Let me ask right at the start: what is Riemann integration really used for?

As far as I'm aware, we use Lebesgue integration in:

  • probability theory

  • theory of PDE's

  • Fourier transforms

and really, anywhere I can think of where integration is used (perhaps in the form of Haar measure, as a generalization, although I'm sure this is vastly incomplete picture).

Let me state a well known theorem:

Let $f:[a,b]\to\mathbb R$ be bounded function. Then:

(i) $f$ is Riemann-integrable if and only if $f$ is continuous almost everywhere on $[a,b]$ (with respect to Lebesgue measure).

(ii) If $f$ is Riemann-integrable on $[a,b]$, then $f$ is Lebesgue-integrable and Riemann and Lebesgue integrals coincide.

(I will try to be fair, we use this result and Riemann integration to calculate many Lebesgue integrals)

From this we can conclude that Riemann-integrability is a stronger condition and we might naively conclude that it might behave better. However it does not; Riemann integral does not well behave under limits, while Lebesgue integral does: we have Lebesgue monotone and dominated convergence theorems.

Furthermore, I'm not aware of any universal property of Riemann integration, while in contrast we have this result presented by Tom Leinster; it establishes Lebesgue integration as initial in appropriate category (category of Banach spaces with mean).

Also, I'm familiar with Lebesgue-Stiltjes integral, greatly used for example in probability theory to define appropriate measures. I'm not so familiar with the concept or usage of Riemann-Stiltjes integral, and I'd greatly appreciate if someone could provide any comparison.

As far as I can tell, the only accomplishment of Riemann integration is the Fundamental theorem of calculus (not to neglect it's importance). I'm very interested to know if there are more important results.

To summarize the question:

Where is Riemann integral used compared to Lebesgue integral (which seems much better behaved) and why do we care?


It seems that it is agreed upon that Riemann integration primarily serves didactical purpose in teaching introductionary courses in analysis as a stepping stone for Lebesgue integration in later courses when measure theory is introduced. Also, improper integrals were brought as an example of something Lebesgue integration doesn't handle well. However, in several answers and comments we have another notion - that of a gauge integral (Henstock–Kurzweil integral, (narrow) Denjoy integral, Luzin integral or Perron integral). This integral not only generalizes both Riemann and Lebesgue integration, but also has much more satisfactory Fundamental theorem of calculus: if a function is a.e. differentiable than it's differential is gauge-integrable and conversely function defined by gauge integral is a.e. differentiable (here almost everywhere means everywhere up to a countable set).

Thank you for all the answers. This question should probably be altered to the following form (in more open-ended manner):

What are pros and cons of different kinds of integrals, and when should we use one over the other?

  • 1
    $\begingroup$ The Riemann integral is useful for evaluating integrals, it is arguably more natural in terms of evaluating the area 'under' a function, and (also arguably) easier to teach. $\endgroup$
    – copper.hat
    Oct 26, 2015 at 0:25
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    $\begingroup$ I think there is some element of 'throwing the baby out with the bathwater'. However, I am an engineer, not a mathematician... $\endgroup$
    – copper.hat
    Oct 26, 2015 at 0:34
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    $\begingroup$ @copper.hat As an engineer you would have appreciated being given a more powerful integral at the time you were trained. Your assumption is that that might have meant learning measure theory and a rather unintuitive approach to integration theory. It has been known, however, for more than half-a-century that there are more compelling ways of teaching the Lebesgue integral. The only reason the Riemann integral survived was as a "teaching integral." Its time for this baby to go. $\endgroup$ Oct 26, 2015 at 0:45
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    $\begingroup$ As far as I know, Kurzweil-Henstock's integral is easier to teach than Lebesgue's integral, has the same powerful theorems, manages conveniently improper integrals, and doesn't require functions to be non-negative.. $\endgroup$
    – Bernard
    Oct 26, 2015 at 0:50
  • 3
    $\begingroup$ @Bernard Indeed! Some argue that one can teach that at about the same level as one teaches the Riemann integral now. Then, in graduate school, you characterize the absolute integral by Lebesgue's constructive methods. In the end you have a better understanding of integration theory on the real line. I would bet that there are at least a few graduate students who do not know whether all derivatives are Lebesgue integrable and do not know the relation between the improper Riemann integral and the Lebesgue integral. $\endgroup$ Oct 26, 2015 at 0:56

8 Answers 8


I have tried to make this point before. Usually I find it best to quote from the "gods," in this case the god-given one:

J. Dieudonné, Foundations of Modern Analysis. (Pure and Applied Mathematics, Vol. X) XIV + 316 S. New York 1960. Academic Press Inc.

"Finally, the reader will probably observe the conspicuous absence of the 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.

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, 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. When one needs a more powerful tool there is no point in stopping halfway, and the general theory of (Lebesgue) integration is the only sensible answer."

In all fairness I should point out the case for the defense to be found here:

Regulated Functions: Bourbaki's Alternative to the Riemann Integral,
S. K. Berberian, The American Mathematical Monthly, Vol. 86, No. 3 (Mar., 1979), pp. 208-211.

  • 3
    $\begingroup$ I also don't see any point in stopping 3/4 of the way, especially with a much more complicated definition than the gauge integral. $\;$ $\endgroup$
    – user57159
    Oct 26, 2015 at 3:02
  • $\begingroup$ Thank you for the quote, I wasn't aware that experts find Riemann integral "obsolete". $\endgroup$
    – Ennar
    Oct 26, 2015 at 17:09
  • $\begingroup$ @RickyDemer, thank you for the link. I've just found out about this kind of integral, and it seems much more satisfactory than Riemann integral, especially the appropriate forms of the fundamental theorem of calculus and Lebesgue differentiation theorem. $\endgroup$
    – Ennar
    Oct 26, 2015 at 17:12
  • 6
    $\begingroup$ @Ennar: I agree with the quote, but Dieudonné certainly didn't speak for all "experts". The overwhelming majority of university professors seems to still believe in the didactic value of the Riemann integral. $\endgroup$
    – Stefan
    Oct 26, 2015 at 18:38

The fact that continuous functions are Riemann-integrable is implicit in the background of many general results involving integration on locally compact groups. For instance, Haar measure is typically constructed via the Riesz representation theorem by specifying what the integral of a continuous function with compact support should be, and a common way to do this is by a construction very similar to Riemann sums (see, for instance, the construction of Haar measure in Folland's A Course on Abstract Harmonic Analysis, Theorem 2.10). It is also useful in various arguments to know that the integral of a continuous function can be approximated by taking sums of values of the function at finite sets of points that get denser and denser (i.e., Riemann sums).

Of course, these results don't use any heavy theory of the Riemann integral per se (and are in fact usually formulated using the language and methods of measure theory), but they still demonstrate the value of understanding that (some) integrals can be computed as limits of Riemann sums, even for highly theoretical purposes.

  • $\begingroup$ All integrals on the real line that I can think of at the moment can be computed as limits of Riemann sums. There is a huge difference between committing to using Riemann sums and committing to a study of the Riemann integral. You have made a defence of Riemann sums, as well you should, but this doesn't defend teaching the Riemann integral as anything other than an historical curiosity. $\endgroup$ Oct 26, 2015 at 1:47
  • $\begingroup$ Sure, I did not mean to defend teaching the Riemann integral as a general framework for integration. My point is merely that even within the modern framework of measure theory, it is still useful to have some passing familiarity with Riemann integrals (or sums, if you prefer). In any case, I don't know that it's any easier to compute integrals as Riemann integrals than as Lebesgue integrals--in most computations, you are just using some formal properties (e.g., the fundamental theorem of calculus) that are easy to establish for both of them, rather than directly computing some limit. $\endgroup$ Oct 26, 2015 at 1:52
  • $\begingroup$ Yes. The definition that one chooses and the sequence of theorems one develops for an integration theory don't have much to do with how to compute an integral. Our calculus students believe for a while that they need to use the definition of the Riemann integral for computations, but in the end learn to rely on the fundamental theorem. In Lebesgue's original presentation he did prove that his integral could be expressed as a limit of Riemann sums. He didn't intend, of course, that that would be a method for calculations. $\endgroup$ Oct 26, 2015 at 1:59
  • $\begingroup$ A valid point on usefulness of Riemann sums. Also, thank you for the reference. $\endgroup$
    – Ennar
    Oct 26, 2015 at 17:06

As there is an abundance of opinions on the topic raised by our questioner here (and the topic has attracted a few viewers) I can hardly resist adding another answer. This letter was distributed to publishers' representatives at the Joint Mathematics Meetings in San Diego, California, in January 1997.


To: The authors of calculus textbooks

From: Several authors of more advanced books and articles -


Robert Bartle, USA <mth [email protected]>
Ralph Henstock, Ireland <[email protected]>
Jaroslav Kurzweil, Czech Republic <[email protected]>
Eric Schechter, USA <[email protected]>
Stefan Schwabik, Czech Republic <[email protected]>
Rudolf Výborný, Australia <[email protected]>

Subject: Replacing the Riemann integral with the gauge integral

It is only an accident of history that the Riemann integral is the one used in all calculus books today. The gauge integral (also known as the generalized Riemann integral, the Henstock integral, the Kurzweil integral, the Riemann complete integral, etc.) was discovered later, but it is a "better" integral in nearly all respects. Therefore, we would like to suggest that in the next edition of your calculus textbook, you present both the Riemann and gauge integrals, and then state theorems mainly for the gauge integral.

This switch would only require altering a few pages in your calculus book. Any freshman calculus book is devoted almost entirely to derivatives and antiderivatives -- how to compute them and how to use them; that material would not change at all. The only changes would be in the more theoretical sections of the book -- i.e., the definitions and theorems -- which take up only a few pages.

The reasons for making this change are twofold: (i) It would actually make some parts of your book more readable. Some definitions and theorems can be stated more simply (and more strongly) if the gauge integral is used instead of the Riemann integral. This is particularly true of the Second Fundamental Theorem of Calculus, discussed below. (ii) It would be a better preparation for the handful of calculus students who will go on to higher math courses. The gauge integral is far more useful than the Riemann integral, as a bridge to more advanced analysis.

The idea of introducing the gauge integral to college freshmen is not entirely new; it was promoted, for instance, in the paper "The Teaching of the Integral" by Bullen and Výborný, Journal of Mathematical Education in Science and Technology, vol. 21 (1990). However, we feel that the idea deserves wider promotion; hence this letter.

Introduction to the integral. If you are not already familiar with the gauge integral, we would recommend the article by Bartle in the American Mathematical Monthly, October 1996; it provides a quick introduction to the subject and gives most of the main references. Perhaps a second introduction would be the book of DePree and Swartz (Introduction to Real Analysis, Wiley, 1988); it is written for advanced undergraduates and is probably the most elementary among the currently available introductions to this subject. Professors Bartle and Schechter have volunteered to make themselves available, at least to some extent, to answer further questions you may have about this subject.

The definition. The gauge integral is a very slight generalization of the Riemann integral. Instead of a constant $\epsilon$ and a constant $\delta$, it uses a constant $\epsilon$ and a function $\delta$. The two definitions can be formulated so that they are nearly identical, and then placed side by side. This would be helpful to the teachers who are using the book and learning about the gauge integral for the first time.

We think that the slight increase in the complexity of the definition will make little difference to the students learning from the book. It has been our experience that, for the most part, freshman calculus students do not fully grasp the definition of the Riemann integral anyway; it is just too complicated for students at that level. The definition of the Riemann integral is included in a calculus book more for completeness and integrity than for teaching. The few students who have enough mathematical maturity to grasp the definition of the Riemann integral will probably have no greater difficulty with the gauge integral, and they will benefit from being exposed to this concept in their calculus book.

Existence and nonexistence of integrals. In recent calculus books it is customary to state without proof that continuous functions and monotone functions on a compact interval are Riemann integrable. The omission of the proofs is unavoidable -- a proof would involve the completeness of the reals, uniform continuity on compact intervals, and other notions that are far beyond the reach of freshmen.

Any Riemann integrable function is also gauge integrable. Many more functions are gauge integrable, as will be evident later in this letter; see especially our remarks about the Dominated Convergence Theorem.

[Remainder of the letter omitted but can be found on the internet. We have since lost Bartle, Henstock, and Schwabik. I hope the others are well.]

  • 1
    $\begingroup$ Thank you for your answer. I've just learned of the notion of gauge integral and I must wholeheartedly agree that Fundamental theorem of calculus is much nicer in this form. I will look into the references as I find this very interesting, and was rusty with formalities of Riemann integration anyway. This seems more worthwhile than rereading about Riemann integration. $\endgroup$
    – Ennar
    Oct 26, 2015 at 17:28
  • $\begingroup$ @Ennar What is it that makes the form of the second fundamental theorem of calculus nicer than the riemann integral version? $\endgroup$
    – snulty
    Oct 14, 2016 at 16:35
  • $\begingroup$ @snulty, if $f$ is R-integrable, then its primitive function is continuous, but not necessarily differentiable almost everywhere. Conversely, derivative of almost everywhere differentiable function needs not be R-integrable. Gague integral takes care of this: (end of paragraph) $\endgroup$
    – Ennar
    Oct 15, 2016 at 11:42
  • $\begingroup$ @Ennar Hmm. Is there a link to simple examples of each? An a.e. Differentiable function who's derivative isn't Riemann integrable, and a Riemann integrable function whose promote isn't differentiable a.e.? $\endgroup$
    – snulty
    Oct 15, 2016 at 11:57
  • $\begingroup$ @snulty, Volterra's function for the first example, but for the second part I was mistaken, primitive of R-integrable function is differentiable a.e. since R-integrable function is continuous a.e. $\endgroup$
    – Ennar
    Oct 15, 2016 at 12:46

Let me turn the floor over to another mathematician who wrote on this subject and published a Monthly article on it. (He is no longer around to have his say and I'd like to remember him here.)

... from Robert G. Bartle's review of the monograph The general theory of integration, by Ralph Henstock. Oxford Mathematical Monographs, Clarendon Press, Oxford, 1991, xi 262pp., ISBN 0-19-853566-X

"In elementary calculus courses we are usually successful in teaching students to evaluate an integral of a suitable function $ f = F'$ on an interval $ [a, b]$, by evaluating $F(b)−F(a)$, but we are often not very successful in connecting this type of integration with Riemann sums and their limits. During their junior/senior year, students who are studying mathematics seriously are then led through a more careful and exhaustive discussion of these ideas. However, they are informed that all of this is only tentative, since when they become graduate students they will replace the outmoded Riemann integral that they have just mastered with the Lebesgue integral.

Of course, it is not completely replaced by this new integral, because there are certain notions, such as 'improper integrals,' that do not fall under this new umbrella and are still of considerable importance; moreover, almost all evaluations of integrals (whether Riemann or Lebesgue) are found by using the $F(b) − F(a)$ method, with a few minor variations. We tell our advanced undergraduates that we would like to introduce them to the Lebesgue integral but cannot do so since it requires a prior study of measure theory and/or topology and is 'too advanced' for them at their present stage of mathematical study. Probably none of us is satisfied by this circuitous procedure.

Suppose that someone came up with an approach to the integral that simultaneously covered the integration of all functions that have antiderivatives, all functions that have Riemann integrals, all functions that have improper integrals, and all functions that have Lebesgue integrals. Moreover, suppose that the definition of this 'superintegral' was only slightly more complicated than that of the Riemann integral, that its development required no study of measure theory, no study of topology, and that this integral had properties that correspond to the Monotone Convergence Theorem and the Lebesgue Dominated Convergence Theorem (among others).

If this mathematical miracle occurred, then wouldn’t this new approach be immediately adopted, at least at the junior/senior level course, and quickly worked into the calculus level? The answer is a resounding: No!

Proof. In fact, such an integral has already been developed and has been around for some time, but its existence has remained largely unknown (except to readers of the Real Analysis Exchange) and it has had very little, if any, educational impact (known to this reviewer)."

See also the article Robert G. Bartle, Return to the Riemann Integral Amer. Math. Monthly 103 (1996), no. 8, 625–632.

  • 3
    $\begingroup$ "In fact, such an integral has already been developed and has been around for some time, but its existence has remained largely unknown (except to readers of the Real Analysis Exchange) and it has had very little, if any, educational impact (known to this reviewer)." - a bold claim. Does the article say what integral this is? $\endgroup$ Oct 26, 2015 at 2:17
  • $\begingroup$ The Monthly article gives a full account and is quite readable. In fact Bartle was awarded the Paul R. Halmos - Lester R. Ford award for expository excellence. It is curious that he wrote this article nearly 40 years after the ideas became known to specialists. $\endgroup$ Oct 26, 2015 at 2:20
  • 2
    $\begingroup$ I don't have a copy of American Mathematical Monthly Volume 103, Number 8, so I can't just read the article. $\endgroup$ Oct 26, 2015 at 2:24
  • 2
    $\begingroup$ It is only a click away or a Google search. I found math.tut.fi/courses/73129/Bartle.pdf but there are others. $\endgroup$ Oct 26, 2015 at 2:29
  • $\begingroup$ Sweet. Nothing useful came up when I searched for the journal issue. $\endgroup$ Oct 26, 2015 at 2:35

I think it's ridiculous to call Riemann integrals obsolete. It's like saying rational numbers are obsolete because we have real numbers and they are complete while the rationals are not.

  1. In complex analysis, contour integrals are defined in a very similar way to the construction of Riemann integrals, not by measure theory. Students do not, and should not, need to learn about measure theory before learning complex analysis.

  2. The notion of equidistribution of a sequence is closely related to Riemann integrals, not Lebesgue integrals.

  3. Many practically-minded people are led to integrals in mathematical models for physical quantities based on the Riemann sum idea, not on measure theory (e.g., the expression of pressure or work as an integral).

  4. Monte Carlo integration, a practical method of computing high-dimensional integrals, is based on generalizing the idea of Riemann integration, not Lebesgue integration.


Try $\int \frac{\sin t}{t}$. This is an old example and is partially bogus. It is valid in that $\int \left| \frac{\sin t}{t} \right| \not < \infty$. However, if we instead evaluate as $\lim_{A \rightarrow \infty} \int_{-A}^A \frac{\sin t}{t}$, we find (after a couple of tricky manipulations) that this is just as Lebesgue integrable as it is Riemann integrable. However, the Riemann integral "cheats" -- it takes the (asymmetric) limit version of the integral as its definition, whereas the Lebesgue integral does not.

I'm pretty sure I can teach a computer, physicist, and engineer to compute Riemann sums and explain how the Riemann integral is the limit of such sums. I can imagine the profound difficulties of teaching them Lebesgue integrals without starting with Riemann integrals.

An analogue of your question in Physics is "why do we teach people Newtonian Gravity when General Relativity is so much better?" And the answer there is similar: Some things are more easily computed via Newton and trying to understand Einstein without grasping Newton just doesn't work.

  • $\begingroup$ I don't follow. The integral of $t^{-1}\sin t$ on $[0,\infty)$ does not exist as a Lebesgue integral, only as an improper Lebesgue integral exactly the way it is interpreted for the improper Riemann integral. Also I am pretty sure that I could teach the Lebesgue integral to someone who had learned the Newton integral or the regulated integral and by-passed the Riemann integral altogether. If you mean, instead, that I need to teach the Lebesgue integral to someone who has never seen any theory of integration ... well I agree it would be a mess. $\endgroup$ Oct 26, 2015 at 8:38
  • $\begingroup$ @B.S.Thomson : In Lebesgue integration $\int_\mathbb{R}$ is atomic and not defined by a limit. In Riemann integration $\int_\mathbb{R}$ is defined as the limit, so a definitional mismatch gives Riemann integration an "advantage". And, yeah, I was thinking of ab initio teaching of Lebesgue integration. $\endgroup$ Oct 26, 2015 at 12:49
  • $\begingroup$ Yes, this is definitely a valid point to use Riemann integral as a stepping stone to Lebesgue, but I wanted to know do we care after that step? Improper integrals are a good example. $\endgroup$
    – Ennar
    Oct 26, 2015 at 17:21
  • $\begingroup$ I'd argue that General Relativity is only better in the sense that it is a more complete theory of gravity. For anything other than pretty much the most fringe cases Newtonian Gravity works pretty much just as well. Also, GR is orders of magnitude more complicated than Newtonian Gravity $\endgroup$
    – Chris
    Nov 13, 2020 at 22:35

This rather open-ended problem has a variety of "answers" and I hope the readers will tolerate some more. I have already given Dieudonne's viewpoint (which likely represents the Bourbaki position). Bob Bartle's response was after he had already written more conventional (successful) textbooks. The Open Letter from the group of research mathematicians should be read too.

Here is my own answer, too long for this web site, but a link is provided for the curious.

Thomson, Brian S. Rethinking the elementary real analysis course. Amer. Math. Monthly 114 (2007), no. 6, 469–490. DownLoad Link

Before clicking the link and reading perhaps we should allow Peter Bullen to have a word. He is a recognized authority on integration theory with many research and expository papers on a variety of nonabsolutely convergent integrals. This is his review of the Monthly paper.

Math Reviews MR2321251 (2008c:26001) 26-01 (26A06 97D80)

This is an important paper for mathematical education and should be read by anyone planning a course or about to write a text on analysis or calculus. Ever since Henstock rather brashly said “Lebesgue is dead!” at the Stockholm meeting in 1962 there have been many hardy souls, colleagues, students and disciples of either Henstock or Kurzweil suggesting the more reasonable and practical “Riemann is dead!”. Many papers have been written that make this point, as well as books by both Henstock and Kurzweil, the author of the present paper, Bartle, DePree and Schwarz, Leader, Lee and Výborný, Mawhin, McLeod, McShane, and no doubt others; as well Dieudonné, in the book that is quoted at the beginning of the present article, made the same point from a completely different point of view. All argue that by now the teaching of both the Riemann and Riemann-Stieltjes integrals should cease; these integrals should be replaced by the Newton and Cauchy integrals in elementary calculus courses and by the generalized Riemann integral in elementary analysis courses, the latter being taught in a way that would lead into Lebesgue theory in more advanced courses. To date the conservative nature of academia has not heard these arguments.

The present paper suggests a very attractive method for the elementary analysis course mentioned above. The author has a rather extreme aversion to the plethora of gauges and tags that are the norm in most approaches to the generalized Riemann integral. Instead he suggests that the basis of the analysis course should be Cousin’s Lemma. This very elementary formulation of the completeness axiom of the real line allows for proofs of all the properties normally deduced from that axiom, including the properties of continuous functions normally considered in basic analysis courses— proofs that are both simple and transparent. It further leads very naturally into the generalized Riemann integral and later to measure theory if that is desired. The paper is very clearly written and is very persuasive but is not an easy read, especially towards the end, and the reader may need the help of a standard book on the generalized Riemann integral. In addition the author shows that a simple recent extension of Cousin’s Lemma will allow some very neat proofs of more subtle properties that may or may not be appropriate for the first analysis course but certainly would allow for the progression to a more advanced course and to measure theory.

It is difficult to change a very well-established academic and publishing tradition but the reviewer hopes that this article will start at least the beginnings of a change. There are already non- or semi- commercial web texts that are moving in this direction, some under the influence of the Dump-the- Riemann-Integral-Project (see http://classicalrealanalysis.info/resources-drip.php) and others less radical being produced by the Trillia Group (http://www.trillia.com). These should all be explored by anyone teaching in this field.

Reviewed by P. S. Bullen

  • $\begingroup$ Thank you for yet another answer. You've really given me material for reading. $\endgroup$
    – Ennar
    Oct 27, 2015 at 12:19

There is one theorem in ergodic theory where Riemann integrability appears to be the right assumption. Let $\alpha$ be an irrational number in $[0,1)$ and consider $\varphi\colon[0,1)\to[0,1)$, the translation by $\alpha$ mod 1, $$ \varphi(x) := (x+\alpha) \text{ mod } 1. $$ Let $R_\pi[0,1]$ denote the set of all Riemann integrable functions $f$ on $[0,1]$ with $f(0)=f(1)$ and consider the Koopman operator $$ R_\pi[0,1] \ni f \mapsto Tf := f\circ\varphi \in R_\pi[0,1]. $$ Then $T$ is mean ergodic, $$ \frac1n \sum_{k=0}^{n-1} T^kf(x) \to \int_0^1 f(t)\,dt $$ uniformly for $x\in[0,1]$.

Note that the restriction of $T$ to $C_\pi[0,1]$ is also mean ergodic, but this space is too small for applications such as the following: the sequence $(n\alpha\text{ mod }1)_{n\in\mathbb N}$ is equidistributed in $[0,1)$, meaning that the relative frequency of sequence elements falling into any given interval $[a,b]\subseteq[0,1)$ converges to $b-a$.

Source: Operator Theoretic Aspects of Ergodic Theory by Eisner, Farkas, Haase and Nagel.

  • $\begingroup$ Thanks. Do you think ergodic theory would suffer much if future generations preferred to describe this as "the space of all bounded a.e. continuous functions?" $\endgroup$ Oct 28, 2015 at 21:12
  • $\begingroup$ Thank you for the answer. I'm not very familiar with ergodic theory, but I appreciate the point well made which directly responds to my question. $\endgroup$
    – Ennar
    Oct 28, 2015 at 21:51
  • $\begingroup$ @B.S.Thomson: I don't claim that my answer could be taken as a reason to teach the Riemann integral, and you're right that ergodic theory wouldn't really suffer. However, the proof of the theorem is straightforward if you work with the definition of the Riemann integral. I didn't think about how it would work with the characterization you give. $\endgroup$ Oct 29, 2015 at 7:13

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