental importance in Quantum Mechanics, such as the Riesz representation theorem (think of Dirac bras and kets) and the spectral theorem (relating self-adjoint operators to measurements), wouldn't take off the ground in the needed generality. In other words, the formalism of quantum mechanics simply wouldn't exist beyond finite-dimensional Hilbert spaces, which would exclude most quantum mechanical systems of physical interest (atoms, molecules, etc.).

Of course, all of the above assumes the "standard" set-theoretical axiom system ZFC. It doesn't hurt, though, to remember that most of the mathematics used in physics so far is more or less "standard". A broader related issue is to which extent we need completeness in Analysis. This has been put into question by intuitionistic mathematicians, who would try to move as far as possible using only finitary algorithms (as many a physicist would like to do) and, obviously, would reject such blatantly non-constructive a premise as the axiom of choice. See Wikipedia's entry on constructive analysis for more on this.

  • $\begingroup$ Thanks for this! However, the most popular answer to the question about the role of completeness in Hilbert spaces (mathoverflow.net/a/36018/50073) actually seems to suggest that although completeness is economical (which I certainly agree), it is probably not physically meaningful, and moreover not theoretically necessary. If someone can substantiate this latter claims, I'll be very glad to know! $\endgroup$ – user21820 May 7 '16 at 4:24
  • $\begingroup$ To be more precise, Piero D'Ancona's answer referred to the completion of a given, dense subspace which may be thought of as "physically meaningful". Indeed, Cantor's completion procedure amounts to grouping all equivalent Cauchy sequences as as a single, "idealized" element which is approximated to arbitrary accuracy by the elements of any Cauchy sequence that represents the element, which we can see as "physical truncations". The crux of the matter is that physical measurements are always approximate, so in this sense only those "truncations" are needed and physically meaningful. $\endgroup$ – Pedro Lauridsen Ribeiro May 7 '16 at 4:51
  • $\begingroup$ Yes that seems very similar to the distinction Terry Tao made in his post about the collection of Lebesgue measurable sets and just the rational elementary sets that approximate these in the mathematical world. When we form the quotient type, its members are no longer of the original type, so it is conceivable that the original type is physically meaningful while the quotient type is merely conceptual, and the fact that the original type embeds into the quotient type is also conceptual. $\endgroup$ – user21820 May 7 '16 at 5:01
  • $\begingroup$ However, which dense subspace should be considered "physically meaningful" is somewhat arbitrary from a mathematical viewpoint (more precisely, it depends on the problem you're studying), and not always clear. The choice may not be (and usually isn't) unique, and you may want to relate results starting from different dense "physically meaningful" subspaces. The machinery of completion takes care of all these issues in a very elegant way. $\endgroup$ – Pedro Lauridsen Ribeiro May 7 '16 at 5:01
  • $\begingroup$ I believe the arbitrariness is an artifact of using ZF. In a constructive setting, the quotient type may not be constructible, and so using it as a complete whole is impossible, but many of the results involving it can be rephrased in terms of the original type and the equivalence relation, so there is a concrete distinction between what seems arbitrary from the point of view of a stronger theory like ZF. Coq for instance uses setoids which are one manifestation of this constructive distinction. $\endgroup$ – user21820 May 7 '16 at 5:27

There are already many nice answers, I will go with a more direct approach.

It is possible to devise a physical setting in which all Lebesgue probability measures on $\mathbb{R}^{2d}$ are physically constructed (more precisely all Radon probability measures on the Borel $\sigma$-algebra of $\mathbb{R}^{2d}$); at least "approximately" as intended by the OP.

It is necessary to accept the following (mathematical) physics facts:

  • there are systems that can be effectively described by a quantum mechanical point particle, i.e. by a non-commutative probability theory of observables and states that encodes the canonical commutation relations (Weyl algebra over $\mathbb{R}^{2d}$ with the canonical symplectic form);

  • there are quantum mechanical states, called squeezed coherent states that have minimal uncertainty;

  • the regime of large quantum numbers (classical physics) has physical meaning.

These three facts are rather commonly accepted, and there is experimental evidence of all of them. In particular it is indeed possible to engineer squeezed coherent states, e.g. by means of lasing.

It has been mathematically proved, and postulated to be true by Bohr at the very beginning of quantum mechanics, that the quantum non-commutative probability theory reduces to a classical probability theory in the classical limit of large quantum numbers, or as commonly intended as $\hbar\to 0$.

In particular, it has been proved that squeezed coherent states converge to classical probability measures concentrated on a point of the phase space $\mathbb{R}^{2d}$. By means of statistical combinations of squeezed coherent states, or using other suitably prepared quantum states, it is then (at least theoretically) possible to obtain any probability measure of $\mathbb{R}^{2d}$ in the classical limit, for linear combinations of point measures are dense in the space of probability measures endowed with the weak topology.

More importantly for the situation at hand, it would also possible to test the faithfulness of such predictions: this would be done by direct measurement of observables on the prepared states, and comparing the statistical distribution of outcomes with respect to the expected probability distribution.

This provides a setting where the emergence of Lebesgue measures in physics as classical states (classical approximations of quantum states) can be directly tested.

  • $\begingroup$ Your approach at first sounds good, but the major problem is that if you cannot devise an experiment to empirically verify the existence of a Jordan non-measurable set (like perhaps a Riemann non-integrable probability density function), then you haven't answered the question at all. Notice that any conservative extension of any physically meaningful theory will not be falsifiable based on sentences in the original language. Just for example suppose that $\mathbb{R}$ is physically meaningful but not the hyperreals. Then you can't detect that based on a first-order sentence! $\endgroup$ – user21820 May 8 '16 at 8:50
  • $\begingroup$ That is why my question explicitly asks for an answer that affirms (1) to provide an explicit Jordan non-measurable set that has physical meaning (namely you can empirically verify its properties). Without such, it is not possible to refute (2). $\endgroup$ – user21820 May 8 '16 at 8:52
  • $\begingroup$ Take a quantum state whose classical limit is a measure concentrated on a Borel non-Jordan-measurable set (it is possible since every classical measure can be reached in the limit). If you can prepare such quantum state, then you can experimentally test observables on it and make measurements that would confirm or falsify the nature of the resulting classical measure. This is a perfectly reasonable physical experiment, that is at least in principle concretely realizable. I don't see any issue here. $\endgroup$ – yuggib May 8 '16 at 10:35
  • $\begingroup$ The issue is that you cannot prepare such a quantum state. Being a limit of preparable quantum states totally does not imply it is preparable. Furthermore, some others here even think that the completeness of the reals may not be physical, not to say limits of sets! $\endgroup$ – user21820 May 8 '16 at 10:42
  • $\begingroup$ The so-called Factorized Fock states (or in general any Fock state) can be seen as a limit of finite linear combinations of squeezed coherent states; nevertheless they're indeed realizable. So being a limit of preparable states does neither imply it is not preparable. Yet from another perspective, with your relaxed concept of "approximate physical meaning", all quantum states obtained by the limit of preparable states are approximately preparable to any given precision. $\endgroup$ – yuggib May 8 '16 at 11:21

For me the trouble with Lebesgue measure is that it is quite specialized.

In probability we might try to measure the probability of an event using other observations at our disposal. Hopefully some combination of unions and intersections, however complicated might give us a meaningful result.

However, there are cases where we can't measure given the information. We can't measure temperature with a ruler.

Maybe... We can use the expansion or contraction with respect to temperature. Then given a prior series of observations we might try to combine them to deduce temperature from length measurements.

If we solve a partial differential equation using Fourier series. We might try to study how much error we created by truncating the series after finitely many terms. The nature of the error might be studied with Lebesgue measure and Lebesgue integral.

Another consideration is a very complicated set. In principle any measurable set will be covered with squares or rectangles or circles. However, we might be limited by how small the circles are, or how many circles we can use, or be forced to use squares instead of circles.

An example, what is the minimum number of squares to cover a circle with 1% error?


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