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A recent answer reminded me of the gauge integral, which you can read about here.

It seems like the gauge integral is more general than the Lebesgue integral, e.g. if a function is Lebesgue integrable, it is gauge integrable. (EDIT - as Qiaochu Yuan points out, I should clarify this to mean that the set of Lebesgue integrable functions is a proper subset of gauge integrable functions.)

My question is this: What mathematical properties, if any, make the gauge integral (aka the Henstock–Kurzweil integral) less useful than the Lebesgue or Riemann integrals?

I have just a cursory overview of the properties that make Lebesgue integration more useful than Riemann in certain situations and vice versa. I was wondering if any corresponding overview could be given for the gauge integral, since I don't quite have the background to tackle textbooks or articles on the subject.

Martin Sleziak
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Chris Brooks
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5 Answers5

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In mathematics, there is a general philosophy (I think due to Grothendieck) that one should work not with a bad category containing nice objects, but with a nice category containing bad objects. (The category of schemes over a base scheme is perhaps one example, as is the category of presheaves on a category.) The statement applied to analysis is perhaps that working with a nice space is more important than the objects in it. For instance, the spaces one may obtain from the Lebesgue integral--namely, the $L^p$ spaces--are wonderful from the point of view of analysis; they are Banach spaces (Hilbert if $p=2$), and with very weak hypothesis have the duality property $(L^p)^* = L^q$ for $p, q$ conjugate exponents (and $p \neq \infty$). By contrast, the Henstock-Kurzweil integral does not lend itself to such nice spaces: to define a norm on some subspace of the space of integrable functions, you would presumably need to consider the integral of the absolute value. But functions $f$ such that $|f|$ is HK-integrable are in fact Lebesgue integrable! So no new information is gained. The fact that the Lebesgue integral can't handle the derivative of every differentiable function is made up for by the niceness of the resulting function spaces.

(If I remember correctly, one can make the space of all HK-integrable functions into a topological vector space, but it's not anywhere near as nice as the Banach spaces that one obtains via the Lebesgue integral.)

Another reason, which has already been given above, is that the Lebesgue integral is fantastically general. The fact that it can integrate functions on euclidean space is only a very limited and special case of its power; it will work on any measure space. To give one example, the Haar integral (one can obtain a translation-invariant measure on a locally compact group; the Haar measure is the Lebesgue integral with respect to this) is frequently used: for instance, in number theory, one wishes to integrate functions over topological groups such as $K^*$ for $K$ a local field, and the Haar integral is the natural way to do this. (Though, as Matt E observes in the comments, in practice one may integrate over $K^*$ or more generally algebraic groups over local fields, fairly explicitly, without need of actually constructing a measure formally (except in the case $K = \mathbb{R}, \mathbb{C}$, where the usual integral on euclidean space suffices).)

Akhil Mathew
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    Dear Akhil, This is a minor point, but for $K^{\times}$, or more generally $G(K)$ where $G$ is a linear algebraic group, one doesn't need any abstract theory to construct the integral: in the case when $K$ is $\mathbb R$ or $\mathbb C$, linear groups embed into Euclidean space in a very explicit way, so one quickly reduces to Lebesgue measure on Euclidean space, while for non-archimedean local fields, the Haar measure is essentially a counting measure. This is not to discount your general point, but just to say that Haar measure on local-field valued points of a linear algebraic group ... – Matt E Mar 22 '11 at 05:21
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    ... is not a case where much abstract theory is required. (Of course, the existence of this abstract theory is a source of comfort even in these more explicit settings, and provides moral justification for various considerations. And perhaps more importantly it suggests ways of thinking and points of view that are not as obvious from a more explicit perspective.) Regards, – Matt E Mar 22 '11 at 05:23
  • @Matt: Dear Matt, that's certainly true; thanks for pointing it out! – Akhil Mathew Mar 22 '11 at 13:50
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    Dear Akhil, Wikipedia tells me that $f$ is Lebesgue integrable if and only if both $f$ and $|f|$ are HK-integrable. – Damien L Mar 28 '13 at 07:47
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I would have written this as a comment, but by lack of reputation this has become an answer. Not long ago I've posed the same question to a group of analysts and they gave me more or less these answers:

1) The gauge integral is only defined for (subsets of) $\mathbb R^n$. It can easily be extended to manifolds but not to a more general class of spaces. It is therefore not of use in (general) harmonic analysis and other fields.

2) It lacks a lot of very nice properties the lebesgue integral has. For example $f \in \mathcal L^1 \Rightarrow |f| \in \mathcal L^1$ obviously has no generalization to gauge theory.

3) and probably most important. Afaik (also according to wikipedia) there is no known natural topology for the space of gauge integrable functions.

Alexander Thumm
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    And on top of that, the space of gauge integrable functions is not a Banach space. – Damien L Mar 20 '13 at 10:41
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  • is not exactly a drawback: it is one of the main reasons why the gauge integral exists, to drop the condition of absolute integrability (the main reason is not that, but is desirable to drop the necessity of being absolutely integrable by the same reason that conditionally convergent series exists)
    • – Masacroso Nov 30 '18 at 07:28