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Many fields in mathematics start from the "dirty" approach. In calculus we do all sort of $\epsilon$-$\delta$ stuffs, until topology gives an elegant formulation using open sets. A first course in linear algebra usually starts with defining matrices, their multiplications, determinants, etc, then the whole theory is built upon that. But if we start from vector spaces and linear operators on them, everything seems to fit in much more cleanly.

What about integrals? There are at least two approaches to define Riemann integrals: upper and lower integrals; or Riemann sums of partitions. None of them looks elegant to me. So, is there a "clean" way to develop the theory of integrals?

Shaun
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miku
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    I'm not sure if that's what you're looking for, but you might be interested in measure theory. – Joel Cohen Apr 07 '12 at 14:41
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    At some point there's no way around $\varepsilon$-$\delta$-stuff! (or at least the alternatives are at least as difficult to grasp) – t.b. Apr 07 '12 at 14:42
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    Actually, $\epsilon$-$\delta$ was the clean approach (vs. the ad hoc and unwarranted manipulation of infinitesimals, which would have to wait until Robinson's work to get on firm logical footing). Topology is not a "cleaning up" of $\epsilon$-$\delta$ proofs, it's a generalization. – Arturo Magidin Apr 07 '12 at 14:59
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    What would be enlightening at this point for everybody (me included) would be that the OP explains as precisely and specifically as possible how they came to think of epsilon-delta proofs as dirty. This is a most serious suggestion. – Did Apr 07 '12 at 16:05
  • If you write \epsilon-delta, with the hyphen INSIDE of $\TeX$, then the hyphen will look like a minus sign thus: $\epsilon-\delta$. But you you exclude the hyphen from $\TeX$, then it's a proper hyphen: $\epsilon$-$\delta$. One could also try \epsilon\text{-}\delta : $\epsilon\text{-}\delta$. – Michael Hardy Apr 07 '12 at 16:15
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    This question is perfectly clear to me. By "dirty" the OP means "messy" and not "not rigorous", and by "clean" he means "elegant" and not "rigorous". – Stefan Apr 13 '12 at 10:09
  • @StefanWalter In case your comment is meant as an answer to mine, I note that you are merely shifting the problem to a proper definition of messy and elegant. (And please use @ to signal your comments.) – Did Apr 14 '12 at 13:02
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    @Didier: My comment wasn't directed at you exclusively. Of course these concepts cannot be defined properly. But I think the OP has done a good job to illustrate them. – Stefan Apr 14 '12 at 14:28
  • @StefanWalter I disagree. Questioning the meaning of these labels (not concepts) and how they are used seems mandatory to avoid the mere, probably unconscious, propagation of prejudices. – Did Apr 14 '12 at 14:42
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    @Didier: Several answerers have already understood the question perfectly well, and I suspect that you do too. I see no point in arguing just for the sake of it. If you want to explain your point of view on what labels would be "correct" (whatever this means), please go ahead. – Stefan Apr 14 '12 at 15:07
  • @StefanWalter Did they? Quote: In any case, the answer to what you want depends heavily (emphasized) in what you mean by "clean". (But please do not put correct between quotation marks since I did not use the word--and for some reasons.) – Did Apr 14 '12 at 15:16
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    @Didier: Still waiting for anything constructive from your part... – Stefan Apr 14 '12 at 15:34
  • @StefanWalter While you are waiting, try to ponder this: to warn against the unconscious use of some representations carrying nontrivial implications is seen (at least in some circles) as more constructive than to insist that these should at no cost be questioned (or do I misrepresent your stance on this?). – Did Apr 14 '12 at 16:04
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    @Didier: My stance: The OP chose words that most people associate with other concepts (see Arturo's comment) than he does, which led to a misunderstanding. The goal of our discussion should be to remove that misunderstanding. Once that is achieved, the question can be answered. It is not necessary to discuss whether the OP's words are incorrect or inferior (or whatever you had in mind), and I doubt that such a question is at all meaningful. However, it is not your insistence on that discussion I find unconstructive, but your failure to contribute anything to it beyond asking others to clarify, – Stefan Apr 14 '12 at 18:40
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    ...,issuing vague warnings, and so on. – Stefan Apr 14 '12 at 18:42
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    I side with Didier. I think that clean is way too subjective. Why would one say that Darboux or Riemann integration is not clean? What is messy about them? Miku, why don't you read Spivaks Chapert 13 in calculus. He introduces integration quite "cleanly" if you must ask. – Pedro May 02 '12 at 20:14
  • This is a more a speculation, less an answer but... does the elegant construction of Riemann integrals from the continuation of simple step functions given above, extend to the construction of Lebesgue integrals as the continuation of countable sums of disjoint step functions? Or would that be too easy? – Rory Allen May 02 '12 at 13:41
  • You may want to read A Garden of Integrals. – lhf May 15 '12 at 10:48

4 Answers4

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Let me say first that I fully agree with the comments posted above.

In any case, the answer to what you want depends heavily in what you mean by "clean" and, more importantly, what you mean by "theory of integrals".

Regarding "clean", your example is open sets as opposed to $\varepsilon$-$\delta$. But when you want to do concrete manipulations with open sets in a metric space you end up dealing with balls, and you prove things exactly by using $\varepsilon$-$\delta$; so it's not "cleaner" at all. Just more general, as Arturo said.

Regarding the "theory of integrals": if you want to obtain an integral that represents what the Riemann integral represents (i.e. "the area below the curve") you will have to deal with some kind of limits of sums. I personally prefer Lebesgue's approach, which is cleaner and more general, but it still implies a fair amount of "dirty".

If you dispense with the need of your integral being "the area below the curve", you might notice that all an integral does is to assign a number to a function in a linear way. So you might think of a locally compact Hausdorff space $X$, consider $C_c(X)$, i.e. the complex-valued compactly supported continuous functions on $X$, and think of all the functionals (i.e. scalar valued, linear functions) $$ \varphi:C_c(X)\to\mathbb{C}. $$ This space is well-understood, and every such functional is an "integral" in some proper sense (the Riemann integral being just a very particular example in the case where $X$ is a closed hypercube in $\mathbb{R}^n$). This last point of view is a lot "cleaner" to my taste, but it is probably not what you are looking for.

Martin Argerami
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  • a bit of clarification needed, what is the type of functions in $C_c(X)$? Is it $X \to \mathbb C$? – Yrogirg May 28 '12 at 11:19
  • Yes, I would say that it is the standard when you consider continuous number functions over a topological space. I'll edit accordingly. – Martin Argerami May 29 '12 at 03:13
  • Just to add a reference for people who might want to know more: the precise content of the statement "every such functional is an 'integral' in some proper sense" is given, for example, at http://en.wikipedia.org/wiki/Riesz_representation_theorem (it's one of the many things called the "Riesz representation theorem") and is proved in many, if not most, functional-analysis-oriented analysis textbooks. – leslie townes May 29 '12 at 04:12
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I don't like very much the distinction clean/dirty which most of time hides a preference for formalism rather than constructivism. And I'm not sure that formalism is the best thing that happened to mathematics. For you, is cleanness just encapsulation or different approach ?

Anyway, the following construction of the Riemann integral should satisfy you.

  1. Let $I$ be a segment of $\mathbb R$. Consider the vector space $\mathcal C(I)$ of functions $I\to \mathbb R$, together with the norm $\| \cdot \|_\infty$. It is a Banach space (i.e. it is complete for this norm).

  2. Now, consider $V$ the sub-vector-space of $\mathcal C(I)$ generated by the indicatrix functions of the form $1_{[a,b]}$, $a,b\in I$. These are the step functions.

  3. Define a linear form $\int : V \to \Bbb R$ by the formula $$\int 1_{[a,b]} = |b - a|$$ At this point, you have to check that this definition is well-founded by checking that if the finite sum $\sum_i \alpha_i 1_{[a_i,b_i]}$ is the zero function, then $\sum_i \alpha_i |b_i - a_i|$ is zero as well.

  4. Check that $\int$ is $|I|$-lipschitzian, where $|I|$ is the length of $I$, that is to say $$\left|\int (f - g)\right| \leqslant |I|\| f - g\|_\infty$$

  5. Remind this wonderful and easy result :
    Let $X$ and $Y$ be metric spaces, and assume that $Y$ is complete. For every $V$ subset of $X$ and $f : V\to Y$ uniformly continuous, then there exists a unique extension of $f$ to $\overline V$ which is uniformly continuous.

  6. Apply this result with $\mathcal C(I)$, $\Bbb R$, $V$ and $\int$, to obtain a linear form $$ \int : \overline V \longrightarrow \Bbb R$$ This is the Riemann integral on $I$

  7. Check that $\overline V$ contains all the continuous functions, and much more, and is complete. The functions in $\overline V$ are called ruled functions, or regulated functions. Note that you don't obtain every Riemann-integrable function this way (cf. Sam L. comment).

How ever, it should be noted that even if this approach simplify the construction in itself, it does not simplify the proofs about the riemann integral any further, and it is very important to have a concrete representation for this completion $\overline V$.

EDIT
I've been kindly asked by email to detail this answer, so here I am ! I don't know any book or other reference having this point of view, I'd be happy if somebody had a one.

Lierre
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    Note that your $\overline V$ does not contain all Riemann-integrable functions. For instance let $C$ be the Cantor set and consider $f(x) = 1_C(x)$. Then $f$ is Riemann-integrable, since it is only discontinuous at points of $C$ (which is of measure 0), but $f$ cannot be approximated uniformly by step functions. – Sam May 15 '12 at 11:12
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    As for references: The approach described above can be found in *Analysis II, Amann/Escher* in the chapter titled The Cauchy-Riemann integral. – Sam May 15 '12 at 11:22
  • Thank you for the reference and for pointing out the error. – Lierre May 15 '12 at 11:25
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One can understand integration as a way of forming means. A very elegant characterization of the integral on $[0,1]$ can be obtained by using some category theory and functional analysis. You can find the characterization in A universal Banach space by Tom Leinster. I don't think the construction gives you much insight though and showing that the object actually exists will still take some time. Also, you don't actually integrate functions but certain equivalence classes of functions in this approach.

[2015-09-07: updated link]

Uday Reddy
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Michael Greinecker
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    The way I understand Leinster's construction is by saying that it is basically coming down to the fact that you have an isomorphism of $[0,1]$ with Lebesgue measure and $C = {0,1}^\mathbb{N}$ with the usual product measure + the fact that $L^1$-space of the latter can be written as a limit of finite-dimensional $\ell^1$-spaces (use the martingale convergence theorem, for example). Another way of seeing the universality described is by using the projections on the subspaces generated by the Rademacher functions in $L^1$. – t.b. Apr 14 '12 at 12:09
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AFAIK, a "polished up" version of the theory of Riemann integration can be crafted by means of the topological concept of net, or generalized sequence (see Moore-Smith Convergence on Wikipedia).

It goes on the following lines: take a function $f\colon [a, b]\to \mathbb{R}$ and introduce the set $S$ consisting of all pairs $(\mathcal{D}, \xi)$, where:

  1. $\mathcal{D}=\{[x_1, x_2]\ldots[x_{\omega}, x_{\omega+1}]\}$ is a partition of $[a, b]$;
  2. $\xi=\{\xi_1 \in [x_1, x_2], \ldots \xi_{\omega}\in [x_{\omega}, x_{\omega+1}]\}$ is a choice of points in $[a, b]$ subordinate to the partition $\mathcal{D}$.

For every pair $(\mathcal{D}, \xi)\in S$ we define the corresponding integral sum:

$$\sigma_f(\mathcal{D}, \xi)=\sum_{r=1}^\omega f(\xi_r)(x_{r+1}-x_r).$$

$f$ is then said to be integrable if $\sigma_f(\mathcal{D}, \xi)$ is convergent as the partition norm $\lVert \mathcal{D}\rVert=\max(x_{r+1}-x_r)$ approaches zero, a phrase which is given precise meaning in Moore-Smith's theory of convergence. If this is the case, then we can define

$$\int_a^b f(x)\, dx=\lim_{\lVert \mathcal{D}\rVert \to 0} \sigma_f(\mathcal{D}, \xi).$$

I believe that this construction is exactly what the OP was looking for. Unfortunately the only reference I know is Italian: Analisi matematica 1 by Antonio Avantaggiati, Ambrosiana editrice, 1994.

EDIT The construction can also be found in Kelley's book General Topology: it is Problem H of chapter 2. (Thanks to Michael Greinecker for pointing this out).

Community
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Giuseppe Negro
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