Why should one still teach Riemann integration? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T20:57:08Z http://mathoverflow.net/feeds/question/52708 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration Why should one still teach Riemann integration? Harry Gindi 2011-01-21T02:25:40Z 2013-03-12T13:42:37Z <p>In the introduction to chapter VIII of Dieudonné's <em>Foundations of Modern Analysis</em> (Volume 1 of his 13-volume <em>Treatise on Analysis</em>), he makes the following argument: </p> <blockquote> <p>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 t o 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.</p> </blockquote> <p>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. </p> <p>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?). </p> <p>Why bother dealing with the Riemann (or Darboux or any other variation) integral in the face of Dieudonné's argument?</p> <p>Edit: The Cauchy integral is defined as follows:</p> <p>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 $g'(\xi) =f(\xi)$ .</p> <p>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 $I$). </p> <hr> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52713#52713 Answer by Pete L. Clark for Why should one still teach Riemann integration? Pete L. Clark 2011-01-21T02:58:43Z 2011-01-21T20:43:54Z <p>Here are some unpolemical facts concerning the Riemann integral:</p> <p>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 <strong>Jordan measurability</strong> 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 <a href="http://math.uga.edu/~pete/NT2009gausscircle.pdf" rel="nofollow">of these notes</a> 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).</p> <p>2) Similarly the Riemann integral and not the Lebesgue integral shows up in the theory of uniform distribution: see e.g. Theorem 7 <a href="http://math.uga.edu/~pete/udnotes.pdf" rel="nofollow">of these notes</a>.</p> <p>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 <em>Principles of Mathematical Analysis</em> 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.</p> <p><b>Added</b>: 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. </p> <p>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 especially fondly remember the <strong>It Doesn't Matter Lemma</strong> from page 6 <a href="http://math.uga.edu/~pete/243integrals1.pdf" rel="nofollow">of these notes</a>. I used it often.) 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 <strong>hard analysis</strong> (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.) </p> <p>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. (<b>Edit</b>: 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. </p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52715#52715 Answer by Thierry Zell for Why should one still teach Riemann integration? Thierry Zell 2011-01-21T03:08:57Z 2011-01-21T03:15:11Z <blockquote> <p><em>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.</em></p> </blockquote> <p>T. Körner: A companion to analysis: a second first and first second course in analysis.</p> <p>I know, this is slightly besides the Riemann vs Cauchy point, but I like this quotation so much I couldn't help myself...</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52716#52716 Answer by user11000 for Why should one still teach Riemann integration? user11000 2011-01-21T03:19:06Z 2011-01-21T03:19:06Z <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52731#52731 Answer by Paul Siegel for Why should one still teach Riemann integration? Paul Siegel 2011-01-21T05:46:22Z 2011-01-21T05:46:22Z <p>From a conceptual standpoint, I think that there are three things one asks of an approach to integration</p> <p>1) An easily accessible geometric interpretation</p> <p>2) A readily available computational toolbox (e.g. the fundamental theorem of calculus)</p> <p>3) A flexible theory</p> <p>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.</p> <p>In line with what others have said in the comments, which of the three criteria you care most about really depends on the audience. </p> <p>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. </p> <p>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).</p> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52733#52733 Answer by Patrick I-Z for Why should one still teach Riemann integration? Patrick I-Z 2011-01-21T05:54:46Z 2011-01-21T05:54:46Z <p>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> <p>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 ;-)</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52739#52739 Answer by Greg Kuperberg for Why should one still teach Riemann integration? Greg Kuperberg 2011-01-21T07:29:03Z 2011-01-21T07:29:03Z <p>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.</p> <p>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.)</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52743#52743 Answer by Stefan Geschke for Why should one still teach Riemann integration? Stefan Geschke 2011-01-21T07:48:29Z 2011-01-21T07:48:29Z <p>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 $\mathbb R$. </p> <p>I found that very elegant, my students found that difficult to digest.<br> 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. </p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52747#52747 Answer by Qiaochu Yuan for Why should one still teach Riemann integration? Qiaochu Yuan 2011-01-21T08:45:21Z 2011-01-21T09:44:44Z <p>I haven't really thought this through, but how does one actually <em>compute</em> 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.)</p> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52756#52756 Answer by Someone for Why should one still teach Riemann integration? Someone 2011-01-21T10:14:36Z 2013-03-12T13:42:37Z <p>hilbertthm90 and Maxime Bourrigan mentioned already in the comments to the question that the <a href="http://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral" rel="nofollow">Henstock-Kurzweil integral</a> offers a good alternative to the Riemann integral (see also the <a href="http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/henstock.pdf" rel="nofollow">lecture notes</a> referred to by Maxime) for a course offered for mathematicians.</p> <hr> <p>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 $\delta = \mathop{max} (x_i - x_{i-1})\;$ approach $0$, one considers for <em>gauge</em> functions $\delta : [a, b] \to \mathbb{R}_{>0}\;$ compatible tagged partitions $(t_i, [x_{i-1}, x_i])_i\;$ with $t_i - \delta(t_i)&lt; x_{i-1}\le t_i \le x_i &lt; t_i +\delta(t_i)\;\;$ and then defines the integral in essential the same way (for details see the <a href="http://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral" rel="nofollow">wikipedia article</a>). </p> <p>Dropping the condition $t_i \in [x_{i-1}, x_i]$ one gets an integral (<em>McShane integral</em>) equivalent to the Lebesgue integral on the line, whereas the Henstock-Kurzweil integral works also for "non-absolutely convergent" cases like $\int_0^1 \frac{\mathop{sin}(1/x)}{x} dx\;$.</p> <p>As one gets all the nice theorems (<a href="http://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral" rel="nofollow">monotone convergence theorem, dominated convergence theorem, second fundamental theorem of calculus, Lebesgue differentiation theorem</a>) 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).</p> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52764#52764 Answer by fedja for Why should one still teach Riemann integration? fedja 2011-01-21T13:33:18Z 2011-01-21T13:33:18Z <p>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. </p> <blockquote> <p>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. </p> </blockquote> <p>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).</p> <p>It is true that you <em>can</em> 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. </p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52770#52770 Answer by Tom Ellis for Why should one still teach Riemann integration? Tom Ellis 2011-01-21T14:35:28Z 2011-01-21T14:35:28Z <p>Because the Riemann integral is an integral, and the Cauchy integral is an anti-derivative.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52779#52779 Answer by Deane Yang for Why should one still teach Riemann integration? Deane Yang 2011-01-21T17:08:01Z 2011-01-22T13:39:11Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52786#52786 Answer by Dirk for Why should one still teach Riemann integration? Dirk 2011-01-21T19:20:24Z 2011-01-21T19:20:24Z <p>When I was introduced to measure theory, the professor chose to use the <a href="http://en.wikipedia.org/wiki/Choquet_integral" rel="nofollow">Choquet integral</a> to obtain the Lebesgue integral. An this uses the "good old" Riemann integral to integrate the pseudo-inverse of the cumulative distriution function (I think, it was <a href="http://books.google.com/books?id=4m9jGLwntswC&amp;printsec=frontcover&amp;dq=non-additive+measure+and+integral&amp;source=bl&amp;ots=12EfwwxyvN&amp;sig=i7J0LOBQRUijWdNEQehBAqpOgNE&amp;hl=en&amp;ei=pdo5Tar5F4fpOdyc4Z0L&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=3&amp;ved=0CCkQ6AEwAg#v=onepage&amp;q&amp;f=false" rel="nofollow">this book</a>).</p> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52789#52789 Answer by Qiaochu Yuan for Why should one still teach Riemann integration? Qiaochu Yuan 2011-01-21T20:29:41Z 2011-01-22T09:15:13Z <p>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.</p> <p><strong>Definition:</strong> A function $f : \mathbb{R} \to \mathbb{R}$ is <em>integrable</em> 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 <em>integral</em> is $\int f = \sum \int f_n$.</p> <p>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. </p> <p>But you <em>do not need any measure theory.</em> In fact you can define the Lebesgue measure from here. I don't know a reference where this is worked out except <a href="http://math.mit.edu/~rbm/18.102/index.html" rel="nofollow">the notes we used</a>.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52792#52792 Answer by Jeff Strom for Why should one still teach Riemann integration? Jeff Strom 2011-01-21T20:52:32Z 2011-01-21T20:52:32Z <p>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. </p> <p>This is how they can understand the presence of integrals in many contexts. Formally, of course, this is a shorthand for Riemann integration.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52810#52810 Answer by Richard Stanley for Why should one still teach Riemann integration? Richard Stanley 2011-01-22T02:01:37Z 2011-01-22T02:01:37Z <p>A small plus for the Riemann integral: evaluation of limits like $$\lim_{n\to\infty}\sum_{k=1}^n\frac{n}{k^2+n^2}.$$</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52832#52832 Answer by Spencer for Why should one still teach Riemann integration? Spencer 2011-01-22T14:57:52Z 2011-01-22T14:57:52Z <p>Although not a direct answer to the question, this may be relevant to the discussion:</p> <p>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.</p> <p>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"?</p> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/53463#53463 Answer by Stefan Waldmann for Why should one still teach Riemann integration? Stefan Waldmann 2011-01-27T08:19:05Z 2011-12-07T14:43:00Z <p>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 :)</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/53532#53532 Answer by roy smith for Why should one still teach Riemann integration? roy smith 2011-01-27T19:48:48Z 2011-01-30T01:06:47Z <p>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.</p> <p>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). </p> <p>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.</p> <p>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.)</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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. </p> <p>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.</p> <p>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.</p> <p>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.<br> 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.</p> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/57621#57621 Answer by Phil Isett for Why should one still teach Riemann integration? Phil Isett 2011-03-07T00:24:27Z 2011-12-08T03:25:28Z <p>One available compromise is to just work with the following definition of a Riemann integral (which works fine in ${\mathbb R}^n$ as well:</p> <p>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</p> <p>$\psi_1 \leq f \leq \psi_2$</p> <p>and</p> <p>$\int_a^b \left( \psi_2(x) - \psi_1(x) \right) dx &lt; \epsilon$</p> <p>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. </p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/57625#57625 Answer by Carl Offner for Why should one still teach Riemann integration? Carl Offner 2011-03-07T02:02:21Z 2011-03-07T02:02:21Z <p>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.</p> <p>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:</p> <p>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 $x$."</p> <p>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".</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/57634#57634 Answer by Michael Hardy for Why should one still teach Riemann integration? Michael Hardy 2011-03-07T04:35:30Z 2011-03-07T04:35:30Z <p>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.....).</p> <p>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.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/68858#68858 Answer by Anil P for Why should one still teach Riemann integration? Anil P 2011-06-26T16:01:04Z 2011-06-26T16:01:04Z <p>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 ageneral 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 intutive. these three definitions are not constructive but Riemann integral has constructive definition.</p> <p>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.</p>