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I've mentioned before that I'm using this forum to expand my knowledge on things I know very little about. I've learnt integrals like everyone else: there is the Riemann integral, then the Lebesgue integral, and then we switch framework to manifolds, and we have that trick of using partitions of unity to define integrals.

This all seems very ad hoc, however. Not natural. I'm aware this is a pretty trivial question for a lot of you (which is why I'm asking it!), but what is the "correct" natural definition we should think of when we think of integrals?

I know there's some relation to a perfect pairing of homology and cohomology, somehow relating to Poincare duality (is that right?). And there's also something about chern classes? My geometry, as you can see, is pretty confused (being many years in my past).

If you can come up with a natural framework that doesn't have to do with the keywords I mentioned, that would also be very welcome.

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Do you know the Lebesgue integral just for R^n or for arbitrary measure spaces? If the former, then the latter might be what you're looking for. –  Ricky Demer Sep 12 '10 at 3:11
    
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I suppose you could say that an integral is a bilinear pairing between $k$-forms on a manifold and $k$-cycles that satisfies some axioms (it is a fun exercise to figure out what those axioms are), but I'm not convinced that that is a very enlightening thing to do. –  Andy Putman Sep 12 '10 at 4:49
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There is no single "natural" theory of integration (as far as I know!) They each have their strengths and weaknesses, depending on what you want them to do. For example, it would be wrong to say that the Lebesgue integral is "better" than the Riemann integral; there are definitely many natural problems where the simplicity of Riemann far outweighs the generality of Lebesgue (e.g. contour integration in Complex Analysis, integration of continuous Banach space-valued functions). So, I think you have to be more specific about what you want integrals for. –  Zen Harper Sep 12 '10 at 14:16
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@Theo: To define “conditionally convergent” Riemann integrals you have to perform one additional step, namely, you have to pass to a limit with respect to the length of the interval. You might as well perform this step after Lebesgue integration and obtain “conditionally convergent” Lebesgue integrals. –  Dmitri Pavlov Sep 13 '10 at 16:43

4 Answers 4

Here is my own favorite construction of the (Lebesgue) integral.

Suppose M is an arbitrary smooth manifold. Denote by Or(M) the orientation line bundle of M. This bundle is equipped with a canonical Riemannian metric. Vectors of length 1 in the fiber of Or(M) over a point p∈M correspond canonically to the two orientations of the tangent space at the point p. The manifold M is orientable if and only if the bundle Or(M) is trivializable. Choosing an orientation of M amounts to choosing an isometric trivialization of Or(M).

The bundle Or(M) together with its natural metric is flat. Hence we can twist the de Rham complex Ω^0(M)→⋯→Ω^n(M) by Or(M) and obtain the following twisted de Rham complex: Ω^0(M)⊗Or(M)→⋯→Ω^n(M)⊗Or(M). (Here by a complex I mean a complex of sheaves.) The line bundle Ω^n(M)⊗Or(M) is called the bundle of densities and is denoted by Dens(M). This bundle has a canonical orientation (hence it is trivializable), but does not have a canonical metric or a canonical trivialization.

The cohomology of the twisted de Rham complex (with compact support) is called the twisted de Rham cohomology (with compact support). We have a canonical map C^∞_cs(Dens(M))→H^n_cs(M,Or(M)). Here C^∞_cs is the space of global sections of a vector bundle with compact support and H^n_cs denotes the nth cohomology with compact support.

The Poincaré duality gives us a canonical isomorphism H^n_cs(M,Or(M))→H_0(M). Finally, the map from M to the point induces a map in homology H_0(M)→H_0(∙)=R.

The composition of maps C^∞_cs(Dens(M))→H^n_cs(M,Or(M))→H_0(M)→H_0(∙)=R gives us a map ∫: C^∞_cs(Dens(M))→R, which is the integration map. Note that the actual integration (over each connected component) happens in the first map. The second map is an isomorphism and the third map simply sums integrals over individual connected components.

The map f∈C^∞_cs(Dens(M))→∫|f|∈[0,∞) is a norm on C^∞_cs(Dens(M)). Completing C^∞_cs(Dens(M)) in this norm yields L_1(M)(=L^1(M)), which can be identified with the space of finite complex-valued measures on M.

The space of bounded measurable functions on M (=L_0(M)=L^∞(M)) can be constructed by completing C^∞(M) in the σ-weak topology induced by L_1(M). Other L_p spaces can be constructed in a similar way to L_1(M) by completing sections of the bundle of p-densities instead of 1-densities (=Dens(M)).

The development of the remainder of measure theory in this approach largely parallels the one explained in one of my previous answers.

I want to stress that these constructions do not rely on any existing integration theory. In fact, they can be used to build integration theory on smooth manifolds from scratch without ever referring to the usual measure theory with its lengthy and technical proofs.

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@Robin: Also, I do not agree with this “in favor of“ ideology. Proofs in measure theory can hardly be useful for anything else. On the other hand, Poincaré duality and de Rham theorem have independent value. –  Dmitri Pavlov Sep 12 '10 at 16:58
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Me too, Victor :-) I don't understand this anti-measure theory ideology either. I'm the first to want to avoid lengthy and technical proofs ('cos I can't understand them), but I prefer not to replace them with even more lengthy and even more technical proofs. :-) –  Robin Chapman Sep 12 '10 at 17:42
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An integration-free proof of Poincare duality can be found in section 3.1 here: math.uni-muenster.de/u/lueck/publ/lueck/ictp.pdf The idea is to first prove Poincare duality abstractly (i.e. for topological manifolds) and then use Hodge theory (which I think is integration-free) to bring differential forms into the picture. It may seem a little perverse to define integration via Hodge theory, but on the other hand surgery theory seems to tell us the Hodge theory is deeply relevant to smooth manifold theory (via the signature theorem). –  Paul Siegel Sep 12 '10 at 17:50
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What does this proof or strategy do in the case where M=(0,1) or [0,1] or the circle? –  Yemon Choi Sep 12 '10 at 22:02
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@Victor: Measure theory = integration theory. @Robin: There is no “anti-measure theory ideology” in my answer. I just point out that measure theory for smooth manifolds can be developed much more easily (modulo existing tools) than in the standard approach (modulo existing tools). –  Dmitri Pavlov Sep 13 '10 at 16:15

As the others have mentioned, integration over a connected oriented smooth manifold $M$ can be characterized (modulo some technicalities) according to the fact that it fits into an exact sequence:

$\Omega_c^{n-1}(M) \to \Omega_c^n(M) \to \mathbb{R} \to 0$

where $\Omega_c^*$ refers to the De Rham complex, the first arrow is the De Rham differential, and the second arrow is integration. In particular, $\int_M$ defines an isomorphism $H^n(M^n) \to \mathbb{R}$. This line of thinking leads to Poincare duality: if one takes a submanifold $V$ representing a dimension $p$ homology class of $M$ (this is always possible ratianally) then the integration isomorphism determines a dimension $p$ cohomology class $[\omega]$, and this can be paired with a cohomology class $[\eta]$ in dimension $n-p$ via the exterior product (cup product):

$([\omega],[\eta]) \mapsto \int_M \omega \wedge \eta$

Poincare duality is precisely the assertion that this pairing is nondegenerate. I think this point of view on integration - as a pairing between homology and cohomology - leads to many of the genuinely non-analytic formulations of integration.


Still, I would not consider this to have the final say as the correct natural definition of integration. For one thing, manifolds are far from the only spaces that one could want to integrate over, and I highly doubt integration has a cohomological formulation in any serious generality beyond manifolds. Second, the theory of measures truly is fundamental to integration and should be involved in an essential way; dynamical systems people often care as much about the measure as they do the integral.

Here is what I think integration is all about. Let $X$ be a locally compact Hausdorff space (this is enough generality for the vast majority of applications of integration that I know of) and consider the space $C_0(X)$ of continuous functions on $X$ vanishing at infinity (in the sense of the one-point compactification). Whatever intuition you have about integration, it must tell you that the integral should be a way of assigning a real number to a continuous function (maybe other functions too) which depends linearly on the function. In other words, it has to be some sort of linear functional on $C_0(X)$. Of course, not every linear functional deserves to be called an integral - if $x \in X$ then $f \mapsto f(x)$ is a linear functional on $C_0(X)$, but it doesn't make much sense to call it an "integral".

So we allow the topology of $X$ to play a greater role. Recall that $C_0(X)$ is a Banach space if it is equipped with the uniform norm, and as such it comes equipped with a preferred collection of linear functionals: the set $C_0(X)^*$ of continuous linear functionals. If we pretend for a moment that we have already worked really hard and built the theory of integration with respect to a Borel measure $\mu$, then assuming the measure is tied closely enough to the topology of $X$ (precisely, it must be a "Radon measure") we would have a continuous linear functional $I_\mu$ on $C_0(X)$ given by $I_\mu(f) = \int f d\mu$.

Riesz Representation Theorem: Let $M(X)$ denote the Banach space of Radon measures on $X$. The map $M(X) \mapsto C_0(X)^*$ given by $\mu \mapsto I_\mu$ is an isometric isomorphism.

Consequently, if we hadn't already invented a notion of integration it would be perfectly possible to simply define $\int_X f d\mu$ to be $I_\mu(f)$. I personally think this is the right way to think about integrals.

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One more thing. You might be interested in the following note: maths.gla.ac.uk/~tl/glasgowpssl/banach.pdf if you would like the Lebesgue integral to appear more natural (in more ways than one). While the punchline may seem profound at first, there isn't actually much serious content. –  Paul Siegel Sep 12 '10 at 17:37
    
I don't know much about the work of Jenny Harrison, but I think that she does have a theory of differential forms, etc., for quite non-smooth spaces. I could be wrong, though; if you (or another reader here) know(s) her work, maybe you can comment on it? –  Theo Johnson-Freyd Sep 12 '10 at 19:16
    
I'm at best marginally familiar with Harrison's work as well, but in my experience calculus over non-smooth spaces generally only works almost everywhere relative to some measure. In particular it requires additional assumptions (e.g. curvature bounds in the sense of Alexandrov geometry) and hard work to prove theorems about the topology of non-smooth spaces. Harrison's techniques are different from what I'm used to though, so I can't be sure. –  Paul Siegel Sep 12 '10 at 19:47

If you are interested in manifolds, then you might be interested in various related notions of measures and distributions. Let $M$ be a smooth manifold with algebra of functions $C^\infty(M)$. There is a very general notion of a distribution, which is that of any linear function $C^\infty(M) \to \mathbb R$. In this framework, a distribution is a measure if it satisfies a positivity condition, namely that it takes everywhere-nonnegative functions to nonnegative numbers.

Another, different definition of "distribution" corresponds to the distribution bundle on $M$, which is a canonical trivializable line bundle on $M$. It can be presented by gluing data and transition amplitudes as follows. Let $U,V \subseteq M$ with $\phi: U \to \mathbb R^n$ and $\psi: V \to \mathbb R^n$ be coordinate charts, and consider the trivial bundles one-dimensional bundles over $U,V$. We glue them together by giving transition data: if $f$ is a section over $U$ of the trivial bundle, on $U\cap V$ we identify it with the section $f \cdot \left| \det \frac{\partial \phi}{\partial \psi}\right|$ of the trivial bundle over $V$. (When $M$ is oriented, this bundle is the same as the determinant bundle $\wedge^{\operatorname{top}} {\rm T}^*M$; the determinant bundle is always a line bundle, and so its square is trivializable, and has a trivializable square root, which is the distribution line bundle whether $M$ is oriented or not.) Note that the transition functions preserve positivity of the sections, and so the notion of "positive distribution" and so on are well-defined.

Finally, if you are interested in a totally algebraic notion of integration for $\mathbb R^n$, you might be interested in the following observation, which in some form is older but nevertheless deserves to be called an observation of Berezin. Namely, the integral, as a linear map $C^\infty_{\operatorname{compact}}(\mathbb R^n) \to \mathbb R$, is uniquely defined up to scalar multiple by the fact that it vanishes on the images of $\frac{\partial}{\partial x_i} : C^\infty_{\operatorname{compact}}(\mathbb R^n) \to C^\infty_{\operatorname{compact}}(\mathbb R^n)$. Here $C^\infty_{\operatorname{compact}}(\mathbb R^n)$ is the algebra of smooth functions with compact support, and $x_1,\dots,x_n$ are the usual coordinate functions on $\mathbb R^n$. There are many situations in which by naming an algebra "of functions" and some "partial derivatives" you can uniquely (up to scalar) determine an "integral". An example is the algebra of de Rham differential forms on an oriented manifolds $M$, and the "partial derivatives" are the de Rham $d$ and the Lie derivatives for all vector fields on $M$. This uniquely picks out the integral that is zero on non-top forms and integrates top forms over $M$ as a canonical "measure" on the "space" whose "algebra of functions" is the differential forms. This is an example of a "superintegral", and it was to motivate a definition of superintegrals that Berezin made the above observation.

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This only allows us to integrate smooth functions with compact support, right? –  Harry Gindi Sep 12 '10 at 7:58
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I really like the observation of Berezin! Do you have a reference? –  Vivek Shende Sep 12 '10 at 11:26
    
@Harry: As I explain in my answer, you can easily complete the space of functions with compact support to the space of all integrable functions. Of course, not every smooth function with non-compact support is integrable. –  Dmitri Pavlov Sep 12 '10 at 15:06
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@Vivek: The observation of Berezin is also known as Poincaré duality. See my answer for details. –  Dmitri Pavlov Sep 12 '10 at 15:08
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One can also view Berezin's observation as an infinitesimal version of the observation that Lebesgue measure is the (unique up to constants) Haar measure on R^n. (If a measure is translation invariant, then it will annihilate derivatives with respect to infinitesimal translations, i.e. partial derivatives.) To me, this is one of the most natural definitions of Lebesgue measure, though it probably does not qualify as "non-analytic" in the sense you are looking for. –  Terry Tao Sep 12 '10 at 17:48

My answer here is realy just a footnote to Paul Siegel's excellent answer, but it has become too long to fit in a comment box. Integrals are siamese brothers to measures; leaving them out seems rather perverse to me. Anyway, here is how I think of integrals. The objective here is to tackle the "categorical" part; the analytical viewpoint will necessarily obtrude. But bear with me a little, this is a somewhat long post, with a punchline at the end.

Fix a Boolean algebra $\Omega$. A map $\nu: \Omega\to V$ with values on a linear space $V$ is finitely additive if $\nu(E\cup F)= \nu(E) + \nu(F)$ for every disjoint $E, F$. Denote the linear space of such maps by $\mathbf{A}(\Omega, V)$.

Theorem 1: There is a linear space $\mathbf{S}(\Omega)$ and a finitely additive map $\chi:\Omega\to \mathbf{S}(\Omega)$ universal among all finitely additive maps. proof: just follow the universal property and do the obvious thing (yeah, I suppose you can use the adjoint functor theorem but why would you?).

The universal property recast in terms of representability gives the natural isomorphism ($\mathbf{Vect}$ is the category of linear spaces)

$$\mathbf{A}(\Omega, V)\cong \mathbf{Vect}(\mathbf{S}(\Omega), V)$$

Before continuing, let me elucidate a little bit of the structure of $\mathbf{S}(\Omega)$.

Theorem 2: Let $f$ be a non-zero element of $\mathbf{S}(\Omega)$. Then there are non-zero scalars $k_n$ and non-zero, pairwise disjoint $E_n\in \Omega$ such that $f= \sum_n k_n\chi(E_n)$. Furthermore, if $\nu$ is a finitely additive map and $\widehat{\nu}$ the map induced on $\mathbf{S}(\Omega)$ by universality, then $\widehat{\nu}(f)= \sum_{n}k_n \nu(E_n)$.

To put it simply, $\mathbf{S}(\Omega)$ is the linear space of "simple functions on $\Omega$" and the map induced by universality is the integral. Now use theorem 2 to put a norm on $\mathbf{S}(\Omega)$:

$$\|\sum_n k_n\chi(E_n)\|= \max\{|k_n|\}$$

Denote the completion by $\mathbf{L}_{\infty}(\Omega)$. On the other hand, put on the linear subspace of the bounded finitely additive maps $\Omega\to B$ with $B$ a Banach space, the semivariation norm (which I am not going to define). Denote this space by $\mathbf{BA}(\Omega, B)$. Then:

Theorem 3: There is a bounded finitely additive $\chi:\Omega\to \mathbf{L}_{\infty}(\Omega)$ universal among all bounded finitely additive maps.

Once again, recasting the universal property in terms of representability, we have a natural isometric isomorphism ($\mathbf{Ban}$ is the category of Banach spaces and bounded linear maps).

$$\mathbf{BA}(\Omega, B)\cong \mathbf{Ban}(\mathbf{L}_{\infty}(\Omega), B)$$

It is illuminating to write down what does the naturality of the isomorphism implies: I will leave that as an exercise to the reader.

Note that $\mathbf{L}_{\infty}(\Omega)$ is a Banach algebra in a natural way (use theorem 2 or juggle the universal property around. Or "cheat" all the way up and use Stone duality) and that $\chi$ is spectral or multiplicative, that is, $\chi(E\cap F)= \chi(E)\chi(F)$. Theorem 3 can now be extended by saying that $\chi$ is universal among all spectral measures (with values in Banach algebras). This extension is trivial given theorem 3.

The case of $\mathbf{L}_{\infty}(\Omega)$ does not need the introduction of measures but of course, this is not so with $\mathbf{L}_{1}$. So fix a finitely additive, positive $\mu:\Omega\to \mathbb{R}$. For the sake of simplification I will assume $\mu$ non-degenerate, that is, $\mu(E)= 0$ implies $E= 0$ (otherwise, you will have to take some quotient along the way). A finitely additive $\nu:\Omega\to B$ with $B$ a Banach space is $\mu$-Lipschitz if there is a constant $C$ such that $\|\nu(E)\|\leq C\mu(E)$ for all $E$. The infimum of all the constants $C$ in the conditions of the inequality gives a norm and a normed space I will denote by $\mathbf{LA}(\Omega, \mu, B)$. On the other hand, endow $\mathbf{S}(\Omega)$ with the norm

$$\|\sum_n k_n\chi(E_n)\|= \sum_n |k_n|\mu(E_n)$$

and denote the completion by $\mathbf{L}_{1}(\Omega, \mu)$.

Theorem 4: There is a finitely additive, $\mu$-Lipschitz $\chi:\Omega\to \mathbf{L}_{1}(\Omega, \mu)$ universal among all such maps.

Before the conclusion let me address a few points.

  1. Measurable spaces are not needed. If you really want them, use Stone duality (that is, points count for nothing in measure theory so why not leave them out, heh?).

  2. Finitely additive measures are really not that much more general than $\sigma$-additive ones. I will leave this cryptic comment as is, and just note that once again, Stone duality is the key here.

  3. I am not advocating this approach to be used in teaching (unless your goal is to flunk and befuddle as many undergrads as humanly possible). For one, you need some functional analysis under the belt (Banach spaces, completions, semivariation, etc.). Intuition is very hard to come by as I have thrown away the measurable spaces without which THE most important example, Lebesgue measure (arguably, the core of a first measure theory course) cannot be constructed. The whole logic of the approach only makes sense after you have seen other instances of categorical thinking at work. I am sure you can think of other objections.

How categorical is this approach? Certainly, the universal properties of the respective spaces are central to the whole business and at least, they make clear that some results are really just a consequence of abstract nonsense. In the words of P. Freyd, category theory is doing what it was invented for: to make the easy things really easy (or some such, my memory is lousy). For example, the Bochner vector integral is obtained simply by taking the projective tensor product. Fubini and Fubini-Tonelli on the equality of iterated integrals are other notable cases of categorical thinking at work. Now pepper with Stone duality and a few more tools (e.g. Hahn-Banach and the compact-Hausdorff monad) and you can get (a slight variation of) the Riesz representation theorem for compact Hausdorff spaces. Use the proper compactifications and generalize to wider classes of topological spaces. Or use Loomis-Sikorski to get Vitali-Hahn-Saks in one line (but this is really "cheating" as the crucial step in establishing Loomis-Sikorski is essentially the same as the one to establish Vitali-Hahn-Saks: a Baire category-theorem application). And a few more.

But once again, how categorical is this approach? Well, the argument is categorical enough to be generalized to symmetric monoidal closed categories. See R. Borger -- A categorical approach to integration, in the 23rd volume of TAC available online. For the modifications needed to internalize the arguments to a topos (and much more) see the delightful Phd thesis of Mathew Jackson "A Sheaf theoretic approach to measure theory" -- this is available online, just google for it. Oh, by the way, you can see (almost) everything I have explained above in volume 3 of D. Fremlin's measure theory 5-volume series, also available online.

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My own hazy recollection is that Freyd's aphorism was along the following lines: "category theory performs the service of showing that trivially true things are trivially true for trivial reasons." I'm sure someone can come along to correct us. –  Yemon Choi Sep 21 '10 at 21:26
    
I really like this presentation of the ideas, since the more compressed versions I've seen presented or alluded to are harder for Bears of Little Brain like myself to follow. Moreover, by going into detail and not just waving some big representability theorem, this pre-empts fears/suspicions of circularity in constructions. –  Yemon Choi Sep 21 '10 at 21:27
    
Link for the Boerger article: tac.mta.ca/tac/volumes/23/12/23-12abs.html –  Yemon Choi Sep 22 '10 at 18:51

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