In the introduction to chapter VIII of Dieudonné's Foundations of Modern Analysis (Volume 1 of his 13-volume Treatise on Analysis), he makes the following argument:

Finally, the reader will probably observe the conspicuous absence of a time-honored topic in calculus courses, the “Riemann integral”. It may well be suspected that, had it not been for its prestigious name, this would have been dropped long ago, for (with due reverence to Riemann’s genius) it is certainly quite clear to any working mathematician that nowadays such a “theory” has at best the importance of a mildly interesting exercise in the general theory of measure and integration (see Section 13.9, Problem 7). Only the stubborn conservatism of academic tradition could freeze it into a regular part of the curriculum, long after it had outlived its historical importance. Of course, it is perfectly feasible to limit the integration process to a category of functions which is large enough for all purposes of elementary analysis (at the level of this first volume), but close enough to the continuous functions to dispense with any consideration drawn from measure theory; this is what we have done by defining only the integral of regulated functions (sometimes called the “Cauchy integral”). When one needs a more powerful tool, there is no point in stopping halfway, and the general theory of (“Lebesgue”) integration (Chapter XIII) is the only sensible answer.

I've always doubted the value of the theory of Riemann integration in this day and age. The so-called Cauchy integral is, as Dieudonné suggests, substantially easier to define (and prove the standard theorems about), and can also integrate essentially every function that we might want in a first semester analysis/honors calculus course.

For any other sort of application of integration theory, it becomes more and more worthwhile to develop the fully theory of measure and integration (this is exactly what we did in my second (roughly) course on analysis, so wasn't the time spent on the Riemann integral wasted?).

Why bother dealing with the Riemann (or Darboux or any other variation) integral in the face of Dieudonné's argument?

Edit: The Cauchy integral is defined as follows:

Let $f$ be a mapping of an interval $I \subset \mathbf{R}$ into a Banach space $F$. We say that a continuous mapping $g$ of $I$ into $F$ is a primitive of $f$ in $I$ if there exists a denumerable set $D \subset I$ such that, for any $\xi \in I - D$, $g$ is differentiable at $\xi$ and $g'(\xi) =f(\xi)$ .

If $g$ is any primitive of a regulated function $f$, the difference $g(\beta) - g(\alpha)$, for any two points of $I$, is independent of the particular primitive $g$ which is considered, owing to (8.7.1); it is written $\int_\alpha^\beta f(x) dx$, and called the integral of $f$ between $\alpha$ and $\beta$. (A map $f$ is called regulated provided that there exist one-sided limits at every point of $I$).

Edit 2: I thought this was clear, but I meant this in the context of a course where the theory behind the integral is actually discussed. I do not think that an engineer actually has to understand the formal theory of Riemann integration in his day-to-day use of it, so I feel that the objections below are absolutely beside the point. This question is then, of course, in the context of an "honors calculus" or "calculus for math majors" course.

  • 25
    Who are the students you have in mind? You don't seriously think that we should teach the Cauchy or Lebesgue integral to engineers and others who need to use mathematics as a practical tool (rather than prove theorems), do you? (This reminds me of a dinner I once had with a French engineer. He spoke quite bitterly of being forced to take overly theoretical math courses that were of no use to him whatsoever.) – Deane Yang Jan 21 '11 at 3:06
  • 62
    @Harry : I suspect that you will change your opinions of how classes should be taught once you have had the opportunity to teach. – Andy Putman Jan 21 '11 at 3:19
  • 30
    @Harry : Dieudonné's books have a very different intended audience than the usual "intro to analysis class" students. My critique of the spirit of your question is the same as my critique of many of your other pedagogical questions, and I think it reflects a certain viewpoint about teaching that will cause you trouble later on (eg I have known of postdocs that decided that their calculus classes should be taught like real math classes, with terrible results both for them and for their students). – Andy Putman Jan 21 '11 at 5:29
  • 28
    (continued) The trouble is that beginning math students are not nearly sophisticated enough to handle the technical baggage underlying the Lebesgue integral. Even if you can get through the material, few of them will be in a position to appreciate why all this stuff has to be done before they can integrate a function! The Riemann integral is easier to handle, largely because its definition is concrete and involves no "black boxes". For goodness's sake, most of these students will have only recently come to grips with a rigorous treatment of limits! – Andy Putman Jan 21 '11 at 5:35
  • 46
    @Andy. While it makes sense to ask someone without teaching experience not to attempt to answer pedagogical questions, I don't think it's fair to jump on somebody asking a question on teaching for not having enough experience. – Dan Petersen Jan 21 '11 at 10:37

24 Answers 24

From a conceptual standpoint, I think that there are three things one asks of an approach to integration

1) An easily accessible geometric interpretation

2) A readily available computational toolbox (e.g. the fundamental theorem of calculus)

3) A flexible theory

The Lebesgue integral is absolutely unrivaled in (3), but it is actually quite obtuse from the other two points of view. Basic results like the Lebesgue differentiation theorem and the change of variables formula are not at all transparent from the Lebesgue point of view, and geometrically it is no better than the Riemann integral. The Cauchy integral is great if you only care about (2), but it is abysmal at (1) and (3). The Riemann integral, for all its faults, strikes a pretty good balance between (1) and (2). It is even known to enjoy an occasional technical advantage over the Lebesgue theory; for instance, one must invent the theory of distributions to make sense of the Cauchy principal value of an improper integral in the Lebesgue theory if I recall correctly.

In line with what others have said in the comments, which of the three criteria you care most about really depends on the audience.

For a class full of engineers and business majors, the question is essentially moot: two students out of a hundred can correctly define the integral of a continuous function at the end of the semester. But in my view the real point of such classes is to help students develop the language and the skills necessary to reason with rates of change. So in the end it hardly matters what the precise definitions were, and to the extent that it does matter the Riemann integral is quite well suited.

For a class full of grad students, on the other hand, the point is to show the students how to prove theorems. So there is no choice but to use the Lebesgue integral, and the Riemann integral can be largely ignored (as it is in most graduate real analysis classes).

The gray area lies in more advanced undergraduate analysis classes. Often these classes are populated by math, physics, and engineering majors who intend to actually do something with mathematics one day. The problem with teaching the Lebesgue integral to such students is that with undergrads you have to spend half the semester on measure theory, detracting from the time that should be spent on the topology of Euclidean space, multilinear algebra, and whatever else belongs in such a course. I can't even imagine how one builds integration in higher dimensions from the Cauchy point of view, short of turning Fubini's theorem into a definition. So the Riemann integral seems like a perfectly reasonable compromise to me. I admit that I found it a little frustrating to learn a theory that I even knew at the time I was probably never going to use, but I lived through the experience.

  • 8
    Dear Paul, For what it's worth, I have never seen Riemann integration in more than one dimension. We skipped that and went straight to measure theory. – Harry Gindi Jan 21 '11 at 6:19
  • 21
    I do not fully agree with the statement that the Lebesgue integral does not have an geometric interpretation. While for Riemann integration of a function $f:[0,1]\to \mathbb{R}$ (say) we divide the domain of $f$ into subdomains and build rectangles from that, for Lebesgue integration we divide the range of $f$ and build rectangles by taking preimages. At least for continuous functions, both approaches are equally easy to grasp geometrically. – Dirk Jan 21 '11 at 7:58
  • 8
    @Dirk: so you agree with Paul's statement that the Lebesgue integral is geometrically no better than the Riemann? – Chris Eagle Jan 21 '11 at 8:48
  • 5
    There's another thing I ask of integration: 4) An interpretation of expected value of a random variable. – Kevin O'Bryant Jan 21 '11 at 14:14
  • 50
    My father's description of the geometry of Lebesgue integration was "That's the one where you add up the rows, instead of adding up the columns?" – Allen Knutson Jan 21 '11 at 15:24

It is frequently claimed that Lebesgue integration is as easy to teach as Riemann integration. This is probably true, but I have yet to be convinced that it is as easy to learn.

T. Körner: A companion to analysis: a second first and first second course in analysis.

I know, this is slightly besides the Riemann vs Cauchy point, but I like this quotation so much I couldn't help myself...

  • 14
    Heh. The version TWK used to use in lectures, IIRC, was something like "I know of some universities in England where the Lebesgue integral is taught in the 1st year of a maths degree, but I know of no universities in England where students learn it in the 1st year of a maths degree" (this is from memory and hence not very accurate) – Yemon Choi Jan 21 '11 at 3:24
  • Who is TWK? (and filler) – Mariano Suárez-Álvarez Jan 21 '11 at 3:54
  • 1
    @Mariano: TWK = Thomas William Körner. – Thierry Zell Jan 21 '11 at 3:59
  • 30
    I went to the university in England that Körner is probably referring to, and, being the swotty type, I really did learn Lebesgue integration in the first year. But I'm not sure it was right to teach it without doing Riemann integration beforehand. I learned the Riemann theory 12 years later, when I had to teach it myself, and I was shocked at how easy it was. My reaction was something like "God, I wish I'd been told this years ago - I didn't realize integration could be so easy". – Tom Leinster Jan 21 '11 at 7:18
  • 1
    Yemon, I remember TWK's comment also, so I also believe this is pretty accurate. I also happen to believe it; I was one of the better Cambridge maths students in my year and specialised in Analysis, yet still found Lebesgue integration very demanding (in my second and third year), despite the fact I found Riemann integration relatively easy in my first year. I dread to imagine how the other students must have felt if they didn't understand Riemann integration either. – Zen Harper Jan 21 '11 at 7:43

I often introduce the Lebesgue integral (say, for positive functions) as a direct generalization of the Riemann/Riemann-Darboux one just saying that the only differences are that we now use measurable sets instead of intervals as partition elements and take countable partitions instead of the finite ones. This is not how it is written in most textbooks but this allows me to capitalize on the knowledge of the Riemann integration and emphasize both similarities and differences in a rather neat way.

The proofs of some of the basic results in measure theory are intricate but also rather boring, and I wonder whether any analysts will stand up to say that they learned something from them and use them in their work.

I will. If it were just for the two covering lemmas alone (Vitali and Besicovich), I would already vote to have the measure theory courses running. Frostman's lemma, Hausdorff dimension, Sarde's theorem (the full version, not the baby one, of course) jump to the head next. Area and Coarea formulae for Lipschitz mappings form the next layer. And so on. As to proofs, the fact that the least monotone classes containing a ring is a sigma-algebra is shocking enough to wake a student up (and the proof is 3 lines). The dreaded Fubini follows from it in 6 more lines once the monotone convergence results are already in place if you do not bother to consider sigma-finite case rather than just finite or pass to the completion of the measure (and those should really be given as exercises).

It is true that you can make the course boring quite easily if you choose to. The surest way to do it is the same as for an English course: instead of reading poetry, spend all the time perfecting the knowledge of the alphabet.

  • 12
    @fedja: Kudos for the last sentence of your post. – Did Jan 21 '11 at 13:57
  • 12
    Thanks for your response. My reaction to many of the things you name is...that I have never heard of them, and certainly not in any first course on measure theory I took (and I was never offered a second one). Can you suggest a reference for things like Besicovich's Lemma, Frostman's Lemma, area and coarea formulas, and so forth? I'm pretty sure these are not in Rudin's Real and Complex Analysis, for instance. – Pete L. Clark Jan 21 '11 at 14:04
  • 11
    @Quaochu And since when isn't "geometric measure theory" "measure theory"? I thought I made it clear enough that you can stay with just the alphabet if you want but why should you? Anyway, take even the Caratheodory theorem. The first thing that you tell your students is that you can visit any old rug shop and bring home an outer measure. The second is that any outer measure is a measure on well-separating sets, which, miraculously, form a sigma-algebra (5 picture proof). Two more steps, and you have the lebesgue measure on the line. I find it rather dazzling and beautiful. – fedja Jan 21 '11 at 19:15
  • 11
    Another problem is that there are very few good textbooks in measure theory in English. One of the best to the date is Fremlin's but it is too long to make a good textbook for an introductory course. Federer is encyclopaedic and written if not by, then, at least, for a computer rather than a human being. I've given the measure theory course 8 or 9 times by now (both graduate and undergraduate) and every time I had to ignore the officially assigned textbook for everything but the table of contents and occasional exercises. Someone still has to write a good short introductory treatise... – fedja Jan 21 '11 at 19:31
  • 7
    @fedja: This sounds very interesting. If you ever get around to turning your student time lecture notes, own lectures notes and the rest of your materials into lecture notes or a book, it seems to me as if you might have quite an audience. :) I kind of shy away from starting to brows through a 5 volume encyclopedia when I don't have a direct application of the learned knowledge in mind. Some yummy course notes, however... ;) – Max Horn Jan 27 '11 at 22:53

Even though it is less general than some other definitions, the Riemann sum definition is very close to the way that integration is interpreted and used in geometry and in applied mathematics. When I set up an integral, say to find the volume of a domain in $\mathbb{R}^n$ or the volume of a manifold, I basically derive the integrand as an idealized (possibly multidimensional) Riemann sum.

For example, say that I want to find the integral of a function $f(\theta,\phi)$ on the sphere $S^2$. Then the limit of a Riemann term is the value of $f$ times a little region in the shape and orientation of Colorado (as if the sphere were the Earth) that subtends angles of $d\theta$ and $d\phi$. The region has height $d\theta$ and width $(\sin \theta)d \phi$, hence the integral is $$\int_{S^2} f(\theta,\phi) (\sin \theta) d\phi d\theta.$$ Admittedly this informal model can be adapted to Lebesgue integration as well as to Riemann integration --- but not especially to Cauchy's trick that Dieudonné promotes. So, Lebesgue integration is a good thing to teach, but it is clearly more complicated than Riemann integration. It's really an upper-division undergraduate topic, or a first-year graduate topic, to explain why Lebesgue integration is an excellent definition and not a gratuitously complicated one. (But it makes sense, when you teach Riemann integration, to briefly state that Lebesgue integration cures particular diseases of Riemann integration.)

  • 10
    +1 for "the shape and orientation of Colorado". – Michael Lugo Jan 21 '11 at 21:48

Here are some unpolemical facts concerning the Riemann integral:

1) The Riemann integral has a geometric interpretation which is different than that of the Lebesgue integral and is certainly useful in some places. For a bounded set $S \subset \mathbb{R}^N$, Riemann integrability of the characteristic function of $S$ is equivalent to the Jordan measurability of $S$. Now the Jordan measurable sets form an algebra but not a $\sigma$-algebra, and there are important geometric implications of this. For instance see Theorem 3 of these notes and the discussion just after, which gives a geometric property of Jordan measurable sets (only!) which is important in number theory (and indeed, these are notes from a first number theory course at the advanced undergraduate level).

2) Similarly the Riemann integral and not the Lebesgue integral shows up in the theory of uniform distribution: see e.g. Theorem 7 of these notes.

3) Another pleasant consequence of the Riemann (and not Darboux, which is easier) theory is that certain sums which come up -- not every day, perhaps, but not so rarely either -- can be evaluated by identifying them as Riemann sums, which would otherwise be hard to evaluate. In fact one of the only gaffes in Rudin's Principles of Mathematical Analysis I know of is the following: he develops the Darboux theory entirely because it's easier, and he likes to keep things succinct and clean at almost any cost. But then at at least one point in the text he uses convergence of Riemann sums to the integral: that's cheating, plain and simple.

Added: with some reluctance, I will wade slightly into pedagogy. Let me say that I taught a second semester real analysis course for undergraduate math majors in which the Riemann integral was one third of the course material. There were two sections for this course, an honors section and a not honors section, and I taught the "not". I was truly curious to see how much of a pain the Riemann integral would be: when you sort of wave at it while teaching freshman calculus, it starts to look terribly complicated.

I actually enjoyed teaching this material maybe most of anything in the course (which I thoroughly enjoyed teaching). I found that there was just the right amount of messiness to it, and especially, with a little foresight, the amount of messiness can be controlled as arguments that are made again and again can be packaged and presented to students as such, if you like. I could have smoothed over the presentation a little more by skipping Riemann in favor of Darboux (again, as Rudin does), but I didn't want to. Permit me to claim that working through some of the bookkeeping of slightly changing a partition and keeping track of the effect on the Riemann sum builds character. Indeed, I think this sort of careful, detail-oriented work with inequalities and such is the essence of a certain very important kind of analytic thinking, which nowadays seems to go under the label hard analysis (as opposed to "soft", not "easy"). (In fact I think that the spirit of hard analysis is mostly absent in the works of Bourbaki and -- not coincidentally! -- Dieudonné, to its detriment. But I really don't want to debate that point here.)

[Added many years later: I no longer have a reasonable copy of the course notes referred to above. I used them as a basis for a treatment of Riemann integration in a later course: please see Chapter 8 of these notes.]

In contrast, everyone I've talked to who teaches a course on Lebesgue integration tells me that the technical details are just as onerous as they remember from their student days -- or more so, depending on how much they've blocked out. (Edit: not anymore. See fedja's enthusiastic answer to this question.) The proofs of some of the basic results in measure theory are intricate but also rather boring, and I wonder whether any analysts will stand up to say that they learned something from them and use them in their work. The instructor of my basic measure theory course (an eminent hard analyst whom I will pay the courtesy of not naming), after writing Fubini's Theorem on the board, paused, sighed, turned to us and said: "The challenge of this proof will be to stay awake." The fact that some analysts have at least semi-seriously advocated alternatives to the Axiom of Choice in which every function is measurable says something to me about the level of technicality of the subject.

  • 7
    I particularly liked "The fact that some analysts have at least semi-seriously advocated alternatives to the Axiom of Choice in which every function is measurable says something to me about the level of technicality of the subject." – Adam Hughes Jan 21 '11 at 19:56
  • About your comment on ''But then at at least one point in the text he uses convergence of Riemann sums to the integral: that's cheating, plain and simple.'' Can't one make the observation that if you know the standard properties of the integral, (namely monotonicity, how it behaves when you chop the interval of integration into two pieces, and the absolute value esimate) it is easy to prove that Riemann sums for, say a continuous function on a compact interval converge to the integral. It's an easy exercise using uniform continuity. I don't think Rudin used something harder than this. – Benjamin Hayes May 1 '11 at 19:52
  • @Benjamin: yes, I agree that for the case of continuous functions, filling the gap would make a nice exercise. So it should appear as an exercise in his book is what I'm saying. I don't mean to imply that this is such a big deal... – Pete L. Clark May 1 '11 at 21:14
  • 1
    @PeteL.Clark: I like that quote "The challenge of this proof will be to stay awake." ^^ Very nice! – C-Star-Puppy Jul 28 '14 at 23:04
  • 1
    @Tommi: Thanks for the tip. Those were very old notes: I looked back, and the latex file that I had needed some work even to compile. I have a (relatively) recent and (relatively) more polished set of notes that includes Riemann integration, so I went back and linked to that. I couldn't find the "It Doesn't Matter Lemma" stated explicitly therein -- I guess it really doesn't matter! – Pete L. Clark Aug 10 at 15:25

I haven't really thought this through, but how does one actually compute integrals? For the Riemann integral one can either prove the fundamental theorem and have a large table of derivatives handy, or one can rely on material which is already present in the calculus sequence (summing sequences) and which is independently useful; that alone, I think, makes it the more pedagogically sane choice. (Related to Pete Clark's answer, Putnam questions also have a tendency to hide Riemann integrals as sums. Slightly more meaningfully, as Pete says in the comments, I can't imagine doing Lebesgue or Cauchy integrals on numerical data.)

For the Lebesgue integral it seems one proves the fundamental theorem or proves a comparison theorem to the Riemann integral (well, or both). For the former, you get essentially the same result (as far as the average student is concerned) with less pain from Riemann integration; for the latter, you have to know something about Riemann integrals anyway.

  • 14
    I agree. In those vast areas of applied analysis where integrals are actually computed and approximated, Riemann integration is absolutely the way to go. I honestly think we are showing our provincialism a bit by claiming that the Riemann integral is useless because of the awkwardness of establishing, say, a Dominated Convergence Theorem. Following Gian-Carlo Rota, I suspect that the role of this sort of thing in the life of most practicing analysts (pure and applied) is greatly exaggerated. – Pete L. Clark Jan 21 '11 at 8:57
  • 13
    Anytime you estimate the mean of a distribution from a sample, you are doing Lebesgue integration on numerical data. – Kevin O'Bryant Jan 21 '11 at 14:09
  • 5
    @Pete: as a practicing (pure) analyst, I disagree that the role of Dominated Convergence is greatly exaggerated. But I agree wholeheartedly with your larger point that the usefulness of the Riemann integral is greatly underestimated in some circles. – Mark Meckes Jan 21 '11 at 18:00
  • @Mark: okay, I take the point. You would know far better than I. – Pete L. Clark Jan 21 '11 at 20:36

A small plus for the Riemann integral: evaluation of limits like $$ \lim_{n\to\infty}\sum_{k=1}^n\frac{n}{k^2+n^2}. $$

  • Do you know of any places where these sorts of limits arise "naturally" (i. e. not in problems posed to a calculus class to see if they get the idea of Riemann integration?) – Michael Lugo Jan 22 '11 at 17:56
  • 3
    I don't know of any such places. The Putnam Competition had at least two problems requiring this technique : 1961 (A3) and 1970 (B1). – Richard Stanley Jan 22 '11 at 23:11
  • 14
    One place is in explaining the Gibbs Phenomenon Look at… to see this worked out in a particular example; the case of a general jump discontinuity can also be reduced to a Riemann sum with more work. – David E Speyer Jan 27 '11 at 20:22
  • 3
    $\sum_{k=1}^n f(k)$ is a Riemann sum of $\int_1^n f(x)dx$. This observation is very helpful, of course, say, it allows to get asymptotics of Harmonic numbers, factorials, generating functions like $\prod (1-x^n)^{-1}=\sum p(n) x^n$ (and hence for $p(n)$ itself) etc. Of course, we need not Riemann integral itself, such estimates are equally obvious in any theory of integration. But they show that mathematical idea behind Riemann sums is natural and useful. – Fedor Petrov Aug 21 '15 at 10:33

I have found most theoretical presentations of integration, whether Riemann, Lebesgue, or something else, to be overkill. I believe every mathematician should know how to construct the Riemann and Lebesgue integrals and have a solid understanding of their properties, but both can almost always be used as black boxes, when proving theorems in, say, PDE's and differential geometry. I in fact try to use only the 1-d Riemann integral as much as possible and get away with this surprisingly often. More sophisticated ideas should be brought in only if absolutely needed.

ADDED: Let me add that I believe strongly that undergraduate and graduate math students (including at the Ph.D. level) should be trained not just in the theoretical aspects of math but also in using math as a practical tool (especially given how many do not end up as research mathematicians). The concept of a Riemann integral and how to approximate it is arguably one of the most useful in mathematics. So although my comment above focused on the usual suspects (engineers), it applies to mathematicians as well.

  • 2
    +1 for your added comment,Deane. I completely agree,but for reasons deeper then those you gave. There's a synergy between mathematics and the physical sciences whose emphasis tends to vary with each generation. Recently,the influence of the sciences-with the explosion of quantum string theory-is once again prominent in many places.The development of the theory of functional analysis was motivated largely by physical problems in quantum mechanics and today,most modern presentations of classical mechanics rely heavily on modern differential geometry.I consider them joined at the hip. – The Mathemagician Jun 26 '11 at 22:04

Paul Siegel mentioned that to teach undergrads the Lebesgue integral you have to spend half the semester on measure theory. This is actually not true. There is a self-contained definition of the Lebesgue integral due to Mikusinski, and I was first taught functional analysis using this definition precisely because the lecturer did not want to spend time on measure theory.

Definition: A function $f : \mathbb{R} \to \mathbb{R}$ is integrable if there exists a sequence $f_n$ of step functions such that $\sum \int |f_n|$ converges (where the integral of a step function is defined in the usual way) and such that $f = \sum f_n$ pointwise whenever $\sum |f_n|$ converges. Its integral is $\int f = \sum \int f_n$.

This definition is somewhat tricky and I have to admit I found it confusing. Working with it is annoying, since to talk about a sequence of integrable functions one has to introduce a sequence of sequences; I guess this is good analysis practice though.

But you do not need any measure theory. In fact you can define the Lebesgue measure from here. I don't know a reference where this is worked out except the notes we used (Internet Archive).

  • 2
    The book "Hilbert Spaces with Applications", by Debnath and Mikusinski, uses this approach. – arsmath Jan 21 '11 at 21:19
  • 4
    I am really confused about how people talk about measure theory and Lebesgue theory as if they were completely unrelated topics. If you have a definition of the Lebesgue integral, then integrals of characteristic functions of sets give you a measure. And properties of the Lebesgue integral immediately translate into the corresponding properties of measure. – Dmitri Pavlov Jan 21 '11 at 23:36
  • 3
    When you say "simple function" in your definition, do you actually mean "step function"? (I was confused about how one can integrate a general simple function without measure theory...) – Alison Miller Jan 22 '11 at 3:44
  • 7
    this is the obvious way to solve the problem that limits of riemann integrable functions are not always riemann integrable. i.e. define a limit of functions to be integrable if their integrals converge. then one must show the integral is well defined. to me this is a very simple solution to the problem of extending integration theory to satisfy the integrability of limits almost tautologically, but analysts I know do not like it. they love measure theory too much. but for me personally, realizing this was also a viable approach gave me once for all a very satisfying feeling. – roy smith Jan 28 '11 at 6:50
  • 3
    The answer to temp's question is the first option: literally pointwise. – Toby Bartels Jan 8 '13 at 20:35

I am going to argue that Dieudonne’ is actually using limits of Riemann sums to define his “Cauchy” integral. Dieudonne’ hides his use of Riemann sums within the proof of existence of primitives of regulated functions. (Of course existence of primitives is exactly where the rest of us also appeal to the Riemann integral of continuous functions.) In section 8.7.2, on page 160 of my first edition of Dieudonne’, he writes down a primitive of a step function, i.e., a Riemann sum for a step function on a sub interval [a,x] of [a,b], which is just a piecewise linear function. Then he appeals to a prior theorem that in case the step functions under consideration converge uniformly to another function f, then these primitives also converge to a primitive of f.

Now if f is continuous, the Riemann sums for f (on intervals [a,x] with variable right end point) are precisely step functions converging uniformly to f. Hence for continuous functions f, he is proving that the Riemann sums, do converge to a primitive of f. Thus the only difference between his integral and Riemann’s is the class of functions to which they apply. A weakness of Dieudonne’s approach to my mind, is the lack of a means of finding the uniformly convergent step functions needed in his definition (for arbitrary regulated functions). In Riemann’s approach the approximating step functions are easy to choose (even for arbitrary Riemann integrable functions).

For any regulated function f, there exist Riemann sum functions that converge uniformly to f, but Riemann observes this is unnecessary, since L^1 convergence suffices. As long as the set of discontinuities has measure zero (which criterion Riemann proved in an equivalent form), then any choice of the Riemann sum functions will converge to the integral. In Dieudonne’s approach, one must find step functions which converge uniformly, which he accomplishes by a non constructive compactness argument in 7.6.1. Of course once these approximating step functions are found, he approximates the integral by their Riemann sum, as in 8.7.2. So the basic idea that an integral is approximated by Riemann sums is common to both treatments. The difference is that Riemann’s approach applies to a wider class of functions, and provides an easier way to choose the approximating sums.

Added later: My hat is off to Harry for this question, and all who have answered. I learned a lot from them, especially short ones like Allen Knutson’s. Building on them, I throw out a possible combined approach. (It helps if the student read Euler’s precalculus book with infinite series but never mind.)

I address those beginning students who care about understanding what is done and why, more than seeing it the absolute “best” way on first encounter. I.e. I suggest the first job is to motivate the best way, not present it in full.

The basic problem is to define a way of adding up, or averaging, values of a function with an uncountable number of values. All competing definitions seem to use limits of functions with a finite set of values, or “simple functions”. Thus the only question is which simple functions to use and what method of taking a limit.

One could begin with the integral of a monotone continuous function, as did Newton, and exhibit the two standard ways of approximating it by integrals of simple functions, namely adding up over columns (Riemann), and adding up over rows (Lebesgue). Here they are almost the same, in that one representative value is chosen for all points in a subinterval of the domain. From a computational viewpoint, Riemann’s method is superior since the length of the domain subintervals is more easily calculated. Thus for elementary approximations, Riemann’s method is worth knowing.

Introducing more complicated functions one observes how the domain “level sets” become more complicated in Lebesgue’s approach, making the definition of their size a significant task, while Riemann’s approximations remain easy to calculate. One could then define the Riemann integral and state its fundamental theorem, and define “negligible sets” and state the Riemann-Lebesgue criterion for Riemann integrability, or just mention the special cases of continuous and piecewise monotone functions, possibly proving the latter, or the former assuming uniform continuity. One could also state and/or prove the convergence theorem involving uniform limits, possibly informally. The characteristic function of the rationals, whose integral should exist for anyone knowing infinite series, but does not in Riemann’s sense, shows one limitation of this approach, whetting the appetite for Lebesgue. It also explains the failure of convergence for pointwise limits.

One might admit that much of the work in the first course will use the technique of antiderivatives rather than integrals, accepting in spirit Dieudonne’s argument, but that Riemann sums are crucial for approximations. For the technique of evaluating integrals by antiderivatives, one can observe the usual proof of the FTC works on continuous functions for any definition of the integral which is monotone and additive over subdivisions of the domain.

Point out it is hard to give a full fundamental theorem for Riemann integration, i.e. to intrinsically characterize indefinite integrals of all Riemann integrable functions. One could mention the advantages of Lebesgue’s definition.

In this context one could discuss a more flexible definition of antiderivative as Dieudonne’ does, at least for step functions, and remark that one can characterize indefinite integrals of bounded functions with only a finite (or countable) set of discontinuities.
Since many instructors in the US omit proofs entirely in beginning calculus, the question of whether to present complete proofs for Riemann integration applies more to a beginning analysis course, where I think it can prove useful practice for many students.

To the argument that presenting the most advanced and abstract version first is more efficient, I would point out that having been taught that way, it took me another 40 years to understand what is going on, hence not very efficient for me.

  • 3
    In elementary calculus, almost all functions are piecewise monotone-continuous, and then there seems to be no difference at all in any of the treatments. The existence of Riemann's intergral in that case is also trivial. – roy smith Jan 27 '11 at 23:35
  • 2
    thus perhaps the only reason to choose one over the other is based on whether one wants to integrate functions with a countable set of discontinuities (Cauchy), a measure zero set of discontinuities (Riemann), or arbitrary discontinuities (Lebesgue). – roy smith Jan 27 '11 at 23:41
  • 2
    see the principia, lemmas 2,3 book 1. for newton's argument. – roy smith Jan 28 '11 at 7:50
  • 2
    @Thierry: since Cauchy and Riemann were both male, I wonder if perhaps the word you were looking for is jingoism? Anyway, that's one of my favorite words, and I recommend it to you highly: I think you'll find it's a lot of fun to say and write. – Pete L. Clark Jan 30 '11 at 1:40
  • 2
    @Pete: Actually, the original chauvinism of Nicolas Chauvin WAS jingoism, specifically French jingoism. Male chauvinism (under that name) came later. – Toby Bartels Jan 8 '13 at 20:13

hilbertthm90 and Maxime Bourrigan mentioned already in the comments to the question that the Henstock-Kurzweil integral offers a good alternative to the Riemann integral (see also the lecture notes referred to by Maxime) for a course offered for mathematicians.

The definition of the Henstock-Kurzweil integral is very similar to Riemann's integral: It is also defined using Riemann sums $\sum_{i=1}^n f(t_i) (x_i - x_{i-1})\;$, but instead of letting $\delta = \max (x_i - x_{i-1})\;$ approach $0$, one considers for gauge functions $\delta : [a, b] \to \mathbb{R}_{>0}\;$ compatible tagged partitions $(t_i, [x_{i-1}, x_i])_i\;$ with $t_i - \delta(t_i)< x_{i-1}\le t_i \le x_i < t_i +\delta(t_i)\;\;$ and then defines the integral in essential the same way (for details see the wikipedia article).

Dropping the condition $t_i \in [x_{i-1}, x_i]$ one gets an integral (McShane integral) equivalent to the Lebesgue integral on the line, whereas the Henstock-Kurzweil integral works also for "non-absolutely convergent" cases like $\int_0^1 \frac{\sin(1/x)}{x} dx\;$.

As one gets all the nice theorems (monotone convergence theorem, dominated convergence theorem, second fundamental theorem of calculus, Lebesgue differentiation theorem) for the cost of a slightly (?) more difficult definition, there shouldn't be any reason (besides tradition) for teaching Riemann integration on $\mathbb R$ to mathematically moderately mature students (and traditions can be satisfied by mentioning that restricting to constant gauge functions gives the Riemann integral).

For more than one dimension (maybe in a second course), one can/should/must then teach Lebesgue integration. Here I have to add from my own experience (I got introduced to integrals equivalent to Lebesgue's in at least five different courses) that a clean separation between measure theory and topology at the beginning like it is done e.g. in Heinz Bauer's "Measure and integration theory" helps enormously for a deeper understanding.

Maybe yet another argument in favor for the Riemann integral: it is fairly easy to generalize it to integration of vector-valued functions. Here the Lebesgue theory becomes more tricky as soon as the target space is something beyond a Banach space. In many situations you want to integrate a not too badly behaved function, say a continuous one, with values in a rather complicated topological vector space (not only Fréchet, but perhaps dual of Fréchet, only sequentially complete, or something like that if you think of distributions). OK, I admit that this is not what you teach in first year calculus, but just be prepared :)

  • 2
    +1. Currently I am writing up some lecture notes on infinite series, and even at this level I cringe a little bit when I see proof exploiting the positive/negative part decomposition of a real number. One should use this only when it is actually necessary, which so far to me means results that are not valid for Banach space valued-things (good example: Riemann rearrangement). Admittedly it is hard to motivate why you are doing this in a course for undergraduates who have not (explicitly) seen a metric space, let alone a Banach space valued series or function. – Pete L. Clark May 1 '11 at 21:33

Although not a direct answer to the question, this may be relevant to the discussion:

I learnt basic measure theory and the theory of Lebesgue integration in a course called "Probability and Measure" at Cambridge. As the title suggests, the course uses probability as the natural motivation for measure theory, which I think is a pretty good idea. As I now understand is quite different to how things are in the US, I've never taken a 'graduate real analysis'-type class. To comment on the discussion following fedja's answer, the book for the course is really the first half of David Williams' Probability with Martingales and is a good English-Language intro to measure theory and isn't too long. Unfortunately, it's called Probability with Martingales so all this isn't terribly obvious, though I've needed no other basic measure theory text and am not a probabilist myself.

The pedagogical aim would be to convince students that measure theory is a good setting for a rigorous theory of probability, which, before that stage, will only have been taught naively. There are plenty of explainable things that could help here: Why can't all sets be measurable? Why do we need countable additivity? What's the deal with "almost surely"?

The need for a rigorous theory of expectation, conditional expectation and various notions of convergence of random variables and so forth leads to the Lebesgue integral relatively naturally, if you ask me. Integrals of simple functions are basically just expectations of Bernoulli random variables... If monotonicity of measures is understood, then the monotone ,and hence dominated, convergence theorems (some of Lebesgue's main bonuses over Riemann) start seeming more natural.

  • 1
    That's a very good answer. Probability theory provides great motivation for developing general measure theory (given that probability is, in a way, measure theory in disguise, although of course no one thinks of it that way) and many measure-theoretic subtleties appear there with a proper context, which makes it easier to learn. – Michal Kotowski Jan 22 '11 at 15:51
  • I completely agree and I only have to add to this that a lot of the power graduate programs in mathematics are beginning to reorganize thier graduate analysis courses in precisely this way-see Columbia. It may be that most programs will follow suite. As for giving a rigorous theory of the integral at the undergraduate level,I think the generalized Riemann (HK integral) is a terrific option that isn't considered enough.Indeed-you can have your cake and eat it too since the Generalized Riemann integral can only be simply defined in Euclidean spaces.(continued) – The Mathemagician Jun 10 '11 at 23:09
  • 1
    (continued) In more general spaces-like a general topological group-some version of measure-theory or Daniell-type integrals are needed. These can be then covered in graduate courses. – The Mathemagician Jun 10 '11 at 23:11

In introductory calculus classes, I work hard to get my students to understand the heuristic: break complicated problem into simple (close enough to constant) pieces; find the answer to the simple pieces; add the results using an integral.

This is how they can understand the presence of integrals in many contexts. Formally, of course, this is a shorthand for Riemann integration.

Although it's true that the Lebesgue integral is technically more complicated than the Riemann integral, I don't think that's really the point. In fact, the Riemann integral is pretty technical already, as far as most undergraduates are concerned. The real point is the "fundamental theorem of the calculus", which is at least conceptual and amenable to a really good plausibility argument in the case of the Riemann integral, but requires some genuinely deep analysis in the case of the Lebesgue integral.

Don't get me wrong -- I really love integration theory. And I've never taught serious undergraduate analysis, so I don't have any actual experience in this area. I do have a funny story, however:

I first learned "real analysis" from a course taught by Richard Brauer. Brauer was of course an algebraist, but at the time, he taught in consecutive years the basic graduate "real analysis", "complex analysis" and algebra courses. He was as close as I felt I would ever get to the great tradition of mathematicians who had a broad view of the field as a whole. Of course the book we used as a reference for measure theory and integration was Halmos. At one point, he remarked, "Halmos is a curious book. When you finish this book you will have a very good understanding of how to integrate the function 1. But you will have no idea of how to integrate the function $x$."

Of course he was referring to the fact that Halmos doesn't even touch on the question of differentiation in Euclidean space, and so doesn't really deal with the "fundamental theorem".

I think if you want to make sense of and motivate numerically integrating a function which is done almost always in numerical analysis, I believe Riemann integration is a better way to motivate than Lebesgue. Also, Riemann integration helps to provide a motivation for Lebesgue integration as to why we need to define Lebesgue integration in the first place.

  • 5
    Motivate? I know of no numerical integration algorithm which is based on anything other than Riemann integral... – Igor Rivin Jan 21 '11 at 3:40
  • 25
    To add to Igor's point, I'll go with yet another quote, by Richard Hamming this time: Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane. – Thierry Zell Jan 21 '11 at 4:01
  • Something about Hamming's quote has always struck me as a straw-man argument. In another context, in Gavin Brown's review of Tom Körner's Fourier Analysis (Bull. AMS Oct. 1989), it is written, "Where is the message that probabilistic intuition and measure theory reinforce each other and that the Lebesgue integral is to be preferred because it is so beautifully simple to use?" In other words, the advantages of the Lebesgue integral are largely conceptual ones. – Todd Trimble Aug 10 at 17:01

In his book on real and complex analysis, Walter Rudin first develops an abstract theory of measure and integration and then introduces Lebesgue measure as the unique measure on $[0,1]$ such that integration wrt to that measure is exactly the Riemann integral on the continuous functions from $[0,1]$ to $\mathbb R$.

I found that very elegant, my students found that difficult to digest.
They were familiar with the Riemann integral but not so much with abstraction in mathematics. Would it have been better to introduce the Lebesgue integral first (say on $\mathbb R^n$)? And then do more abstract measure theory later, never mentioning the Riemann integral? I have no idea.

  • 4
    Definitions of the type "the unique [thing] that [makes it work]" are very bad pedagogically, even if they are nice for theory-building. The obvious fix in your case would be to use the usual definition (you know, length of intervals etc.). – Ketil Tveiten Jan 21 '11 at 10:15
  • 2
    When I first read Rudin's Real and Complex Analysis I found his "construction" of Lebesgue measure on $\mathbb{R}^n$, where he essentially applies the Riesz Representation Theorem to the functional given by the Riemann integral, hard to understand. I personally find Folland's approach of constructing Borel measures on $\mathbb{R}$ from increasing, right continuous functions and then taking products to get Lebesgue measure on $\mathbb{R}^n$ more intuitive and less "magical." – Keenan Kidwell Jan 21 '11 at 16:18
  • 1
    Since we only need integrals of continuous functions, we could equally well use the Cauchy integral. – Toby Bartels Jan 8 '13 at 20:39

One available compromise is to just work with the following definition of a Riemann integral (which works fine in ${\mathbb R}^n$ as well:

A bounded function $f$ on $[a,b]$ is Riemann integrable if and only if for every $\epsilon > 0$ there are step functions $\psi_1, \psi_2 : [a,b] \to {\mathbb R}$ such that

$\psi_1 \leq f \leq \psi_2 $


$\int_a^b \left( \psi_2(x) - \psi_1(x) \right) dx < \epsilon$

I think using this definition is easy and geometrically intuitive, and on the other hand working with this definition prepares you conceptually for the Lebesgue integral where you juggle "simple" functions instead of step functions. Thus, you already get a nice piece of the Lebesgue point of view.

The definition also demonstrates the broad principle that to construct an object in real analysis which should be a real number, one often needs a good way to overestimate/underestimate the object you're going for, and there are plenty of examples of that in real analysis outside of integration (liminf and limsup being the simplest) -- after all, this is just one point of view on how the real numbers are constructed.

To give an example of the ease of use: to prove the fundamental theorem, suppose $F$ is continuous with a Riemann integrable derivative $F' = f$ and let $\psi_2$ be a step function above $f$ which induces the partition $a \leq x_1 \leq x_2 \leq \ldots \leq x_{n-1} \leq x_n = b$. Then the total change of $F$ from $a$ to $b$ is the sum of the "small" changes $F(b) - F(a) = \sum_{k=1}^n (F(x_n) - F(x_{n-1}))$ which is less than or equal to $\int \psi_2(x) dx$ by the mean value theorem. Similarly for the other inequality.

Thus, one can go pretty far with this definition, but the results which separate the Riemann integral from the Lebesgue integral (e.g. that $g(f)$ is Riemann integrable for $g$ continuous and $f$ Riemann integrable or the fact that Riemann sums converge to the Riemann integral), one needs to use the observation that step functions can't be close in terms of area without being uniformly close on all but a small set. You could think this feature is either a clarification or a disadvantage. One certain disadvantage is that it is not the right point of view for integrating vector-valued functions. So you might decide it's better to define the integral in terms of Riemann sums in the first place (giving more of a "metric space" point of view and less of an "ordered space" point of view). Or you might even decide to skip some of these other topics, depending on your point of view and time available.

Because the Riemann integral is an integral, and the Cauchy integral is an anti-derivative.

  • Could you elaborate on this? I have always seen the Riemann integral as a weird way of defining an antiderivative, and thus have failed to see the "fundamentality" of the fundamental theorem – Guido Kanschat Jul 29 '13 at 5:54
  • 3
    The Riemann integral is defined to be the area under a curve, when that area can be bounded between upper and lower sums. Thus it is interesting and non-trivial that the Riemann integral is actually an antiderivative. The Cauchy integral is defined to be the antiderivative. Thus it is interesting and non-trivial that it is also the area under a curve. – Tom Ellis Aug 1 '13 at 16:28

The Riemann integral has a good geometrically motivated definition. So one teaches it but disadvantages are that even in elementary analysis it is not a complete tool. Lebesgue integral has the advantage that it is defined in a general set up and can handle multiple integration very well. the book by Apslund and Bungart used Mikusinski's definition and one can define integral quite quickly .It is reasonably intuitive but not as intuitive as Riemann.the approach does not require measure theory Daniel's approach does not require to define measure but again the class of upper functions and extension process is not intuitive. these three definitions are not constructive but Riemann integral has constructive definition.

But the gauge integral (Henstock-Kurzweil) covered nicely in C. Swartz book is intuitive and a nice generalization of Riemann. Every derivative is integrable. There are no improper integrals . All improperly integrable functions are gauge integrable as well. the space L! is precisely the space of absolutely integrable mappings.the proof of fundamental theorem of calculus ( part Ii) is beautiful and elegant. We just use a gauge rather than mesh of a partition or a NET based o refinements.The integral is super Lebesgue yet definition is almost verbatim similar to Riemann integrable We can have a nice uniform convergence theorem. For graduate courses one can have The usual dominated convergence theorem( slightly more general version than for Lebesgue integral). Further the differential calculus in Banach spaces can be carried out without any restrictive assumption like continuity of derivative and we can have nice integral version of mean value theorem . the version is same as in along (analysis 1) but better as Lang restricts to Cauchy integral. On real line the integral is the perfect integral a satisfactory theory exists on euclidean spaces as well . one can define measure using this integral . about generality there is a question which I have asked on this forum. Professor Buck in his book "Garden of integrals) Terms this integral as integral of 21 st century.

  • may i know why i got -3 marks. if you compare all answers mine is the most complete answer. it says why rienmann integration should be taught in a subtle way. i think each one who has voted negative should explain me why ? the community should not be biasedf! – Anil P Jul 1 '11 at 17:03
  • the answer of Stefan waldman is good and points out why it is necessary to use Riemann integral. – Anil P Jul 16 '11 at 16:54

In Cartan calculus, integrating a $k$-differential form $\alpha$ on a $k$-cubic chain $\sigma$ doesn't need more than the Riemann integral: $$ \int_\sigma \alpha := \sum_{i=1}^N c_i \int_{I^k}\sigma_i^*(\alpha) = \sum_{i=1}^N c_i \int_0^1dx_1 \cdots \int_0^1dx_k f(x_1,\ldots,x_k) $$ where the $\sigma_i$ are standard smooth cubes and $$ \sigma = \sum_{i=1}^N c_i \sigma_i \mbox{ and } f(x_1,\ldots,x_k) = \sigma_i^*(\alpha)_x(e_1,\ldots,e_k) $$ And with that you can do a lot already if not almost everything: Stokes' theorem, Cartan formulae, variation of integral of forms on chains (variation calculus) etc. (It is BTW what I used to extend Cartan calculus to diffeology, where there is no Lebesgue measure). Of course, differential calculus is not integration theory. But I would not over-teach and use tools where they are far beyond the needs. So the choice depends, as usual, on our goal :)

P.S. By the way Dieudonné is not the example I would take for a course, if his books can be regarded as references, his pedagogy can be used for the definition of infinitesimal ;-)

  • 3
    Dieudonné repeatedly explained that Bourbaki's Éléments de mathématique were, first and foremost, pedagogical treatises. – Did Jan 21 '11 at 7:35
  • 3
    After it was clear that the reason for that was that the people in charge wanted to make the idea of vector spaces of continuous function more acceptable. And if you begin with $\bf R^2$ it is more difficult (maybe) to extend your mind toward infinite dimensional vector spaces? But we were 15 years old when we were introduced to these maths (three years before the baccalaureate), switching brutally from classical maths to these set theoretical approach. Well, well... Dieudonné (together with Lichnerowicz) were the names attached to this experiment, for the good the bad and the ugly ;) – Patrick I-Z Jan 21 '11 at 9:12
  • 8
    Patrick: Yes. I simply wanted to mention the (admittedly huge) gap between Bourbaki's intentions and the reality of their influence on pedagogical matters. And with respect to my own field of expertise, you can imagine I personally have somewhat mixed feelings about Dieudonné's famous stance that Radon measures on locally compact spaces were the only kosher approach to integration, thus in effect ejecting the whole field of probability theory out of the realm of la mathématique. (cont'd) – Did Jan 21 '11 at 9:36
  • 4
    (cont'd) And I owe you some thanks because your posts gave me an occasion to reread Laurent Schwartz's account in his memoirs of the outrageous behaviour of Bourbaki members at the conference Doob gave in Paris. – Did Jan 21 '11 at 9:37
  • 3
    An excerpt from Schwartz's account can be found, as with so many things it seems these days, via Google:… – Yemon Choi Jan 22 '11 at 3:41

In many first-year calculus courses, one should not teach Riemann integration nor Lebesgue integration nor any other theory of integration. An integral is an area under a curve, but also any of various other sums of infinitely many infinitely small quantities. The expression $\int_a^b f(x,y)\,dy$, the "$dx$" serves not only to identify which variable one is integrating with respect to, but also what the units are (e.g. of $f(x,y)$ is in meters per second and $dx$ is in seconds, then ..... etc.....).

I teach what I call the "boundary rule" on the first day of a calculus course: $$ \left[\text{size of boundary}\right] \times \left[\text{rate of motion of boundary}\right] $$ $$ = \left[\text{rate of change of size of bounded region}\right]. $$ With concrete examples, of course. That's half of the Fundamental Theorem, and the other half can be given later, after one begins to talk about integrals. The product rule is a corollary: think of a rectangle with two moving sides. Proving that the area of a circle is $\pi r^2$ is another corollary. So are some other things that are less important but are good exercises.

I guess one point that hasn't been made is that Riemann integral may have pedagogical value precisely because it's awkward and difficult. Look at the definition of the Riemann integral: $$ \forall \varepsilon >0 \;\exists \delta>0 \; \forall n\; \forall t_0<x_0<t_1<x_1<\dots<x_{n-1}<t_n \in \mathbb{R} \;\\\text{such that} \; a=t_0,\; b=t_n,\;\text{and}\;\forall i<n, \; |t_{i+1}-t_i|<\delta $$ followed by an inequality involving part of this data. There are more quantifiers in this expression than in any other expression that the students have seen in any of their courses by this point, plus the quantifies are over contrived sets like the set of $n$-tuples. In my experience, this is precisely what makes it the most difficult definition in the analysis course, and even reasonable students struggle with it. Usually, even the treatment of Lebesgue integral does not contain conditions that are logically that complicated - rather, the definition is split into many simple steps.

But, at some point, better earlier than later, math students have to learn how to make sense of complicated statements with many nested quantifiers. Why not when learning Riemann integral?

When I was introduced to measure theory, the professor chose to use the Choquet integral to obtain the Lebesgue integral. An this uses the "good old" Riemann integral to integrate the pseudo-inverse of the cumulative distribution function (I think, it was this book).

As a student I enjoyed this approach because I really knew what the Riemann integral was about and also I had an understanding of the problems with it - but I was really confused by the way we had had the Lebesgue integral at the first place.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.