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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.

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5 Answers 5

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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.

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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

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).)

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+1. Thanks for the answer; this is a very interesting way of looking at it. –  Daenerys Naharis Mar 22 '11 at 1:33
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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 5:21
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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 5:23
    
@Matt: Dear Matt, that's certainly true; thanks for pointing it out! –  Akhil Mathew Mar 22 '11 at 13:50
    
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 7:47

This is more of a comment than an answer, but it is too long to be a comment. Take my opinion with a grain of salt.

You ask what properties make the gauge integral "less useful" than the others. This is really several questions: "useful" in terms of pedagogy? in terms of how its ideas generalize? in terms of the theorems it provides in the context it was designed for (I think, standard calculus in one variable or in $\mathbb{R}^n$)? And "usefulness" is such a fuzzy and contentious notion... but we don't need to go there.

In the context for which it was designed, is certainly less well known, less widely documented in textbooks, and certainly less taught than the Riemann or Lebesgue integrals, and it seems to me that your "usefulness" question is really about this. You're guessing, quite reasonably, that there must be some mathematical or pedagogical reason why Calculus I/II/III or Analysis I/II/III at University X is almost guaranteed to study another integral and not the gauge integral.

Well, I don't think there is any logical reason for this.

But there isn't a pressing need for the gauge integral, either. Its benefits, whether technical (in getting theorems with fewer hypotheses) or pedagogical (some feel it is easier to learn or to teach), do not seem to be enough to outweigh historical tradition. And it is really just a question of tradition.

It's a bit like asking why the USA doesn't use the metric system. In each case, there isn't any abstract reason why people don't do it, and if history could be replayed with different initial conditions, things might have been done differently.

The strength of tradition would be more of a surprise if the "technical advantages" of the gauge integral were more pronounced in relation to what is already in wide use. A huge factor in the adoption of the Lebesgue integral was all of its nice properties (e.g. completeness of the $L^p$ spaces, the dominated convergence theorem) that made analysis involving integrals possible, and on a firm logical footing, in ways that it really wasn't before. Speaking informally, I think 99% of what humanity wants out of integrals was taken care of with these advances. Of course the next thing that comes along can't offer as dramatic a change, so nobody cares to switch.

The chief "added advantage" of the gauge integral (or at least the one most often mentioned) is its more general "fundamental theorem of calculus." IMHO this sounds much better than it actually is. I'm an analyst and every time I've ever used the FTC, theorems about older integrals were enough; there was simply no need for more.

Unless you feel "having the best FTC possible" is a key property an integral should have... but analysis is full of statements that are easy theorems under restrictive hypotheses, and harder theorems under more general hypotheses, and often there is an epsilon of stuff left over, not covered by the standard tricks, where the statement is still true. Statements about interchanging limiting operations (e.g. differentiation under the integral) are classic examples; the truth boundary is so often unknown--- or it is so difficult or unrewarding to formulate useful "if and only if" conditions under which the statement is a theorem, that nobody bothers to do it. From this view, it's not surprising that there is a better FTC than the Lebesgue FTC, and it's not surprising that there is not a movement to correct this "defect".

So that's my take: people don't use the gauge integral because it is our heritage to use other integrals. These other integrals do most of the same things, and the gauge integral isn't so much better that people switch. And once we know why most experts don't use it... we know why it's generally not taught to anybody. [The pedagogical case for or against any integral of your choosing is a completely separate issue.]

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My understanding is that the gauge integral, also known as the Henstock-Kurzweil integral, is actually not more general than the Lebesgue integral in the sense that it does not generalize to higher dimensions spaces more exotic than subsets of $\mathbb{R}^n$. The Lebesgue integral, on the other hand, generalizes to the integration of a measurable function with respect to a measure on a measure space, which is enormously general and useful.

Keep in mind that part of the reason we prefer the Lebesgue integral to the Riemann integral is not just that we can integrate more functions, it's that the theorems are nicer.

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Alexander contradicts you concerning generalizations to higher dimensions. –  Yuval Filmus Mar 21 '11 at 6:32
    
Fair enough. Corrected. –  Qiaochu Yuan Mar 21 '11 at 6:41
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I didn't say anything about whether dominated convergence holds for the Henstock-Kurzweil integral because I didn't know. All I said is that the ability to integrate more functions is not the only reason people prefer the Lebesgue integral to the Riemann integral. –  Qiaochu Yuan Mar 21 '11 at 15:27
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Ah, sorry. I just wanted to say: besides integrating more functions we should indeed care about theorems, and also about difficulty of proofs. If you want to integrate over R then HK integral wins. Fundamental thm of calculus (a limited form) is a difficult theorem in Lebesgue's theory (LT). In the HK theory it's trivial. Dominated convergence thm is easy in LT, and it's also easy in HK theory. Definition of the Lebesgue's measure is not easy, definition of the HK integral is very easy. I fully agree though that measure theory is much more important than the HK integral. (...) –  user8268 Mar 21 '11 at 20:04
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(...) Both because spaces other than $R^n$ and because of the unity of the subject: all "reasonable" measure spaces (of the same total measure) are isomorphic (e.g. Wiener measure is isomorphic to Lebesgue's on $[0,1]$). HK integral should not be a replacement of Lebesgue's theory. The best suggestion I know is, rather, to replace Riemann integral by HK integral. The theory is the same and theorems much better (often with easier proofs). Sorry for this long comment :) –  user8268 Mar 21 '11 at 20:11

The gauge integral is as general as Lebesgue integral. It can be defined over arbitrary spaces. In fact it is absolutely necessary when we simplify calculus. There are simple proofs of Hakes theorem the study of improper integrals need not be carried out separately Feynman path integrals can be more naturally and easily defined using gauge integration rather than lebesgue and it is also easier to use gauge integral for Weiner integration.

The question regarding space of gauge integrable functions not being useful are irrelevant as one can always use the space $L^1$ for absolutely convergent gauge integrals. It is the same $L^1$ space, but gauge integrals give you added benefits of having better fundamental theorem of calculus , and Stokes theorem is given simpler proof. Fubini’s theorem proof is simpler and more natural. Further though monotone convergence theorem has simpler prof in lebesgue the simplicity is deceiving as first one has to extend measure to sigma algebra. In Henstock automatically measure is generated. It is simply the human inertia of a generation.

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To improve the readability I recommend reading your answer through before you post it. –  AndreasS Nov 16 '12 at 18:00

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