# When is Lebesque integration useful over Riemann integration in physics?

Riemann integration is fine for physics in general because the functions dealt with tend to be differentiable and well behaved. Despite this, it's possible that Lebesque integration can be more powerfully used even in physical situations that can be solved by Riemann integration. So my questions is:

In solving physics problems, when is Lebesque integration useful over Riemann integration?

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## 3 Answers

An important example in quantum mechanics is e.g. the Hilbert space

$$H~=~L^2(\mathbb{R}^3)$$

of Lebesgue square integrable wave functions $\psi$ in the position space $\mathbb{R}^3$. The Lebesgue square integrable functions (as opposed to just the Riemann square integrable functions) are needed to complete the Hilbert space with respect to the square norm

$$||\psi||_2~:=~\sqrt{\int d^3x ~ |\psi(x)|^2}.$$

Concerning completeness, see also this Phys.SE post.

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For physicists a little rusty on their analysis and who find that wiki link dense, I'll add: "completeness" means that given a sequence of vectors (square-integrable functions from $\mathbb{R}^3$ to $\mathbb{C}$ in our case) getting arbitrarily "close" to one another, there exists a limit they are approaching, and that limit is itself in our space. Thus we can write infinite sums and know we're talking about something well-defined. – Chris White Oct 1 '12 at 4:44

In theory: Lebesgue integrable functions form a Banach space, whereas Riemann integrable functions do not. This causes problems in, e.g., Quantum mechanics if we try to work with Riemann integrable functions instead of Lebesgue integrable functions...

(We want the functions to form a Banach space so we can use good old fashioned linear algebra to solve problems!)

When does it matter? Take $$\chi_{\mathbb{Q}}(x)=\begin{cases}1 & x\in\mathbb{Q}\\ 0&\mathrm{otherwise}\end{cases}$$ It is Lebesgue integrable but not Riemann integrable.

Riemann integration cannot happen over infinite intervals, e.g., $\int^{\infty}_{0}f(x)\,\mathrm{d}x$ is illegal for Riemann integration.

On the other hand, $\mathrm{sinc}(x)$ is Riemann integrable but not Lebesgue integrable.

These are all measure theory statements that are irrelevant to physical explanations, but mathematicians are conscious of. Consequently...

In Practice: For everyday physics, the "symbolic integration" you learn in calculus is perfectly fine. Even when physicists say "We work with $L^{2}(X)$..." we think about integration in the "symbolic manner".

It's only if you work on making the path integral rigorous that you need to be careful about "What you mean with integration...".

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Wouldn't $\int_0^\infty f(x)\mathrm{d}x$ be definable as an improper Riemann integral, though, i.e. $\lim_{a\to \infty}\int_0^a f(x)\mathrm{d}x$, for a reasonable function? – David Zaslavsky♦ Oct 1 '12 at 0:29
For "most reasonable functions" (e.g., impose continuity or something), that's true. So to answer your question, (a) for measure theorists, no; (b) for physicists, yes. It's just...I answered as a measure theorist since it's a measure theoretic question :S – Alex Nelson Oct 1 '12 at 0:33

I would like to point out a completely practical example here, too. You can also need to change the order of integration and summation, or integration and derivative is some calculations, i.e., $\int dx\sum_n \to \sum_n \int dx$, or $\int dx\, \partial/\partial t \to \partial/\partial t \int dx$. While this is a problem with Riemann integration, it works for the Lebesgue integral, under certain assumptions, which are, in physical systems, usually fulfilled.

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