# Is the notion of Lebesgue Measure a necessary construct for statistical physics? [duplicate]

In chat last night a user and I were discussing the "physical" meaningfulness of the notion of lebesgue measure. In particular, we were curious as to whether physicists can "make do" without it. I mentioned that the dominated convergence theorem is needed to prove certain theorems in statistics that would be needed in areas like statistical thermodynamics, where you want to know that when dealing with a huge quantity of particles things like velocity/energy are approximately normally distributed (Central Limit Theorem). We were then surprised to find a proof the CLT that no only was free of the DCT, but formulated entirely in terms of the Riemann Integral.

My question is: Are there any specific areas in physics that rely on the notion of the lebesgue measure? (either directly or indirectly via theorems for which this notion is needed to prove). To the point of being necessary and not merely useful?

• Lebesgue integration is used in basically all areas of physics. Example: the $L^2(\mathbb{R}^3)$ Hilbert space in QM should be complete. Possible duplicate: When is Lebesgue integration useful over Riemann integration in physics?. – Qmechanic Jan 18 '18 at 18:49
• Most of statistical mechanics is not as mathematically rigorous as you may think... – valerio Jan 18 '18 at 18:50
• The distinction between Lebesgue and Riemann integration almost never matters in physics in the sense that we usually treat calculus as an algebraic system rather than actually caring about epsilons and deltas. For example, you can just prove that $\int x^n = (1/(n+1))x^{n+1}$ in whatever sense of integration you care about and then derive all other integrals from that (using e.g. Taylor series) without worrying whether or not the order of the limits of the summation and the integration can be switched. – DanielSank Jan 18 '18 at 19:09
• That said, there are important cases in physics which switching the order of limits does cause problems. These cases are essentially those where we have a system with a continuum of normal modes, such as a wave supporting medium of infinite extent. If the system has no damping, then attempts to compute a Green's function lead to divergences. One then must introduce damping and take the limit as the damping goes to zero after computing the integrals. From a mathematical point of view, this is a way to regularize the integral so that limits can be switched, and I think it requires Lebesgue. – DanielSank Jan 18 '18 at 19:13
• @DanielSank: Concerning your comment about damping, I am quite confident (but don't know enough about this to say for sure) that any explicit limiting integral arising in physics can be proven without Lebesgue integration by simply finding hard bounds and then applying the generalized squeeze theorem. This is based on my own experience in logic and measure theory. Also, the divergence before 'regularization' is a problem with the physical model, and so is a separate matter from the integral. – user21820 Jan 19 '18 at 5:26

Unpopular opinion here: no, you don't "need" the Lebesgue measure to do physics. You don't need any kind of functional analysis, nor any distribution theory, nor any math beyond what a high schooler knows.

None of these things are essential to describing what Nature does; the content of the postulates of quantum mechanics, or special relativity, or statistical mechanics are physical, not mathematical. It is true that you can add in fancy mathematical objects to make the postulates look nicer, as was often done decades or centuries after the fact. But that's a different thing entirely from the core task of finding out how Nature behaves.

Consider a single particle with the Hilbert space $L^2(\mathbb{R}^3)$. Absolutely nothing experimentally visible changes if I put in a momentum-space cutoff, e.g. a discrete lattice that the particle hops on, say at the Planck scale. Neither does anything change if I put the particle in a large but finite box, say, the size of the observable universe. But now we're in a finite-dimensional vector space and there's no need for fancy math.

We still need calculus, but even this can be removed; just discretize time, performing timesteps like every numerical simulation ever written. Then you're down to just elementary arithmetic. We've lost all the math, but we still have quantum mechanics, because the physical postulates of superposition, unitary evolution, the Born rule, etc. are all still intact. Similarly, in statistical mechanics, the thermodynamic limit $N \to \infty$ doesn't exist; you're always dealing with a perfectly finite number of particles, reducing the situation to classical mechanics. This setup works even in relativistic quantum field theory; it is regularly used in lattice QCD, where it produces results that can't be found any other way.

I don't disagree that mathematical tools can be elegant and useful, but I strongly object to conflating them with the physics itself. We know how Nature works once we find the right set of laws and show they agree with experiment -- not when we write them with full mathematical rigor.

• I agree with much of this post, but "To my knowledge, an after-the-fact mathematical refinement of a physical postulate has never led to new knowledge about how Nature works." seems unreasonable. We can come up with the basic idea of quantum field theory, but it's only by computing various scattering cross-sections that we're actually able to come up with something experimentally testable! I think experimental confirmation is a important aspect of "new knowledge about how Nature works." Further more, it requires math to come up with stuff like the Higgs and propose how to even look for it. – DanielSank Jan 18 '18 at 21:27
• @DanielSank True! But I count computing scattering cross sections as good ol' physics, not math being added on top; it's just cranking out the predictions of a new physical theory in the most direct way possible. The additional math structure I mean is stuff like algebraic quantum field theory or turning everything into a fibre bundle. (i.e. the difference between 'let's try describing nature with $\phi^4$ theory' and 'let's rigorously construct the path integral for $\phi^4$ theory'. Of course this whole post betrays my bias as a phenomenologist.) – knzhou Jan 18 '18 at 21:40
• @DanielSank You're right. I conflated two things: when math is useful and when math is necessary. Most of my post argues almost no math is necessary and I stand by that. The last sentence makes the additional claim that additional math beyond what is useful for calculations does not tend to yield new physical laws, but that's a different claim entirely. – knzhou Jan 18 '18 at 22:03
• "The last sentence makes the additional claim that additional math beyond what is useful for calculations does not tend to yield new physical laws, but that's a different claim entirely. " Indeed, and I believe it to be incorrect. A good mathematical packaging of our knowledge is incredibly useful for gaining non-mathematical intuition! Math and physics help each other. I think it's foolish to argue that one has a meaningful existence without the other. – DanielSank Jan 18 '18 at 22:09
• @DanielSank Frederic Schuller makes the point that the inability to cast an idea about the physical world into rigorous mathematical form tends to be an indication that the idea (and by extension, its implications) is not fully understood. In that sense, nit-picky mathematical physics provides a measure of how well we actually understand what it is that we're trying to say. – J. Murray Jan 19 '18 at 1:33

edit I have edited the answer to deal with some of the criticism in the comments

To the extent that the Lebesgue measure is needed to define Lebesgue integration, it is central to Quantum Mechanics: in general we require wavefunctions, as a function of position, to be Lebesgue square integrable.

More specifically, in QM states respond to rays in a Hilbert space. Hilbert spaces are complete inner product spaces, and the Lebesgue integral is required to complete the relevant Hilbert space, see When is Lebesgue integration useful over Riemann integration in physics?

It is true that there is nothing special about the position basis, but this requirement cannot be escaped: the Lebesgue measure is necessary to define an appropriately square normalisable wavefunction in the position basis.

• Can you clarify why it is necessary for Quantum Mechanics to use Lebesgue integration as oppose to Riemann Integral? – stochastic Jan 18 '18 at 19:12
• @stochastic it's not. People saying you need Lebesgue to do quantum mechanics are thinking of wave functions in the position basis in a domain of infinite extent. First of all, quantum mechanics is not even usually done with position space wave functions. Second, the idea of an infinitely extended domain is unphysical. Third, even in an infinite domain, you don't actually need Lebesgue integration to calculate stuff. – DanielSank Jan 18 '18 at 19:18
• I'd agree that it's essential that Hilbert spaces are complete and the Lesbegue integral is important there. – Mozibur Ullah Jan 18 '18 at 20:32
• Is the spectral theorem the end goal for completeness here? – David Reed Jan 18 '18 at 20:40
• @DavidReed I am not sure what the deep reason is for it. I think it may have something to do with the fact that observables correspond to Hermitian operators in a $C^*$ algebra: if a sequence of operators in the algebra converge to an operator $A$ we want $A$ to also be in that algebra. – Martin C. Jan 18 '18 at 20:56

The integral was first defined by Newton as part of his method of Fluxions, ie the calculus.

Riemann put the integral on a rigorous footing using the methods of real analysis. It was quickly understood that it didn't have many good properties under taking limits.

(To note how useful this property is one might recall one of the stories that Feynman related in his books where he would swap move the derivative operator past the integral sign).

The definition that was finally adopted was the Lesbegue integral which did - for example it had the monotone and dominated convergence theorems and the Fubini theorem on swapping limits in a double integral. Also it was capable of being defined on abstract spaces, for example groups, and in fact, locally compact groups have an avatar of the Lesbegue measure, called the Haar measure, which is the unique translation invariant measure. Finally, square integrable functions formed a complete inner product space, aka a Hilbert space.

In so far as physics is or cares to be formalised (which isn't always the case, for example one only need note the Dirac delta function or the path integral) it has proven useful to define appropriate function spaces and so on.

• @David Reed: You're welcome. Yes, I know that. But when it was first used it wasn't defined rigorously and that is what I'm alluding to. Furthermore, even after it had been defined formally, one does not need to know its formalised definition to use it so long as one obtains a practical understanding of how it can be used. This is how the Path integral is still used, practically, rather than rigorously. – Mozibur Ullah Jan 18 '18 at 20:41