# Why does Lebesgue measure undergird wavefunctions?

My understanding (and correct me if I am wrong, because I am not a physicist) is that wavefunctions are all normalized members of the Hilbert space of square-integrable functions over some subset of $\mathbb{R}^n$, where the integration is typically done with respect to Lebesgue measure (though in some contexts may be done with respect to a counting measure or so forth). Given that Lebesgue measure more or less corresponds to classical notions of length, area, volume, etc., on what grounds is it used rather than some other Borel measure, or even some non-Borel measure?

• What is different about the notions of length, area, and volume in quantum mechanics than in classical mechanics? – Daddy Kropotkin Sep 17 '18 at 18:26
• None of this measure-theoretic stuff ever has any real-world consequences. It's just a mathematical convenience. – user4552 Sep 17 '18 at 18:52
• Possible duplicates: physics.stackexchange.com/q/41719/2451 , physics.stackexchange.com/q/38761/2451 and links therein. – Qmechanic Sep 17 '18 at 19:12
• @user4552 Way to miss the pitch. Our mathematical descriptions of the real world can't be justified formally without this measure-theoretic stuff. Having physical descriptions of the universe that you can't prove to be consistent seems pretty inconvenient. – Andrew Dec 26 '20 at 22:12

The relevance of the Lebesgue measure arises from the fact that as soon as you have a pair of abstract selfadjoint operators satisfying standard canonical commutation relations of position and momentum (with some further technical requirement like irreducibility), the Hilbert space turns out to be isomorphic to $L^2(R, dx)$ where $dx$ is, in fact, the Lebesgue measure over the real line. Here, the position operator and the momentum one take the standard form. This way, given a normalized vector $\psi$, $|\psi(x)|^2 dx$ has the properties of a probability measure over the spectrum (the attained values) of the position operator.
All relevant measures used to represent $L^2$ spaces as the quantum space are, in fact, Borel. This is because they are defined over the spectrum or the joint spectrum (thus subsets of $R^n$) of commuting observables and the measure function is construted out of measures associated to vectors and the PVMs of those observables. These measures are Borel.
• But every separable Hilbert space is isometrically isomorphic to $L^2(\mathbb{R},dx)$. So you're saying that it's special because the position and momentum operators take a desirable form there? – Jeh Sep 17 '18 at 18:17