Math people:

I think it is a good idea to teach beginning calculus students the Riemann Integral (I refer to what calculus books call the "Riemann Integral" and ignore any controversy about whether it should be called the Darboux integral, etc.), since it gives a rigorous definition to the idea of "area under a curve" and you don't need measure theory to define it. But is the Riemann integral ever used in "real" mathematics by mathematicians who know Lebesgue integration? Is there any purpose in proving difficult-to-prove properties of the Riemann integral (say, in a real analysis class)? The Lebesgue integral is more powerful and has those properties, plus more.

The only uses I can think of are: (i) using it as as trick to evaluate certain infinite sums or limits and (ii) a jumping-off point for learning the Riemann-Stieltjes integral.

EDIT: I just discovered a similar, unanswered question at Are Specific Facts about the Riemann Integral Logically Required? . If you read it, you may understand my question better.

  • 3
    $\begingroup$ But that is a silly example: why would they want to integrate the Dirichlet function, whose only purpose is to exhibit a simple example of a function which cannot be integrated using Riemann's integral? $\endgroup$ – Mariano Suárez-Álvarez Aug 30 '13 at 16:14
  • 1
    $\begingroup$ Riemann integral can change the world. $\endgroup$ – Kaster Aug 30 '13 at 16:26
  • 8
    $\begingroup$ The Riemann Integral and its relatives, such as the Trapezoidal Rule, are fundamental for estimation. $\endgroup$ – André Nicolas Aug 30 '13 at 16:34
  • 2
    $\begingroup$ @AndreNicolas Excellent point, Andre $\endgroup$ – Stefan Smith Aug 30 '13 at 16:40
  • 2
    $\begingroup$ On the other hand, there are functions whose Lebesgue integral doesn't converge, like $\frac{\mathrm{sin} x}{x}$ but they have improper Riemann integral. This is why Lebesgue integrable and Riemann integrable functions aren't subsets (with Dirichlet function being the other counter example). Indeed both classes belong to measurable functions. $\endgroup$ – NumberFour Aug 30 '13 at 17:14

I think it is a bit of a shame that the standard pedagogical motivation for the Lebesgue integral seems to involve "dumping on" the Riemann integral.

There is (of course) a sense in which the Lebesgue integral is stronger: the collection of Lebesgue integrable functions properly contains the collection of (properly!) Riemann integrable functions, so the Lebesgue integral is "better".

As Mariano has pointed out in the comments, this is not necessarily very convincing: the standard examples of bounded, measurable, non-Riemann integrable functions look rather contrived.

In my opinion, most of the true advantage of the Lebesgue integral over the Riemann integral resides in the Dominated Convergence Theorem. This all-important result is much harder to prove directly for the Riemann integral. In part of course it is hard to prove because it is not true that a pointwise limit of Riemann integrable functions must be Riemann integrable, but again that's not where the crux of the problem lies. In the setting of the DCT if we add the hypothesis that the limit function is Riemann integrable then of course the theorem holds for the Riemann integral...but try to prove it without using Lebesgue's methods! (People have done this, by the way, and the difficulty of these arguments is persuasive evidence in favor of Lebesgue.)

I honestly think that in many (certainly not all, of course) areas of mathematics, it is the DCT (and a couple of other related results) which is really important and not the attendant measure theory at all. Thus I wish the approach via the Daniell integral were more popular: e.g. I can imagine an alternate universe in which this is part of undergraduate analysis and "measure theory and Lebesgue integration" was a popular "topics" graduate course rather than something that every young math student cuts her teeth on and many never use again. If measure theory were more divorced from the needs of integration theory one would naturally be tempted to either introduce more geometry or make explicit the connections to probability theory: either one of these would be a major livening up of the material, I think.

Right, but I'm meant to be answering the question rather than ranting.

There is another sense in which the Riemann integral is stronger than the Lebesgue integral: since Riemann's definition of Riemann integrability is a priori so demanding, knowing that a function is Riemann integrable is better than knowing it is Lebesgue integrable. It can be used to evaluate certain limits, yes, but this is not just a trick! Rather, the fact that an incredibly broad range of "interpolatory sums" associated to e.g. an arbitrary continuous function all converge to the same number is incredibly useful. As I have said before and others have said here, the entire branch of analysis known as approximation theory sure looks like it is founded upon the back of the Riemann integral, not the Lebesgue integral. In this branch of mathematics one is interested in various interpolatory schemes closely related to Riemann sums, and often one looks for a good tradeoff between convergence rates, efficiency and so forth in terms of the amount of smoothness of the function. An approximation scheme which worked for every $C^2$ function, for instance, would be regarded as quite general and useful. Does a numerical analyst ever meet a non-Riemann-integrable function?

It helps to fix ideas to restrict to the characteristic function $1_S$ of a bounded subset $S \subset \mathbb{R}^n$. Then $1_S$ is Lebesgue integrable iff $S$ is Lebesgue measurable. A general Lebesgue measurable set can be quite pathological. On the other hand, $1_S$ is Riemann integrable iff $S$ is Jordan measurable; this is a less well-known concept but is both technically useful and in some respects more natural. The fact that the volume of a Jordan measurable set can be computed as a limit of lattice-point counting is a key idea linking discrete and continous geometry. Just as an example, this came up (in a very standard and well-known way) in a paper I wrote recently: see Proposition 3.7 here. Geometric facts like these fail for, say, the characteristic function of the rational points in $[0,1]^d$.

Here is a somewhat related instance of Riemann integrability: a sequence $\{x_n\}$ in $[0,1]$ is uniformly distributed iff for all Riemann integrable functions $f: [0,1] \rightarrow \mathbb{R}$, $\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N f(x_n) = \int_0^1 f$. (See e.g. Theorem 7 of these notes.) On the right hand side it (of course) doesn't matter whether you take the integral to be in the sense of Riemann or Lebesgue, but if $f$ is not Riemann integrable then nothing good needs to happen on the left hand side. This is another instance in which Riemann integrable functions are better.


I can think of a few reasons, most of which have already been mentioned in the comments.

  1. It provides some level of intuition for what it means to integrate. Sure, it can't integrate many of those crazy functions mathematicians love to construct, but I for one cannot visualize the Lebesgue integral nearly as well as the Riemann integral. Pedagogical value is mathematical value, in many respects.

  2. It is identical to the Lebesgue integral for continuous functions on compact subsets of Euclidean space anyways, so it seems very much worth it to prove that attempts to calculate the Riemann integral for such situations are not efforts to calculate the "wrong" value.

  3. Numerical integration is always done using a variant of the Riemann integral formulation. All numerical schemes involve some sort of partition of the domain of integration, after which the integral in each element of the partition is estimated by approximating the graph of the function by something simpler. I think more than a handful of people would agree that numerical integration is good for something...

Edit: Potato has a great point that the Fundamental Theorem of Calculus is much easier to prove (for nice functions) using the Riemann integral formulation, whereas generalizations to the Lebesgue case like the Lebesgue Differentiation Theorem requires more machinery, like maximal functions, etc. Surely you wouldn't want calculus students to wait to that point to learn such an important concept.

  • $\begingroup$ Chris, does (2) mean "It is identical to the Lebesgue integral for continuous functions whose domains are compact subsets of $\mathbb{R}^n$"? $\endgroup$ – goblin GONE Aug 30 '13 at 16:52
  • 1
    $\begingroup$ Haha, yes, it just dawned on me that "Euclidean domain" means something else in algebra - I will change that. $\endgroup$ – Christopher A. Wong Aug 30 '13 at 17:02
  • 2
    $\begingroup$ Yeah I was just like, "Wow, ideas from abstract algebra are showing up in elementary analysis. Wait...." $\endgroup$ – goblin GONE Aug 30 '13 at 17:03
  • 1
    $\begingroup$ Good answer. I would add that it's much easier to show the FTC and start actually computing things ($\int_0^1 x^2$, etc) with the Riemann integral than with the Lebesgue integral. There's a lot of pedagogical value in seeing this (and things like integration by parts) in a simple case (that suffices most of the time!) first. $\endgroup$ – Potato Aug 30 '13 at 17:44

The Riemann integral is good for at least two things:

  • Definition to be used in logical arguments; and
  • Numerical integration.

In courses where logical rigor is not expected, I'd omit Riemann's definition of the integral. One could merely say it's the area under a curve.

But one approach to intuitive definition says that rather than using an integral to find an area, you use and area to approximate an integral. An integral is a sum of infinitely many infinitely small quantities $f(x)\,dx$, and $f(x)$ could be in units of force and $dx$ in units of distance so that $f(x)\,dx$ is in units of energy, and it's not about area under a curve at all. But draw the graph of $f(x)$ as a function of $x$, and use the picture to estimate the area, and you've got an estimate of the energy. Riemann's definition is obviously not involved in all this.

Riemann's definition gives criteria for when a number is too small or too large to be the integral. Rather than defining infs and sups, I might say the integral is defined as the only number (if there is only one) that is neither too large nor too small.

(In the same way, Lebesgue's definition, applied to non-negative functions, says the integral is the smallest number that's not too small. And if every number is too small, then it's $\infty$. The criterion for when it's too small is simple. The complicated part is only constructing the measure and showing it's countably additive, etc.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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