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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?
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    See The Mathematics Of Quantum Mechanics by Prugoveki. Lebesgue is necessary throughout.2016-03-26

4 Answers 4

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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 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.

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    @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.2011-07-25
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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.

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    Hard to know what context a question assumes...2011-07-26
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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 $\lim_{n\rightarrow\infty} \int_{a}^{b} f_n(x) dx = \int_{a}^{b} f(x) dx$.

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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.