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As someone who has taken two semesters of real analysis, having been exposed to the rigorous definition of the Riemann-Stieltjes integral - why should I learn Lebesgue integration? The Riemann integral already furnishes us with a means of integrating continuous, bounded functions and functions with finitely-many jump discontinuities. What compelling advantage does the Lebesgue formulation offer to Riemann's?

My interest is in an answer possessing utility in the sciences or engineering fields.

Mikhail Katz
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zaccandels
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    ''Dirichlet function or any other bizarre, exotic function with no utility in the sciences or engineering fields''...you do understand that finance, for example, is almost entirely based on Brownian motion and stochastic processes ? – Hamdiken May 27 '24 at 10:15
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    Two reasons: probability, and we also get the exchange of limit and integral with the help of some machinery. But someone more qualified than me can expand on that. – Sean Roberson May 27 '24 at 10:16
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    According to Lebesgue, the measure of irrational numbers on $(0,1)$ is $1$; according to Jordan, it does not exist. The Riemann integral is additive, Lebesgue sigma-additive, etc., so, Lebesgue have more wide possibilities. – zkutch May 27 '24 at 10:19
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    @Hamdiken not sure how that is a counterpoint to what the OP said, however. – Andrew May 27 '24 at 10:31
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    Imagine someone who knows no analysis asks what is the point of the real numbers, why can’t we live with just the rationals? The answer is that the rationals are not complete, but then said person will rightfully ask so what? The unfortunate fact of the matter is that you only realize the power of the completion once you have studied enough analysis. In a similar vein, the perhaps unsatisfactory answer to your question is that you will have to wait and see why we care about the Lebesgue integral, which is in some sense, the completion of the Riemann integral. – Andrew May 27 '24 at 10:41
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    If only looking for utility in some branch of science or engineering, probably no reason to study it at all. It's a bit of an empty question though as it could be asked about any topic. Life is short, do it if you think you will enjoy it. – Paul May 27 '24 at 11:08
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    @Andrew Brownian motion is an example of an "exotic" function, according to what OP said. – Hamdiken May 27 '24 at 12:12
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    @Paul I really don't think it's an empty question. I'm deeply interested in mathematics and I want to know why the Lebesgue integral has supplanted the Riemann integral as the integral par excellence in mathematics. Surely there's some compelling reason. – zaccandels May 27 '24 at 12:50
  • @Andrew Ok, fair enough - this is sufficient motivation for me to study the Lebesgue theory of integration. Now I'm curious though, how do you feel about the finitist position in mathematics? – zaccandels May 27 '24 at 12:51
  • The boundary values of analytic functions (in the upper half plane say $\lim_{\varepsilon \to 0}F(\lambda+i\varepsilon) $) are Lebesgue measurable but can take all real values in any finite interval. Such limits arise naturally in the solution of second order differential equations and the spectral theory of operators. – Paul May 27 '24 at 13:31
  • @Hamdiken The OP means exotic function in the context of Riemann integral not being able to handle it. $\mathscr R\int_a^b B_t,dt$ is well-defined. If you mean $\int_a^b B_t,dB_t$, well not even the Lebesgue integral can handle this... – Andrew May 27 '24 at 14:08
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    The concept of Riemann–Stieltjes integral is fundamentally bound to $\mathbb{R}$. Of course it is not impossible to extend this to $\mathbb{R}^d$, but this necessitates us to define a multivariate version of increasing function that is less satisfactory to work with. On the other hand, measure theory deals with this task very naturally and elegantly. This enables us to deal with (multidimensional) Dirac delta and other singular measures supported on curves and surfaces, which is crucial in physics and other applications. – Sangchul Lee May 27 '24 at 14:58
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    @zaccandeis: This site is not designed for mathematical meandering that wanders from the Lebesgue integral to finitist positions. This is a question and answer site. – Lee Mosher May 27 '24 at 14:59
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    Some of the comments above are more informative than others, but the question is perfectly reasonable and not opinion-based. Furthermore, user Vercassivelaunos provided a perfectly reasonable answer. – Mikhail Katz May 27 '24 at 15:05
  • I need to understand Ergodic Theory, which is doable (albeit difficult) with Lebesque; without him I think it would be impossible. – Simon Crase May 28 '24 at 01:14
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    Consider reading my answer to Why do we need Lebesgue Integration and what are the limits to Riemann Integrals? The summary is that we get complete function spaces, which is crucial for PDEs (100% Science, and particularly, E&M motivated); next it deals with limits properly; analysis is the study of limits at various levels so this is crucial. Finally, the Radon-Nikodym Theorem, which allows us to reduce abstract questions of duality to more concrete questions about functions. Lebesgue theory is NOT about integrating more stuff. – peek-a-boo May 28 '24 at 06:11
  • The answer depends on whether you would like to prove mathematical statements or use mathematical statements: if you want to prove things, then knowing more theory allows you to use more theorems as lemmas. And if you just want tu use mathematical statements, you don't need to study their proofs, neither do you need them to be true! Physicists assume false mathematical statements all the time, and this does not make them necessarily wrong. – Plop May 28 '24 at 08:20
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    You are basically looking for this paper: https://mast.queensu.ca/~andrew/notes/pdf/2007c.pdf – Pilcrow May 28 '24 at 08:51
  • @SangchulLee Thank you for this response – zaccandels May 28 '24 at 09:58

4 Answers4

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If you want something that's useful to the sciences: In quantum mechanics, the wave function $\psi$ of a particle in some space $S\subseteq\mathbb R^3$ is an element of the space $L^2(S,\mathbb C)$, that is, the complex Lebesgue square integrable functions on $S$. Or more specifically, it's not a function, it's an equivalence class of functions, where two functions are equivalent if the square of their difference has Lebesgue integral $0$. It is vital to the theory of quantum mechanics that this is a Hilbert space, which it is only if we're basing it on the Lebesgue integral, since the corresponding space based on Riemann integrals is not complete (certain limits of Riemann integrable functions do not have a Riemann integral equal to the limit of the Riemann integrals of the sequence whose limit is under scrutiny).

CiaPan
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To me, the most immediate benefit of Lebesgue integration over Riemann is taking limits. If you want to exchange a limit with a Riemann integral, you generally need uniform convergence of the integrands, which is a strong condition. With the Lebesgue integral, you have really useful theorems (Monotone, Dominated, Fatou) that let you know when you can exchange a limit with an integral despite only having pointwise convergence.

And yes, this is relevant to science and engineering. Taking limits is about approximation, and approximation is at the heart of physical science.

user1149748
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Your comments seem somewhat contradictory:

My interest is in an answer possessing utility in the sciences or engineering fields.

I'm deeply interested in mathematics and I want to know why the Lebesgue integral has supplanted the Riemann integral as the integral par excellence in mathematics. Surely there's some compelling reason.

The first quote suggests that you only care about applications of Lebesgue integration and measure theory, while the second quote suggests that you are interested in the pure-mathematical perspective.

Anyway, I will offer answers to both, as a non-expert:

If you use somewhat advanced probability theory (stochastic processes, perhaps particularly continuous-time ones) in your applied math, then measure theory makes probability theory a more coherent subject. The measure-theoretic perspective does away with the discrete-vs.-continuous dichotomy of random variables and is (in my experience), basically mandatory for understanding basic properties of something like Brownian motion.

From the pure-mathematical perspective, I think the standard answer often strongly implies that the Lebesgue integral is superior because it (mostly) generalizes the Riemann integral, and lets you integrate weirder functions like the Dirichlet function. There are also spaces of functions associated with the Lebesgue integral that are "complete", which is a nice property in analysis. However, I will offer another answer which is that the Lebesgue integral is just another kind of integral, better in some cases and worse in others. There are a number of different types of derivatives, so if you're interested in analysis, it's not too strange that there might be different types of integrals you should learn.

Lastly, the pure-mathematical reasons to study measure theory are basically, i) you think it's cool, and/or ii) if you think any more advanced analysis is cool, like functional analysis or PDEs, then measure theory is probably mandatory.

Novice
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The thing about measure theory and the Lebesgue integral is that it all works in huge generality. While you were provided with reasons why we should look at the Lebesgue integral on $\mathbb R^n$, I'd like to point out that this is just a small fraction of what measure theory gives you. Indeed, we get nice integration theory on much more complicated spaces than $\mathbb R^n$, which are actually useful.

If you can accept as an "application" that certain Lie groups like $SU(2)$ arise in physics (Application of SU(2) in physics ; I however am no physicist) then it suggests itself that it would be of interest to better understand $SU(2)$ in order to better understand physics.

Like the Fourier transform on $\mathbb R$ and Fourier series on the circle $\mathbb S_1$ [i.e. Fourier series of periodic signals], there is Fourier theory on $SU(2)$ allowing us to decompose "signals" $f\colon SU(2)\to \mathbb C$ into "frequencies". In order to be able to even formulate such things, we need a notion of integration on $SU(2)$, handsomely provided by the Lebesgue integral with respect to the so-called Haar measure (a measure "equally distributing" mass among all points of $SU(2)$).

The point I'm trying to make: Integration is not only useful on $\mathbb R^n$.

Note: Of course you could embed $SU(2)$ as a submanifold in some $\mathbb R^n$ and try to work with coordinates but carrying all this with you around will hinder you significantly when trying to see the "bigger picture"; also there are lots of interesting spaces to integrate on which we can't embed into a $\mathbb R^n$...

GhostAmarth
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    Re: the last sentence, could you give an example of an interesting space to integrate on which can't be embedded in $\mathbb{R}^n$? – user196574 May 27 '24 at 18:51
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    @user196574 What about the natural numbers $\mathbb{N}$ with the counting measure? The Lebesgue integral gives you the theory of all of the $\ell^p$ spaces essentially as a special case. – csch2 May 28 '24 at 00:47
  • @csch2 Maybe a Dirac comb could capture that, so that it could be viewed as living in $\mathbb{R}$? – user196574 May 28 '24 at 01:08
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    @user196574 The unit ball in some infinite dimensional Banach space is an example of something that cannot be embedded in $\Bbb R^n$. Integrating on subsets of a infinite dimensional space has many applications, e.g. Choquet theory. – BigbearZzz May 28 '24 at 15:56
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    @user196574 When decomposing $f\colon SU(2)\to \mathbb C$ into frequencies, the set $\widehat{SU(2)}$ of frequencies appearing is still quite easy (we can identify it with $\mathbb Z$). However, we may study similar Fourier decompositions (in sense of a Plancherel theorem) for a large class of groups $G$. The Plancherel measure (which on $SU(2)$ is just the counting measure on $\mathbb Z$) tells us how important certain "frequencies" are in the decomposition. For general type-I groups, the (Fell-) topology on $\widehat G$ is messy and the Plancherel measure will look like the the counting – GhostAmarth May 28 '24 at 16:30
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    measure in some places and like the Lebesgue measure in other places. We can talk about how "important" certain frequencies are thanks to the Plancherel measure (which is a Borel measure with respect to the Fell topology on $\widehat G$) but this complicated topology makes $\widehat G$ look drastically different from your typical $\mathbb R^n$ as for example $\widehat G$ need not to be Hausdorff – GhostAmarth May 28 '24 at 16:32
  • Addendum: A situation as described above occurs for example in the case $G=SL(2,\mathbb R)$, where $\widehat G$ has discrete and continuous parts... – GhostAmarth May 29 '24 at 08:57