Why does quantum theory use the Lebesgue integral?

In quantum mechanics, one comes across the formula for the expectation value $\langle \hat{M} \rangle$ of a Hermitian operator $\hat{M}$:

$$\langle \hat{M} \rangle = \langle \psi | \hat{M} | \psi \rangle = \int_{-\infty}^\infty \psi^* \hat{M} \psi \,d^3r$$

And it turns out that the integral in this formula is not the Riemann integral, but the Lebesgue integral. Why is this so? Why do we use Lebesgue integration instead of the familiar Riemann integration?

The standard answer is that Lebesgue integration is a generalisation of Riemann integration. Indeed, if $f$ is a Riemann-integrable function¹ then $f$ is also Lebesgue-integrable, and the values of the two integrals coincide:

$$\int_{R} f = \int_{L} f$$

(This is a slight abuse of notation since the subscripts have nothing to do with the domain of integration: here, $\int_{R}$ denotes the Riemann integral and $\int_{L}$ denotes the Lebesgue integral).

Additionally, there are certain functions² that are Lebesgue integrable but not Riemann integrable. The standard example is the function $f:[0,1]\to \mathbb{R}$ that is $1$ at every rational number and $0$ at every irrational number:

$$\qquad f(x)=\chi_{\mathbb{Q}}=\begin{cases}1, \text{if }x\in \mathbb{Q}\\0, \text{if }x\notin \mathbb{Q} \end{cases}$$

Thus, Lebesgue integration generalises Riemann integration.

But when we look under the hood, we realise that the kinds of functions that are Lebesgue integrable but not Riemann integrable (such as the characteristic function of $\mathbb{Q}$ that we saw above) are rather contrived functions, which we would not expect to describe any physical system.

So a more general method of integration seems like a useful thing to want... but if the only extra functions the new method allows you to integrate are weirdly contrived functions, then why bother? One could see why mathematicians would like such a general method (after all, mathematicians do like studying even weirdly contrived functions), but why do physicists use it to model quantum theory?

Perhaps the generalisation argument is not the whole picture...³

Indeed, there are other reasons why we use the Lebesgue integral in quantum theory.

$L^p$ spaces are complete under Lebesgue integral but not Riemann

Lebesgue integral gives weaker convergence criteria (Riemann needs uniform convergence)

Footnotes

1. There is a theorem that a function $f$ is Riemann integrable if and only if it is bounded and its set of discontinuity points has zero Lebesgue measure. ↩

2. ↩

3. There is a wonderful paper by Andrew D. Lewis titled "Should we fly in the Lebesgue-designed airplane?—The correct defence of the Lebesgue integral", which explores what is missing in the generalisation argument, and presents other, better arguments.(https://arxiv.org/pdf/2309.08908) ↩