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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!

Definitions

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?

Remarks

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".

user21820
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    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"? – user21820 May 06 '16 at 12:17
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    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? – Alex Degtyarev May 06 '16 at 12:54
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    @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] – user21820 May 06 '16 at 13:22
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    [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] – user21820 May 06 '16 at 13:22
  • Maybe the distinction can be found in probability. I suppose it will be questionable how physically meaningful probability spaces are. But of course they are very important in physics. – Monroe Eskew May 06 '16 at 13:23
  • @MonroeEskew: Please tell me more about any potential distinction that you can find. Probability theory was one of the major motivations behind my question, since I'm often told that the Lebesgue measure is the most crucial in deriving the fundamental results. I've no doubt about that, but I don't know whether it is because the fundamental results are in fact stronger than what is necessary for the physical world. – user21820 May 06 '16 at 13:25
  • @AlexDegtyarev: [continued] What might be happening is that the true structure underlying a physical path in spacetime (in the locally zero curvature approximation) is elementarily equivalent to $R$ but not the same as the one I have defined based on $\mathbb{R}$, which in turn is ultimately based on the power set of $\mathbb{N}$. That is fine, and still means that the theory of $R$ is physically meaningful, even if the structure is not, and most of my question is still valid for the theory. – user21820 May 06 '16 at 13:28
  • @AlexDegtyarev: (2) I left open the possibility that even $Th(R)$ is not physically meaningful in my last part of (), in which case I would gladly welcome anyone who shows me even a little evidence that our physical world is so strange. However, I believe most of us agree that the field of computable* reals have physical meaning that can be empirically verified (up to instrument precision or quantum scales, whichever blocks first), since we can more or less handle them using Turing machines. Of course, many second-order properties no longer hold, but may be hard or impossible to test. – user21820 May 06 '16 at 13:48
  • Speaking about spacetime, path of a photon, and such, you are merely trying to embed $\mathbb{R}$ to your current model of the physical world. – Alex Degtyarev May 06 '16 at 13:48
  • @AlexDegtyarev: I'm not embedding into my own model of the world. The "path of a photon" (on everyday scales) is something that every physicist understands, even if they have different opinion of how it really happens 'underneath'. To be more concrete. I can just say "the places where you can detect a photon that has been emitted by this focused laser pointing in some fixed direction relative to my lab! Also, most physicists believe in the correctness of general relativity at large scale, which does rely on the properties of $\mathbb{R}$, at least the way it is conventionally formulated. – user21820 May 06 '16 at 13:52
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    I am thinking (a) it takes some serious machinery to give a formal foundation for i.i.d. random variables. (b) One can easily get events by taking vey simple and natural Borel sets, that one would want to know the probability of, but when viewed geometrically are not very pretty and not Jordan measurable. I'd have to think a bit harder to make this convincing. – Monroe Eskew May 06 '16 at 13:53
  • @AlexDegtyarev: Unless we say that physicists are not really talking sense when they say the believe in GR at large scales, there must be something they are trying to get at. I'm trying to make that intuitive notion precise in my question. Sorry for this long discussion, but I feel we are solidly getting somewhere in teasing out what I'm trying to ask. – user21820 May 06 '16 at 13:55
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    For a slightly odd definition of "physical intuition", Goldblatt constructs the Lebesgue measure using non-standard methods in Lectures on the Hyperreals, Chapter 16. It corresponds essentially to assigning an infinitesimal constant weight to each point of an infinitesimal-width lattice $L$ on the reals. For each point in the set $X \subseteq \mathbb{R}$, we then take the points of $L$ infinitesimally close to $X$, and sum up the contribution of all the weights; this yields a non-standard real $r$, and we take its standard part to obtain the Lebesgue measure of $X$. – Patrick Stevens May 06 '16 at 16:19
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    You misunderstood me. I didn't mean your personal model; I meant the current model in which most/some physicists may believe. But it will always be a model, not the "real world". So, the question about the "real world" is meaningless mathematically; the question about a particular model has no physical meaning :) – Alex Degtyarev May 06 '16 at 16:26
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    @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. – Eric Towers May 06 '16 at 16:37
  • Church–Turing–Deutsch principle: "The principle states that a universal computing device can simulate every physical process." – Count Iblis May 06 '16 at 17:05
  • @CountIblis I don't think this is relevant, since there is a field of study called computable measure theory. And of course they mean Lebesgue measure. – Monroe Eskew May 06 '16 at 17:42
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    The process of doing math is in itself a physical process, otherwise we could not practice math. This means that mathematics can always be given a physical meaning :) . – Count Iblis May 06 '16 at 23:48
  • For the purpose of pointing out what I would consider holes in the philosophical foundations of this question: The statement of the question itself requires some philosophical theory of reference that allows a mathematical object to correspond directly with (or refer directly to) a physical object. You need a way of saying that a tuple of three real numbers (constructed from some model of the reals in ZF) actually means a certain point in space (given certain local coordinates). If, like me, you think that there is no connection except in the minds of humans, this question is nonsense. – Alexander Woo May 06 '16 at 23:57
  • @AlexanderWoo: Nope in my question I never required any mathematical object to correspond or refer to any physical object. If you read my definitions carefully, you find that between the mathematical objects and the physical objects there is a bridge, which I called an embedding only to invoke the idea of structure-preservation, not at all to suggest that the range of the embedding is a mathematical object. This means we indeed can only empirically test sentences that are somewhat first-order but couched in natural language. That's fine, such a translation is evident given the embedding. – user21820 May 07 '16 at 02:34
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    I confess I really and truly feel that taking "physically meaningful" seriously has to lead to at least one of "we can have that all mathematical objects are physically meaningful", "the set of real numbers is not physically meaningful", and "whether something physically meaningful is not a yes/no proposition". –  May 07 '16 at 04:52
  • @Hurkyl: I'd be very glad to hear more of your view if you don't mind sharing. Your first option is definitely invalid; it is impossible for the universe to contain every consistent mathematical structure, because itself is a single structure, unless you're claiming the universe is not only self-similar but incompatible with ZF. Personally, I really and truly feel that $\mathbb{R}$ is actually physically meaningful in the sense I defined in my question. You're right that physical meaning is not a purely mathematical concept, but we have no choice; mathematics alone cannot refer to reality. – user21820 May 07 '16 at 05:16
  • @EricTowers: I think you understand my question quite well, and if you have anything to say, even partially, I look forward to your answer! – user21820 May 07 '16 at 05:19
  • @user21820: I do not understand how any non-mathematical object can have a structure in anything resembling the sense that you need. As far as I am concerned (and I am aware there are others who disagree), the physical world has no structure that is not constructed by human minds. (This does not mean human minds are free to construct whatever they wish.) To be particular, I see no reason to deny the possibility of two (or more) non-isomorphic but equally correct theories of the physical world, one of which requires Lebesque measure and the other of which does not. – Alexander Woo May 08 '16 at 03:29
  • Ought this to be tagged [tag:math-philosophy]? – Nate Eldredge May 08 '16 at 16:50
  • @NateEldredge: It's a related tag, but let's leave it out; when I chose the tags for the question "philosophy" never once occurred to me, as I was solely concerned with empirical verification, not philosophical claims that cannot be falsified. I even tried making it so that, whatever the answer, it would be concrete. For (1), the answer should give an explicit Jordan non-measurable set with physical meaning. For (2), the answer should show that physically meaningful theorems proven using the Lebesgue measure can be proven without, and I think Terry Tao claims this but with some handwaving. – user21820 May 08 '16 at 17:09
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    Fair enough. But from my point of view, the whole question of what it means for a claim to be "empirically verifiable", and how we should understand the relationship between mathematical constructs and the real world, is a philosophical one. – Nate Eldredge May 08 '16 at 17:20
  • I am trying to vote to close this question but the software is not cooperating. This question has too many obvious conceptual holes, not to mention that it is NOT a mathematical question. – Włodzimierz Holsztyński May 08 '16 at 18:00
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    @WłodzimierzHolsztyński: (a) You, like the downvoters, didn't give a single concrete example of what is wrong with the question. (b) It is a question solidly based on logic and the scientific method. If you don't understand the question because you lack knowledge in proof theory and model theory at all, that's not my fault. (c) Terry Tao has clearly understood the question, even if his answer doesn't explicitly answer it. – user21820 May 09 '16 at 09:09
  • I'm not a downvoter, but a legitimate reason to downvote is "not useful", which in terms of this question could be restated as saying, "answering this question will not add a reasonable amount of meaning to our current understanding of mathematics". Sometimes questions that are very long (as this one is) are the opposite of pithy and thereby possibly lower-quality. And of course, there's no requirement that downvoters justify their downvotes. One view one can take of downvotes is to be self-critical and attempt to understand what others might find in one's question that is a poor fit for MO. – Todd Wilcox May 09 '16 at 16:04
  • @ToddWilcox: Thanks for giving a possible explanation! However I'm not sure the downvoters can justify that this question is not useful, given that it is clearly a question a lot of people are interested in, and that the answers are quite enlightening, and that my question itself gives a very concrete way one might go about establishing a better understanding of the interface between mathematics and physics. PA has widespread acceptance precisely because it has been widely empirically verified. So has real analysis. I'm simply attempting to extend this goal of verification further. – user21820 May 09 '16 at 16:14
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    I feel that the continually moving goal posts in this question, its answers, and the comments to the question and the answers, make it clear on a mathematical basis that this question is unclear. Accordingly I have voted to close. – Lee Mosher Nov 06 '16 at 15:55

8 Answers8

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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.

Michael Hardy
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Terry Tao
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  • Thanks; this is very interesting! So you're saying that if we simply take Cauchy sequences of rational elementary sets, we can recover most if not all the physically meaningful results that are commonly done using the full Lebesgue measure, while avoiding the set-theoretic questions such as AC and CH? If so, can I say that the answer to my question is probably (2), in the sense that we can restrict to first-order definable sequences of rational elementary sets, which then only requires the theory of the Jordan measure to manipulate? (Unless of course true spacetime is discontinuous.) – user21820 May 07 '16 at 03:33
  • Do you mind addressing my comment, as well as clarifying: (a) exactly which physically meaningful theorems can be proven over $L/~$? ; (b) how exactly set-theoretic axioms are irrelevant to physics, and why it is not coincidental? – user21820 May 08 '16 at 17:13
  • The answer talks about "idealised physical observables," but the question never uses the word "idealised." The question seems to me to be about real-world physical observables, so I don't think this answers the question. –  May 09 '16 at 14:45
  • @BenCrowell: I interpret Terry's "idealised" to loosely correspond to what I defined in my question as mathematical objects that have an approximate physically observable manifestation (approximate here meaning under some asymptotic condition). This is similar to the ideal points and lines in Euclidean geometry that have approximate physical manifestations. But, until he clarifies, I can't tell whether he is trying to affirm option (2) as answer to my question or not. =) – user21820 May 09 '16 at 14:55
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    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.) – Terry Tao May 09 '16 at 17:25
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    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. – Terry Tao May 09 '16 at 17:28
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    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 https://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 https://en.wikipedia.org/wiki/Strong_measure_zero_set . – Terry Tao May 09 '16 at 17:31
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    However, many of the more well known statements about measurable sets and functions that are sensitive to these axioms are no longer encountered when one works purely in ${\mathcal L}/\sim$ rather than ${\mathcal L}$. For instance, the question of whether every subset of ${\bf R}$ is measurable does not have any obvious formulation when working just with ${\mathcal L}/\sim$ as the metric completion of elementary sets, unless one introduces a bit of set theory or higher order logic. – Terry Tao May 09 '16 at 17:33
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    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 ${\mathcal L}/\sim$. – Terry Tao May 09 '16 at 17:35
  • Thanks; your comments clarify a bit some of my questions, and perhaps they should be included in your answer post? Specifically, could you give precise statements about what can be done using just $L/\sim$, such as what you mean by "mathematical foundations of QM for non-relativistic particles"? Also I think your comment about set theoretic axioms' impact on 'low-level' mathematics is very relevant to partly answering my question. As you say, Borel's conjecture is not a first-order axiom, and I think countability can't even be stated over my structure. – user21820 May 11 '16 at 12:45
  • As for your last comment, I didn't get what you meant by the Lebesgue measure on $L/\sim$ being analogous to the metric on $\mathbb{R}$. Did you mean the measure on $L$? I'm already convinced by your answer that the ontological status of $\mathbb{R}$ carries over to rational elementary sets and hence to first-order definable rational elementary set approximations (to represent sequences). I guess if one adds sequences as an extra sort then it carries over to $L/\sim$, though I've no idea whether that extension is conservative or not. But I don't think it carries over to $L$? – user21820 May 11 '16 at 13:01
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    As I said in the post, one can use Jordan measure to create a metric $d$ on the space ${\mathcal E}$ of rational elementary sets (technically one has to work with ${\mathcal E}/\sim$ since some rational elementary sets can differ by sets of measure zero). Passing to the metric completion, one obtains a metric $d$ on the space ${\mathcal L}/\sim$ which agrees with the Lebesgue measure of the symmetric difference: $d(E,F) = m( E \Delta F)$. In particular, the Lebesgue measure of a set $E$ in ${\mathcal L}/\sim$ is equal to the distance of that set to the empty set in this metric. – Terry Tao May 11 '16 at 15:27
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    To continue the preceding analogy, the Lebesgue measure on ${\mathcal L}/\sim$ is then analogous to the absolute value function on the reals, i.e. the distance function to the origin. As for QM, the point is that to model $n$ non-relativistic particles one needs the Hilbert space $L^2({\bf R}^{3n})$ (or slightly fancier versions thereof in the presence of spins, etc.), and this can be constructed using ${\mathcal L}/\sim$ (square integrable measurable functions on ${\mathcal L}/\sim$ in $3n$ spatial dimensions). – Terry Tao May 11 '16 at 15:29
  • Ah okay I see. Do you mind including these details in your post, so that other readers won't have to go through the comments? (By the way I wasn't notified of your responses to my comments.) – user21820 May 12 '16 at 02:03
  • Note my usage correction here. The notation $\mathcal L/\sim$ is typographically different from $\mathcal L/{\sim}$ for a reason that should be clear to anyone who understands why TeX was created in the first place. The former is coded as \mathcal L/\sim and the latter as \mathcal L/{\sim}. The former has an amount of space to the left of $\text{“}{\sim}\text{”}$ that is appropriate between a binary relation symbol, as in $a\sim b,$ and the expression to its left. But that spacing is wrong in this case because of the way in which the symbol is being used. – Michael Hardy Feb 15 '22 at 01:37
  • @WillieWong : $\qquad \uparrow \qquad$ N.B. $\qquad$ – Michael Hardy Feb 15 '22 at 01:39
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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.

KConrad
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    That is exactly one of the quotes I had in mind, but didn't want to add unnecessary side remarks to my question. – user21820 May 06 '16 at 13:23
  • 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. – user21820 May 06 '16 at 13:33
  • 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. – user21820 May 06 '16 at 13:40
  • (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. – user21820 May 06 '16 at 13:44
  • (4) did you see my comments about the computable reals? They address your point that fixed-precision reals have bad properties. – user21820 May 06 '16 at 14:08
  • Using computable reals sounds inconsistent with saying you only want to use vectors having length bounded away from 0 -- no matter what positive bound you choose, going down that route means your set of numbers is not a field (can't keep dividing by 2 arbitrarily many times). – KConrad May 06 '16 at 19:12
  • I think it's pretty clear from the varied responses so far that the answer to your question is (3). Are you not convinced of that yet? – KConrad May 06 '16 at 19:15
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    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? – KConrad May 06 '16 at 19:21
  • (a) You misunderstood my definition of approximation. If you refer to my toy example of the vector-space $\mathbb{R}^3$, I didn't restrict the domain to reals bounded away from $0$, but rather I modify all theorems to approximation versions. You can still keep dividing by $2$, but the approximation means that you expect more deviation for small scalars and vectors. – user21820 May 07 '16 at 03:04
  • (b) Also, approximation is perfectly compatible with computable reals; we can just replace all quantification over $\mathbb{R}$ with quantification over the computable reals. We can't empirically distinguish this structure from the original based on first-order properties alone, because both are real closed fields. So we need second-order empirically testable sentences. Do you have one? As of now, I'm convinced that my structure defined for computable reals is definitely physically meaningful, but I believe my original structure is too, so I believe the correct answer is (2) not (3). Honestly. – user21820 May 07 '16 at 03:05
  • (c) I still think your answer has not addressed my question in the sense of my comments 2 to 5. Let me try to rephrase my main reason: Let's say $S$ is a mathematical structure actually underlying the physical world. It may be possible to extend one or more domains of $S$ (and the interpretations of functions and predicates as well) such that the new structure has the same theory, but then this extension is almost never physically meaningful, since it has 'non-physical' elements, unless you believe that the world is fractal and has inner models... hahaha! [continued] – user21820 May 07 '16 at 03:16
  • So in this question I'm not looking an elegant mathematical structure but one that has no 'non-physical' elements (in some interpretation). Eric Towers's comment also says that quantum fields seem to be continuous-valued, but unfortunately I know almost nothing about quantum field theory. If the world truly has discontinuous spacetime, then you would be right that the answer to my question is (3). – user21820 May 07 '16 at 03:24
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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...

orbitals

Gerald Edgar
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    Why would you say that quantum mechanics has no physical meaning? – Nik Weaver May 06 '16 at 14:54
  • 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? – user21820 May 06 '16 at 15:20
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    Questions on QM belong in http://physics.stackexchange.com not here. – Gerald Edgar May 06 '16 at 15:22
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    @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 – Steve Huntsman May 06 '16 at 17:50
  • @SteveHuntsman: I agree that Lebesgue is practically very useful, but my question really is about whether or not it is "physically essential". – user21820 May 07 '16 at 03:38
  • By the way, I'm pretty sure a physicist would say that electron orbitals (in general even) are Jordan measurable! – user21820 May 07 '16 at 05:45
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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.

Timothy Chow
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    Of course ZF is not consistent with "all sets are Jordan-measurable!" The set of rational points is not Jordan-measurable. – Monroe Eskew May 06 '16 at 17:48
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    @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. – user21820 May 07 '16 at 03:48
  • I think RCA0 is weaker than what is necessary to describe physical reality. If we replace $\mathbb{R}$ in my structure with the computable reals, we get a structure with a different second-order theory but I'm sure its theory still interprets second-order arithmetic, far outstripping those subsystems in reverse mathematics. So using your idea, do you have any explicit theorem of second-order arithmetic or my structure (whether using computable reals or otherwise) that you believe is not physically meaningful? – user21820 May 07 '16 at 04:09
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    @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? – Timothy Chow May 08 '16 at 18:35
  • I definitely welcome your probing! The main difficulty in answering my question comes from the incredibly varied aspects, one of which is the boundary of computability, which appears if one is committed only to first-order arithmetic. I could ask essentially the same question as you. If we accept PA shouldn't we accept Con(PA)? Con(PA+Con(PA))? All the way? Con( the union of the sequence )? We could iterate through all the computable ordinals, and still it won't prove its own consistency! The theory of reality of course affirms its own consistency, so it must have uncomputable proof validity. – user21820 May 09 '16 at 09:36
  • To clarify, the collection of halting programs is a well-defined notion but may not have a physical meaning if you embed it into say lattice points in space. We're so used to embeddings meaning "essentially the same" that we easily assume that physical meaningfulness is preserved under embeddings. It isn't, unless the ambient structure supporting our embedding is itself physically meaningful! For example, the physical meaningfulness of the collection of rational numbers encoded as strings is witnessed by the program that accepts it, but not as a subset of the ordered reals or so I think. – user21820 May 09 '16 at 09:48
  • Oh and when I said "theory of reality", of course this is assuming there is a (higher-order) theory of reality. If there isn't, my question is moot, but so are all scientific experiments (based on reproducible empirical verification of boolean assertions) and all applied mathematics and even all analytical philosophy, and it is like denying the existence of the internet whose technological underpinnings is a huge mountain based on this assumption. So I readily admit the assumption, but see nothing wrong with it. =) – user21820 May 09 '16 at 09:56
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    @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. – Timothy Chow May 09 '16 at 15:10
  • Okay. In my opinion Brouwer's fixed-point theorem is empirically true because even if the fixed-point is not computable, the theorem can be verified because at the fixed-point, no matter what precision we have, the continuous function's output is indistinguishable from its input. The fixed-point itself can be considered to be given by an oracle. And even better still, for any given nonzero precision we can deterministically find a fixed point (at that precision). I guess you'll then say that I shouldn't restrict to computable reals... let me think about that some more... – user21820 May 09 '16 at 16:23
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    @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. – Timothy Chow May 09 '16 at 20:33
  • My proposal is definitely okay for standard reals. But your point about a continuous function on the computable points in the unit square that has no computable fixed point again touches at the distinction between different embeddings. Is such a function physically meaningful? As a program it is, but as a physical entity that acts on some unit square in space I don't think so. Since the structure in my question does not include functions, I don't think my proposal is so easily dismantled. In any case I do think the standard structure $\mathbb{R}$ is physically meaningful. – user21820 May 10 '16 at 08:53
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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.

  • Thanks for this! However, the most popular answer to the question about the role of completeness in Hilbert spaces (http://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! – user21820 May 07 '16 at 04:24
  • 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. – Pedro Lauridsen Ribeiro May 07 '16 at 04:51
  • 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. – user21820 May 07 '16 at 05:01
  • 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. – Pedro Lauridsen Ribeiro May 07 '16 at 05:01
  • 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. – user21820 May 07 '16 at 05:27
  • I think that, intuitively, many physicists would prefer to use constructive analysis on such grounds, as I remarked in the last paragraph of my answer, despite the subject being (still?) sort of a straitjacket... I recall that constructive analysts would even avoid using measure theory at all, giving preference to other generalizations of the Riemann integral such as the Denjoy-Perron or the Kurzweil-Henstock integrals. The space of KH-integrable functions is not complete but is ultrabornological, which is enough for the Banach-Steinhaus, the open mapping and the closed graph theorems to hold. – Pedro Lauridsen Ribeiro May 07 '16 at 15:52
  • So you're saying that although the KH integral is more general than the Lebesgue integral for functions in general, it is more constructive and hence can be used in a constructive setting on constructively obtained functions, unlike the Lebesgue integral? What happens if the gauge function is required to be Riemann integrable? Perhaps it might be worth including in your answer, especially if you've any citations of physicists' explanation of preference for constructive analysis and constructive alternatives to the Lebesgue measure. – user21820 May 08 '16 at 02:05
  • Indeed, the KH integral is known to be just as constructive as the Riemann integral, provided one defines gauges directly as nonvoid set-valued functions. If one restricts oneself to lower semicontinuous gauges, the KH integral reduces to the Riemann integral. I don't know what happens if one requires this only almost everywhere (particularly, if one requires gauges to be Riemann-integrable). On the other hand, Pfeffer showed in 1988 that one may restrict to gauges which are upper semicontinuous almost everywhere in order to recover the KH integral. – Pedro Lauridsen Ribeiro May 09 '16 at 00:45
  • As for my general impression on physicists' potential preference for constructive analysis, I should qualify it a bit. The quote from "Sure You're Joking, Mr. Feynman" in the math.SE question I've cited in the last paragraph of my answer can be read as expressing Feynman's informal distaste on physical grounds for "infinitary" mathematical procedures such as the Banach-Tarski paradox. Constructive analysis in the strict sense (e.g. using intuitionistic logic) is mostly unknown in the physics community, apart from a few researchers trying to reformulate parts of (quantum) physics using topoi. – Pedro Lauridsen Ribeiro May 09 '16 at 01:06
  • I see. I'm not so cynical as Feynmann is portrayed, because just because we don't have arbitrary precision in practice does not imply anything about the underlying structure, which could be arbitrarily precise but not fully accessible to us. Also, by constructive analysis I didn't mean all the way to intuitionistic logic. About the KH integral, what you say suggests that perhaps, under suitable definitions, the constructive fragment of the theory of KH integrable functions is none other than that of Riemann integrable functions, which if shown would be a strongly justified option (2). – user21820 May 09 '16 at 09:26
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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.

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    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. – user21820 May 07 '16 at 05:43
  • @user21820- does a chaotic attractor in classical mechanics physically exist? I'm pretty sure this breaks before getting down to the Planck scale – Steve Huntsman May 07 '16 at 14:44
  • 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. – user21820 May 07 '16 at 15:14
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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.

yuggib
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  • 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! – user21820 May 08 '16 at 08:50
  • 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). – user21820 May 08 '16 at 08:52
  • 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. – yuggib May 08 '16 at 10:35
  • 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! – user21820 May 08 '16 at 10:42
  • 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. – yuggib May 08 '16 at 11:21
  • I honestly don't think you understood my definitions. The way I defined approximation is not at all what you're saying here; it must be correct under some asymptotic condition. Please read carefully my toy example of the vector space and look at the precise formal sentence that the approximation should be like; it is not approximation of objects at all. Otherwise just having each rational being physically meaningful would automatically imply that each real is physically meaningful, which is not the case. Terry Tao's answer for example distinguishes between approximations and their limit. – user21820 May 08 '16 at 11:38
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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?

john mangual
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