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I am studying Lebesgue integrals, or to be more general, Measure Theory. I am really having difficulties with understanding why we would go over to the Lebesgue measure. The books I'm studying are maybe to technical. Could you please elaborate a bit on the need for measure theory. Why would we want functions to be measurable? Why do we need Lebesgue Integrals? What are the limits of Riemann Integrals, and why wouldnt you maybe introduce Lebesgue Integrals first, if they are better?

Thanks in advance.

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    kinda. I checked that before but didnt find it satisfying, at least for me. – mathsdummie1234 Feb 13 '23 at 11:11
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    Riemann can’t do probability theory – FShrike Feb 13 '23 at 11:13
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    For many strong theorems about Lebesgue integrals like the dominated convergence theorem you need just pointwise convergence of the sequence of functions involved, for Riemann integrals you usually need uniform convergence...just one aspect – Peter Melech Feb 13 '23 at 11:16
  • Also relevant: https://math.stackexchange.com/questions/1497662/how-much-do-we-really-care-about-riemann-integration-compared-to-lebesgue-integr – Hans Lundmark Feb 13 '23 at 11:40
  • @HansLundmark Thanks that gave more insights. I believe youre German, I just bought the book by Heinz Bauer – mathsdummie1234 Feb 13 '23 at 11:58
  • Swedish, actually. (But Hans is a German name, of course.) – Hans Lundmark Feb 13 '23 at 13:56
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    Riemann integral is better for finding antiderivatives and explicit computations,, but the property of being Riemann integrable is not preserved under taking pointwise limits, so it's awkward in a more general/abstract setting (so no Fatou lemma, no convergence theorems unless the convergence is uniform). It also fails if we want to integrate over any measurable space (f.i. in probability theory), since we generally can't partition the space into cubes. But the two integrals are often equal, even in multidimensional cases. – Jkbb Feb 14 '23 at 00:55
  • It's also worth noticing that a common misconception is that we don't have the FTC for Lebesgue integrals; but we actually do have the Lebesgue differentiation theorem. – Jkbb Feb 14 '23 at 01:02
  • The true question should be: "why would we need the Riemann integral?" Indeed, we don't. The Lebesgue integral supersedes it completely. Riemann contributed to mathematics in many ways, he won't be mad. Actually, the Riemann integral is by far not his most important contribution. – Giuseppe Negro Feb 02 '24 at 13:41

2 Answers2

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What I’m about to present is an answer, from only one perspective (which is the version I tell myself, but keep in mind there are several other ways of answering your question), taking hindsight into account, not Lebesgue’s original motivations.


The ”Riemann integral is bad with Limits” route:

One of the common responses to this is that Riemann integrals cannot handle functions like $\chi_{\Bbb{Q}\cap[0,1]}$ (the function defined on $[0,1]$ which is $1$ on the rationals and $0$ otherwise). And sure, this function is not Riemann-integrable, but I wouldn’t consider this alone as a true failure. So what if we can’t handle a pathological case, maybe it’s just so bad it doesn’t deserve to be handled.

I’d say the more significant reason is that various function spaces equipped with integral norms are not complete. For instance, the space $C([a,b])$ (similar issue with $C^k([a,b])$) of continuous real-valued functions on $[a,b]$ equipped with say the inner product $\langle f,g\rangle=\int_a^bfg$ does not define a complete space. Likewise, the larger space of Riemann-integrable functions $\mathcal{R}([a,b])$ is not complete for this inner product. This was a huge issue in the study of PDEs, particularly the Laplace’s/Poisson’s equation; if you’re interested, you should read up on the history of the Dirichlet problem to see the various ups and downs (Dirichlet introduced the idea, Riemann used it in complex analysis, then I believe it was Weierstrass who pointed out a flaw in the argument which essentially boils down to lack of completeness, and then in the 20th century this was finally fully handled with I think Hilbert, and later on with Sobolev spaces etc).

The above historical illustration with PDEs is just an example. Another historical example where there used to be lots of handwaving was Fourier analysis (which I don’t want to get into now). Anyway, the general message is that completeness, and so taking one step further, good behavior under limiting processes, is EXTREMELY important in analysis. So much of higher level analysis explicitly relies on it. Even at the very basic level, we have recognized that the rational numbers are insufficient, and that’s why we came up with the concept of real numbers. If our very number system is expected to be complete, why not our fancy (actually not really fancy… they’re very naturally occurring) function spaces?

Ok, so at this stage, one might say that we could just consider the abstract completion of these spaces with respect to appropriate norms, but that is unsatisfactory because the completion is abstract, and not realized as a set of functions. Furthermore, it doesn’t fully address our concerns with the Riemann integral. The Riemann integral plays very poorly with pointwise convergence as the resulting function usually fails to be Riemann integrable. So, in this sense, widening the definition to Lebesgue integrals remedies this failure. So, with Lebesgue integrals, we get (relatively easily, thanks to the works of giants):

  • The famous convergence theorems: monotone convergence, Fatou’s lemma, dominated convergence.
  • Various complete function spaces, $L^p(\Bbb{R}^n)$ for $1\leq p\leq \infty$, and going a step further, the Sobolev spaces $W^{k,p}(\Bbb{R}^n)$ (an essential collection of spaces when studying PDEs)
  • Lebesgue’s differentiation theorem, and hence the Lebesgue version of the two Fundamental theorems of Calculus (vast generalizations of those in the Riemann setting)

The first two bullet points address the previously mentioned defecits in the Riemann integral. The third says that the Lebesgue integral (with respect to the Lebesgue measure) is a strict generalization of the Riemann integral. So, anything Riemann’s integral can do, Lebesgue can do that and more. This is just the overview of the ‘main benefits’ of the Lebesgue integral with respect to Lebesgue measure, when we come at things from the perspective of ‘Riemann integral is bad with limits’.


The ”what is volume?” route:

Now, having arrived at the concept of the Lebesgue integral and measure, one can say hey the concept of Lebesgue measure can obviously be studied from a more abstract perspective. So, rather than focusing so much on the integral (a linear map which assigns to each suitable function, a number), one focuses on the measure itself. This is also a very natural starting point of investigation. This gives us measure theory. We just take some basic primitive notions as our starting point (a set, a sigma algebra so we can conveniently play around with the sets, and a measure so we can assign numbers to these sets).

This untethers us from the real line or $\Bbb{R}^n$, and allows us to ask about volumes in a more general context. And soon, we can realize that the concept of ‘measure’, while it may have initially been motivated as a generalization of ‘volume’, it is far from limited to that interpretation. Very natural things like mass/charge density all behave like (signed) measures. So, many natural things which people have always been interested in can be studied more formally, precisely, systematically and generally once measure theory is introduced. The entirety of modern probability is built on this foundation: sample spaces, probability measures etc.


Summary: Pros and the one Con of Lebesgue Integrals.

So, we now have a trifecta:

  • measures: a vastly generalized notion of volume, and can be applied in a whole bunch of situations which previously we had to handle in an adhoc manner.
  • Lebesgue integral with respect to arbitrary measures: this gives us the three famous convergence theorems
  • Lebesgue function spaces defined using the Lebesgue integral: we now have complete spaces. So this first of all gives us a vast collection of examples of Banach spaces, and secondly, it means that many practical problems one might face in analysis can be reformulated in the Banach space context, where we can appeal to the spectacularly powerful theorems of functional analysis.

And, contrary to what many people might say, the concept behind Lebesgue and Riemann integrals are nearly identical. For Lebesgue, you take your simple functions, define their integral in the obvious manner, and then by taking an appropriate limit, define it for a much broader class of functions. For Riemann/Darboux integrals: partition your interval, consider upper/lower/Riemann sums, and then take a suitable limit (supremum of upper sums or infimum of lower sums). If one talks about Riemann integrals rigorously, then one necessarily has to introduce $\epsilon,\delta,\sup,\inf$ etc so in that regard the Riemann integral isn’t really that easy either.

The main technical difficulty with Lebesgue integrals on the real line is that because we’re allowing for a much larger collection of sets (Lebesgue/Borel measurable sets, which arise from a $\sigma$-algebra) we have a lot more stuff to deal with. As such, one of the main technical difficulty there lies in proving the existence of the Lebesgue measure (or more generally, it is in Caratheodory’s extension theorem). It is this bit of technicality (and also its vast generality) that (very understandably so) prevents us from introducing Lebesgue integrals first.

The Lebesgue integral (and the accompanying measure theory) is better than the Riemann integral in every mathematical regard; the only thing holding it back is that student’s usually aren’t ready for it yet. One has to understand that as a first year calculus student, they are not yet mature enough to appreciate the need for general measures, and completeness (in fact it is only after a full year of my calculus course using Spivak that I started to get glimpses of the importance of limits and completeness). This difficulty is compounded immensely by the fact that some students don’t even understand Riemann integrals (a thing whose sole purpose is to ‘add many things’): they conflate the Riemann integral with the FTC which facilitates its computation, thinking that “$\int_a^bf(x)\,dx$ is defined as $F(b)-F(a)$ where $F’(x)=f(x)$”.


Edit: Another Reason

Another reason why Lebesgue’s integral is good is because of the Radon-Nikodym theorem which allows us to nicely/conveniently apply the general principle in functional analysis that one should study not just a given topological vector space (e.g Frechet or Banach space) but also its dual. Rather than me talking further, here’s part of the introduction to Dieudonne’s Treatise on Analysis, Vol II, Chapter 13 (Integration):

Nowadays the purposes of a theory of integration are very different from what they were at the beginning of this century. If the aim was only to be able to integrate “very discontinuous” functions, integration would hardly have gone beyond the rather narrow confines of the “fine” theory of functions of one or more real variables. The reasons for the importance that Lebesgue’s concept of integral has acquired in modern analysis are of quite a different nature. One is that it leads naturally to the consideration of various new complete function spaces, which can be conveniently handled precisely because they are spaces of functions (or of classes of “equivalent” functions) and not just abstract objects, as is usually the case when one constructs the completion of a space. Another is that the theorem of Lebesgue-Nikodym and the properties of measures defined by densities $(13.15)$ give us a method for dealing with denumerable families of measures on a locally compact space, by fixing a basic measure and working with the densities relative to this basic measure (hence again with functions); this again proves to be extremely convenient. Here the modern point of view emerges: given a $\mu$-integrable function $f$, what is important is not the values taken by $f$ so much as the way in which $f$ operates on the space of bounded continuous functions by means of the linear mapping $g\mapsto \int fg\,d\mu$ (this mapping depends only on the equivalence class of $f$ and therefore does not change when we modify $f$ at the points of a set of measure zero). The development of this point of view will lead in Chapter XVII to the theory of distributions, which is a natural generalization of the notion of measure on differential manifolds.

Perhaps one final remark I can make, regarding the last sentence, is that Dieudonne (as do Bourbaki) defines (rather than proving, as many other texts do) a measure to be an element of the dual of the space of bounded continuous functions; this is why in the last sentence he says distributions (which live in the dual of the space of smooth compactly supported functions) are natural generalizations of measures on smooth manifolds.

peek-a-boo
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    You write "so at this stage, one might say that we could just consider the abstract completion of these spaces with respect to appropriate norms, but that still doesn’t fully address our concerns with the Riemann integral." What was left out (at least explicitly) is the desire that the completion be expressible as actual functions, which is what the $L^p$-spaces achieve. – KCd Feb 20 '23 at 04:57
  • @KCd weird, I was thinking it, but you’re right I didn’t actually write it. Thanks for pointing it out. I’ve edited my answer to reflect your comment. – peek-a-boo Feb 20 '23 at 04:58
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    You write that Lebesgue integrals are "better than the Riemann integral in every mathematical regard". Not quite: (i) contour integrals in single-variable complex analysis and the Ito and Stratonovich integrals in stochastic analysis are defined like Riemann integrals (subdivide a 1-dimensional domain into small parts and pass to a limit as a mesh tends to 0) and (ii) equidistribution of a sequence in $[0,1]$ is equivalent to the limiting average of functions along that sequence being the integral of the function on $[0,1]$ for Riemann integrable (not Lebesgue integrable) functions. – KCd Feb 20 '23 at 05:11
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    While not a mathematical issue, physical models need Riemann integrals: the slicing of various domains to model work, pressure, etc. leads to an integral as a limit of Riemann sums. So the reason that physics and engineering students need to learn about integrals is due to the way Riemann integrals are built up as a limit. – KCd Feb 20 '23 at 05:14
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    @KCd like I said in my third last paragraph, the ideas behind the RIemann and Lebesgue integral are the same. If we wanted, we could restrict arbitrary measurable sets to intervals in the definition of the Lebesgue integral and we’d essentially get a Riemann sum. So I maintain that the Lebesgue integral is better theoretically, and conceptually, it is just as easy/hard as Riemann integrals (depedning how you want to view it). “Sum basic things up, and take a limit” – peek-a-boo Feb 20 '23 at 05:28
  • Regarding say contour integrals, we define $\int_{\gamma}f(z),dz:=\int_a^bf(\gamma(t))\cdot\gamma’(t),dt$. The RHS can certainly be interpreted as a Lebesgue integral of a complex-valued function on $[a,b]$ with no problem. – peek-a-boo Feb 20 '23 at 05:30
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    Regarding equidistribution, it is a condition that the limit of a certain average of a function be equal to the integral of the function. Of course we start with a primitive collection of functions (indicators of intervals) and then we can ask how far can we push this. And sure, we may only get to Riemann integrable functions (equivalently continuous functions on the circle, or trigonometric polynomials). But I don’t see how the fact that the identity may not hold for a larger class of functions is a failure/sign of inferiority of the Lebesgue integral. You’re comparing different things here. – peek-a-boo Feb 20 '23 at 05:31
  • While contour integrals can be interpreted as Lebesgue integrals, they don't need to be for a first course in complex analysis, where the natural mode of convergence for analytic functions is uniform on compact subsets (which behaves nicely with Riemann integrals). It means students can start learning the beautiful subject of complex analysis without having to take a course in measure theory as a prerequisite. I am not disputing that Lebesgue integrals are superior to Riemann integrals for many purposes, but not for all purposes even just within mathematics. – KCd Feb 20 '23 at 05:41
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    @KCd sure, and I’m not saying don’t teach the Riemann integral either. In fact I said quite the opposite in my last paragraph. When I say ‘better mathematically’, I mean literally that: everything a Riemann integral can do, so can the Lebesgue integral. I’m not talking pedagogy here at all when I say ‘better mathematically’. – peek-a-boo Feb 20 '23 at 05:41
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    The answer as well as discussion in comments are really great. +1 – Paramanand Singh Feb 20 '23 at 06:42
  • Thanks, really appreciate your time you put in!!! – mathsdummie1234 Feb 20 '23 at 09:56
  • "everything a Riemann integral can do, so can the Lebesgue integral" That's actually not true, consider the improper integral of the sinc function. – Jürgen Sukumaran Feb 02 '24 at 14:32
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    @JürgenSukumaran that has to be the most commonly cited “counterexample”, but it’s pretty meaningless. Sinc is not Riemann-integrable on the real line (and sure its not Lebesgur integrable either). It is only improperly Riemann-integrable, i.e a limit of Riemann integrals exists; so obviously the corresponding limit of Lebesgue integrals exists too. Perhaps Lebesgue originally intended to be able to define his integral such that $\sin x/x\in L^1(R)$ so perhaps he ‘failed’ in that regard, but like I said in the beginning I’m not really considering Lebesgue’s original motivations. – peek-a-boo Feb 02 '24 at 19:29
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    Using the sinc argument is like comparing apples and oranges. It is a theorem that if a function is Riemann integrable, then it is Lebesgue integrable and the two integrals are equal (again recall this means a bounded function on a compact rectangle in $\Bbb{R}^n$ such that supremum of lower sums equals infimum of upper sums… or an equivalent characterization using Riemann sums). If you now want to do extra things with the Riemann integral (e.g take a limit according to some exhaustion of the domain) then you can do it with the Lebesgue integral too and you’ll get the same value. – peek-a-boo Feb 02 '24 at 19:37
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    also, (I know you didn’t say this, but still this is a common point brought up… but I don’t feel like bumping the post again with an edit): it is often said that the Riemann integral is more ‘practical’ in that it allows us to actually ‘compute things’. People who say this are so used to Riemann integration that they forget that what actually facilitates computations in practice is the fundamental theorem of calculus, and for this Lebesgue’s version is more general, so even in that regard Lebesgue integral is better than the Riemann integral mathematically. – peek-a-boo Feb 02 '24 at 19:44
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Analysts would like to interchange the order of limits and integration as often as possible, so they want practical conditions that can tell them it is legitimate to say $$ \lim_{n \to \infty} \int_S f_n(x)\,dx = \int_S \lim_{n \to \infty} f_n(x)\,dx, $$ where $S$ is the domain of integration (maybe $S = [a,b]$, maybe $S$ is more complicated). There are two issues that occur when asking if the above equation is true.

(1) Even if each $f_n$ is integrable, $\lim_{n \to \infty} f_n$ might not be integrable. (There are many notions of limit for functions: pointwise limits, uniform limits, etc., but back in the 19th century people only cared about limits that made sense as pointwise limits, even if they were also uniform limits.) A pointwise limit of continuous functions need not be continuous, a pointwise limit of piecewise continuous functions need not be piecewise continuous, and a pointwise limit of Riemann integrable functions need not be Riemann integrable. But if $f(x) = \lim_{n \to \infty} f_n(x)$ for all $x$ and each $f_n$ is measurable, then $f$ is measurable. That's fantastic! So even if a pointwise limit of continuous functions is not continuous, at least we can say it is measurable.

(2) Before measure theory was developed, the only general criterion known that justified the interchange of a limit and integral was uniform convergence, and that might be hard to verify or might not even be true in cases where we want to make that interchange. Even when all $f_n$ and their pointwise limit are Riemann integrable, it could be hard to justify that interchange in the setting of Riemann integrals without assuming uniform convergence. Lebesgue integration gives us powerful convergence theorems: under conditions weaker than uniform convergence that are also easy to verify in practice, the interchange of limit and integral is justified. The dominated convergence theorem is a great example of this, and it has no analogue for Riemann integration. As a concrete example, suppose each $f_n \colon [0,1] \to \mathbf [0,1]$ is continuous and $\lim_{n \to \infty} f_n(x) \to 0$ for all $x$ in $[0,1]$: does $\int_0^1 f_n(x)\,dx \to 0$? It is trivial to show the answer is "yes" using the dominated convergence theorem, but I don't know if this can be done using Riemann integrals alone (if it can, I think it is quite challenging).

The answers to another MSE question on Lebesgue integration here are worth reading for more perspectives on why the Lebesgue integral is so much nicer than the Riemann integral for limit theorems.

KCd
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  • Thanks, really appreciate your time you put in!!! – mathsdummie1234 Feb 20 '23 at 09:56
  • Could you explain why, intuitively speaking, the set-based approach to the Lebesgue integral leads to better interplay with limits? This is something I haven't been able to find a resource for. – nootnoot Feb 02 '24 at 06:50
  • @nootnoot what do you mean by “set-based” approach? Riemann integrals also involve subsets of the domain of integration in its definition. – KCd Feb 02 '24 at 13:13
  • @nootnoot I added a new paragraph at the end with a link to another MSE post on Lebesgue integrals. – KCd Feb 02 '24 at 13:37