70
$\begingroup$

This question has puzzled me for a long time. It may be too vague to ask here. I hope I can narrow down the question well so that one can offer some ideas.

In a lot of calculus textbooks, there is usually a chapter about "applications" after the one for Riemann integral. Students can do a lot of calculations and appreciate the power of Riemann integration --- they solve many problems in physics and geometry.

While learning the Lebesgue integral, or more generally, integration on measure space, I cannot appreciate the power of this kind of integration util I learn some modern PDE. One the other hand, I found that there are much much more inequalities when doing Lebesgue integration than equations when applying Riemann integral. Instead of calculating something, people do estimation with the convergence theorems.

Here are my questions:

  • How do people apply Lebesgue integration theory? If putting the methods into categories, can I say that it primarily deals with the problems related to convergence?
  • What's the fundamental difference between applying these two different integration techniques?
  • Is there an example such that people solve some problem which may be very hard (but still can be solved) when using Riemann integral but relatively easy with Lebesgue integral?
$\endgroup$
9
  • 3
    $\begingroup$ For the third question I would look at the field of Fourier series. For example, it is relatively easy to show that every square-summable function on the unit circle has a unique Fourier series converging to the original function in $L^2$-norm. This, however, relies on Hilbert space techniques which in turn rely on Lebesgue integral, to ensure completeness of $L^2(\mathbb{T})$. How to establish an analogue result with Riemann integrals? $\endgroup$
    – Giuseppe Negro
    Jul 22, 2011 at 17:09
  • 2
    $\begingroup$ In the land of Riemann integrals, you need to show uniform convergence in order to justify switching limits and integrals, which can be rather painful to work out. It's just a whole lot easier (so easy that its fun) to do it with Lebesque integration and it's convergence theorems. $\endgroup$
    – John M
    Jul 22, 2011 at 17:28
  • 8
    $\begingroup$ The fundamental difference is that the space of Riemann integrable functions is not complete, i.e. not closed under taking limits, which makes analysis hard to do. The $L^p$ spaces are complete, which allows us to take advantage of the fact, for example, the smooth, compactly supported functions are dense in $L^p$, so we can define and show things on smooth functions, then take limits to extend our result to any $L^p$ function. All this is impossible in the smaller space of Riemann integrable functions. $\endgroup$
    – John M
    Jul 22, 2011 at 17:34
  • 2
    $\begingroup$ This old paper is mostly devoted to exactly what you're asking about: Ralph Palmer Agnew, Convergence in mean and Lebesgue integration, American Mathematical Monthly 44 #1 (January 1937), 4-14. $\endgroup$
    – Dave L. Renfro
    Mar 26, 2014 at 20:54
  • 1
    $\begingroup$ See The Mathematics Of Quantum Mechanics by Prugoveki. Lebesgue is necessary throughout. $\endgroup$
    – DanielWainfleet
    Mar 26, 2016 at 13:20

4 Answers 4

Reset to default
38
+150
$\begingroup$

For the second question, you have that whenever the Riemann integral is defined, so is the Lebesgue integral. The only thing I'm aware of that you have for the Riemann integral that you don't for the Lebesgue integral is the fundamental theorem of calculus. This is more because the functions that Riemann-integbrale functions are nice, and possess more features than $L^1$ functions.

In some sense, what Riemann integration lacks is not the inability to integrate functions that you care about (because most of the functions that people are concerned with are at least piece-wise continuous), but rather that it is easy to take the limit of continuous functions and get discontinuities, and so you lack a lot of nice formal properties becasue you cannot work by going from nice functions and passing to limits. Therefore, if something is true for a function, you must be able to show it directly. For Lebesgue integration, the fact that you can take limits means you get completeness of the $L^p$ spaces, and thus that you can bring Hilbert/Banach space techniques to bear.

Here are a few applications:

  • The use of function spaces ($L^p$ spaces, sobelev spaces, etc). As you mentioned, weak derivatives allow you to solve PDE's in a two step process: show that a weak solution exists, and then prove some sort of regularity result that the weak solution is actually a real function. However, there are many many more uses of these function spaces, and they allow you to do some pretty neat things.

  • Fourier transforms. There are a number of results on convergence of Fourier transforms, and function spaces are necessary to prove them. It has been said that putting Fourier transforms on solid footing was the first major triumph of Lebesgue integration

  • Probability. The modern point of view of probability is based on measure theory: a probability space is just a measure space of measure $1$, and an "event" is just an element of a sigma algebra. Even for a simple problem like flipping a coin an infinite number of times, this allows you to make sense of the fact that some "events" just aren't talked about, because they aren't part of the sigma algebra, which lets you escape seeming paradoxes related to infinity. Once you have recast the definitions or probability, you can then phrase many results in terms of integration, where their proofs become rather easy, at least compared to the traditional proofs. It also allows for discussing more complicated stochastic processes, like stochastic differential equations, and I honestly don't even know if there are fully satisfactory ways to talk about these things without the measure theoretic point of view.

$\endgroup$
10
  • 15
    $\begingroup$ Also, I don't understand what you mean by saying that you don't have the FTC for Lebesgue. Sure you have! You even have a complete characterization of the functions for which it holds. See e.g. my answer here $\endgroup$
    – t.b.
    Jul 22, 2011 at 18:13
  • 6
    $\begingroup$ @Quinn: No. The function $f(x) = \frac{\sin x}{x}$ is not Riemann integrable on $(0,\infty)$, but rather improperly Riemann integrable on $(0, \infty)$. Riemann integration is only defined for bounded functions on bounded intervals, so Aaron is correct. Improper Riemann integration is a separate process. (see also my answer here) $\endgroup$
    – Jesse Madnick
    Jul 22, 2011 at 19:06
  • 9
    $\begingroup$ @Mark: I've never seen anyone actually compute a non-trivial Lebesgue integral from its definition without resorting to Riemann integration theory at some point. I don't really understand what you're trying to tell me and why, I didn't say anything on Riemann integration and its merits. $\endgroup$
    – t.b.
    Jul 22, 2011 at 19:18
  • 2
    $\begingroup$ @Jesse Okay, thanks. Though I'm not sure that most people intend to exclude the improper integral when they speak of the Riemann integral. Do you think they do? In any case, when a statement like "...whenever the Riemann integral is defined, so is the Lebesgue integral." is made, it is probably important to ensure that the reader/listener understand that Riemann integral does not include the improper Riemann integral. $\endgroup$
    – Quinn Culver
    Jul 24, 2011 at 16:30
  • 2
    $\begingroup$ @Noah Yes. And whether you want to consider such improper integrals depends on context: $\int_0^{\infty} \sin(x)/x dx$ makes sense if you are asking for long term behavior of a system or the value of a certain contour integral, but not if you are asking "what is the area under the curve." In higher dimensions where the space can be exhausted in different ways, there may be no canonical way to integrate a non-$L^1$ functions. However, because $\mathbb{R}$ is ordered, we can "integrate" things we can't with a purely global theory. The result just may or may not be garbage. $\endgroup$
    – Aaron
    Jul 25, 2011 at 23:06
8
$\begingroup$

Here's an example which shows the power of Lebesgue integration theory.

The following theorem is sometimes useful in calculus(only continuous functions are involved). It is an easy special case of Lebesgue dominated convergence theorem. However, it's difficult to prove within Riemann integration theory.

Let $M$ be a real number such that $0 \le M < \infty$. Let $(f_n)$ be a sequence of continuous functions on a finite interval $[a, b]$ such that $|f_n| \le M$ for every $n$. Let $f(x)$ be a continuous function on $[a, b]$. Suppose $\lim_{n\rightarrow\infty} f_n(x) = f(x)$ for every $x \in [a, b]$.

Then $\lim_{n\rightarrow\infty} \int_{a}^{b} f_n(x) dx = \int_{a}^{b} f(x) dx$.

$\endgroup$
7
$\begingroup$

The Riemann integral depends heavily on the structure of the line, where the general process of integration does not. Because of the way the Riemann integral is constructed, continuity of integrands is very important. We now have two abstractions "siamese twinned": integration and topology. The Lebesgue integral is an excellent abstraction that gives a minimal structure in which to do integration. This is cleaner and better.

$\endgroup$
7
  • $\begingroup$ I don't understand this at all. Lebesgue theory is built on the Borel sigma algebra and so requires a topology. Or one can view it as a Haar measure on a topological group. Again, topology. In what sense is the Lebesgue theory minimal and Riemann integration isn't? Perhaps you were talking about general measure theory (although there is also an approach to this through continuous functionals) but how is that relevant to the question which is explicitly concerned about the real line then? $\endgroup$
    – Marek
    Jul 25, 2011 at 22:26
  • 2
    $\begingroup$ Sometimes "Lebesgue theory" is used to refer not only to the tangible Lebesgue integral on $\mathbb R^n$, but also to the abstracted version that uses any $\sigma$-algebra, defines measurable function, defines integrals for positive real-valued as sup of the obvious sums attached to finite positive linear combinations of characteristic functions of measurable sets, etc. The ambiguity doesn't help a discussion such as this! But, yes, it was not obvious in 1900 that integration could be separated from topology. Not that we want to do so in applications... :) $\endgroup$
    – paul garrett
    Jul 25, 2011 at 22:35
  • $\begingroup$ @paul: you mean there are people who call all of measure theory a Lebesgue theory? That would indeed be unfortunate. However, I've never heard anybody use term Lebesgue theory that way (I am mostly familiar with measure theory from probability theory literature though). $\endgroup$
    – Marek
    Jul 25, 2011 at 23:00
  • 1
    $\begingroup$ When I refer to Lebesgue theory, I mean the general integral on a abstract measure space. These spaces have no topology. This achieves a separation of abstractions. $\endgroup$
    – ncmathsadist
    Jul 26, 2011 at 0:11
  • 1
    $\begingroup$ @marek, yes, indeed, there are many people who (as a convenience? out of laziness? imitating others?) call all measure theory "Lebesgue theory". Historically, this is pretty inaccurate, in several ways, but that's never stopped anyone... :) And/but as ncmathsadist says, the point (apart from names) is that a large part of measure theory can be separated from topology, and this possibility is technically enlightening. $\endgroup$
    – paul garrett
    Jul 26, 2011 at 0:26
3
$\begingroup$

Lebesgue Integral is used for the Birkhoff Ergodic Theorem. From these notes

Let $(X, \mathcal{B}, \mu)$ be a probablity space and let $T: X \to X$ be an ergodic measure-preserving transformation. Let $f \in L^1 (X, \mathcal{B}, \mu)$ then $$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} (f \circ T^k)(x) = \int f \, d\mu $$ For $x$ almost everywhere in $X$.

That's Lebesgue not Riemann. That convergence is not guaranteed to occur at any particular point.

$\endgroup$

You must log in to answer this question.