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Are Riemann integrals special cases of Haar integrals? Why do we need the invariant property under some actions of groups in the definition of Haar integrals? For example, if we have a group of real matrices $G$ and we have an inner product $\langle\>\ ,\ \rangle: G \times G \to \mathbb{R}$, for any $A, B$ in $G$, $C$ in the subgroup $O:=\{C\in G: C^{T}C=I\}$, $\langle A, B\rangle=\langle O^{T}AO,O^{T}BO\rangle$, then we can define an integral $\int_{G} f d\mu$ over $G$. Why do we need the property: $\langle A, B\rangle =\langle O^{T}AO,O^{T}BO\rangle$? Thank you very much.

Edit: I am having a course about random matrices. The course is related to the lecture notes: http://arxiv.org/abs/0801.1858. On page 2, equation (2.4), I think that the integral is a haar integral. I don't know why we need (2.5) in the paper. Thank you very much.

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For the differences between Riemann and Lebesgue integrals, see math.stackexchange.com/questions/7436/lebesgue-integral-basics/… –  Arturo Magidin Jan 11 '11 at 19:53
    
The Lebesgue measure on R is a special case of the Haar measure. The Riemann integral is constructed differently. –  Qiaochu Yuan Jan 11 '11 at 19:57
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Could you give a little context on what you mean by Haar integral? And possibly a reference? I am not an expert here, but my take on Haar measure is that it is a measure defined on a locally compact abelian group for which translation invariance is (essentially) its characteristic property. So at this level, my answer would be "We need the translation invariance because it's part of the definition." Obviously this is not very satisfactory, which is why I've asked for more information. –  Pete L. Clark Jan 11 '11 at 20:12
    
@Pete: In the locally compact setting there is no need to distinguish between Radon (= sufficiently regular) measures and positive linear functionals on the continuous functions with compact support, because there is a bijective correspondence between them (this is usually called the Riesz-Markov theorem). Moreover, Haar measure is defined on any locally compact group, no need for commutativity. –  t.b. Jan 11 '11 at 21:01
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I looked at the link and indeed, (2.4) describes Haar measure on the unitary group. The formula (2.5) is a consequence of (2.4). In fact, Haar measure on a compact group is automatically invariant under both left and right translations. See also Molinari's notes: wwwteor.mi.infn.it/~molinari/NOTES/haar.pdf –  t.b. Jan 12 '11 at 3:44

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up vote 8 down vote accepted

Lebesgue integration, in the sense of integration with respect to Lebesgue measure on $\mathbb{R}^n$, is a special case of Haar integration, because Lebesgue measure is the Haar measure on the abelian group $\mathbb{R}^n$ with addition and the usual topology (normalized so that the measure of a unit $n$-cube is $1$). Riemann integration is closely related to Lebesgue integration, but a fundamental difference is that the Riemann integral is not directly related to a measure, and as a result there are far fewer general results applicable. For example, a pointwise limit of uniformly bounded Riemann integrable functions on a bounded interval need not be Riemann integrable, even if the convergence is monotone.

Lebesgue integration can be seen as a completion of Riemann integration. If $(R)\int f$ denotes the Riemann integral of a continuous function with compact support on $\mathbb{R}^n$, then $f\mapsto(R)\int f$ is a positive linear functional on $C_c(\mathbb{R}^n)$, and the Riesz representation theorem yields a positive Borel measure $\mu$ on $\mathbb{R}^n$ such that $(R)\int f=\int fd\mu$ for all $f\in C_c(\mathbb{R}^n)$. The completion of $\mu$ is Lebesgue measure. For (proper) Riemann integrable functions, the Riemann and the Lebesgue integral are the same, but there are many functions that are "nice" from the measure theoretic perspective but not Riemann integrable, such as the characteristic function of the rationals on $\mathbb{R}$. On the other hand, there are functions whose improper Riemann integral exists but whose Lebesgue integral does not exist due to the positive and negative parts not being integrable, such as $\frac{\sin(x)}{x}$ on $\mathbb{R}$. In such cases, you could define an "improper" Lebesgue integral by taking limits as the domain increases to obtain the same result as with improper Riemann integrals, so there is no real loss in only considering Lebesgue integrals.

I recommend reading pages 195-197 of Körner's A companion to analysis for a nice discussion of the motivation of going beyond Riemann integration.

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You can do Lebesgue integration on domains other than $\mathbb{R}^n$ with the Lebesgue measure, and not every measure is a Haar measure, surely. So I assume your first sentence is meant to refer exclusively to Lebesgue integrals with the Lebesgue measure on $\mathbb{R}^n$, or close variants thereof? –  Arturo Magidin Jan 11 '11 at 20:16
    
@Arturo: Yes, by Lebesgue integration I meant integration with respect to Lebesgue measure on $\mathbb{R}^n$. Thank you for pointing out the lack of clarity, and I will edit to fix it. –  Jonas Meyer Jan 11 '11 at 20:18
    
I'll note that this is intended to partially answer the title question. Like Pete L. Clark, I am not sure exactly what is being asked about Haar integrals in the body of the question. –  Jonas Meyer Jan 11 '11 at 22:32
    
thank you very much. –  user Jan 11 '11 at 23:32

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