Talk:Lebesgue integration

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[edit] Initial comments

Hello! I think this article is looking very good. Over the past several months many improvements have been made. I wonder if it would be better titled "Lebesgue integral", since (if I'm not mistaken) that term is rather more common in text books. Certainly "Lebesgue integral" is more common than "Lebesgue integration" on the web as shown by Google searches. Likewise "Lebesgue-Stieltjes integral" is more common on the web than "Lebesgue-Stieltjes integration". There do exist Lebesgue integral and Lebesgue-Stieltjes integral in WP but these are redirects. However, Riemann integral is an article and Riemann integration is a redirect; also integral is an article and integration is a disambiguation page (there is no integration (mathematics) article). So: in summary, I propose that we move Lebesgue integration to Lebesgue integral and likewise with Lebesgue-Stieltjes. I look forward to your comments. Regards & happy editing, Wile E. Heresiarch 16:57, 21 Jun 2004 (UTC)

On the whole, page names migrate towards the more abstract term. I think this is more suitable for WP, really: so that e.g. nerve is thought of under nervous system first (haven't checked the actual status of those). So I'd be happy to leave it as Lebesgue integration. Charles Matthews 18:08, 21 Jun 2004 (UTC)
Hmm, I guess I'm not convinced. The Lebesgue integral is the object of interest, and the purpose of Lebesgue integration is to construct the Lebesgue integral, is it not? What other items are considered under the general heading of Lebesgue integration? Perhaps I don't get out enough, of course; I wouldn't be at all surprised. Happy editing, Wile E. Heresiarch 23:31, 22 Jun 2004 (UTC)
Maybe one way would be to have an article like integration theories in mathematical analysis, as a sort of umbrella; and then call the articles on particular theories Riemann integral, Lebesgue integral and so on. Well, such an article would be useful in itself, as a survey. Charles Matthews 17:05, 23 Jun 2004 (UTC)

This seems to me like another instantiation of the "thing", "thing theory" problem. MarSch 15:20, 11 Mar 2005 (UTC)

Hello. About notation, under "Equivalent formulations" there are L1 and Cc. Do these want to be italicized as L1 and Cc or maybe Cc ? I'm just wondering how these are conventionally presented. Happy editing, Wile E. Heresiarch 14:14, 13 Apr 2004 (UTC)

I think Lang and Rudin both use L1 and Cc. Loisel 06:21, 16 Apr 2004 (UTC)

Hello, I've reworked the "Introduction" section to situate the Lebesgue integral historically, and to summarize the important differences with the Riemann integral. Then there are definitions and properties, and then there are the more detailed discussion sections. I hope this organization address the concerns with the "Rudinesque" previous revision. FWIW I'm not really happy with the "Lebesgue vs Riemann" aspect of this article, which might suggest there are exactly 2 defns of the integral; it's more "Lebesgue vs every predecessor". Also, I think "failure of the Riemann integral" gives a mistaken impression -- the Riemann integral has been a big success overall. Well, hope this helps. Wile E. Heresiarch 20:09, 8 Feb 2004 (UTC)

All I know is, after I studied the Riemann integral as an undergrad, I thought it was pretty hot stuff, so when I heard about this thing called Lebesgue integration, it took quite a bit of argument to convince me that this new-fangled thing was even necessary in the first place. The introduction is a good start, but I would take select parts of the "discussion" and merge them. As for giving the impression there are 2 defs of "integral", this is more a misunderstanding of math in general...people think there is some Platonic concept called "the integral" that mathematicians go out and "discover" its "true nature"...when there are just various definitions and theories that just are what they are. It's very difficult to explain this. Maybe it should be pointed out that the Riemann and Lebesgue integrals aren't the only theories of integration that have been developed -- just the most popular and widely used. Revolver 22:56, 8 Feb 2004 (UTC)

I don't think the discussion should come after the formal definitions and results. If someone is unfamiliar with the Lebesgue integral (and most will be who read the article) then the formal definition will seem pointless without some kind of motivation or discussion. As the article is now, someone who doesn't know the Lebesgue integral is likely to just give up part-way through the formal derivations and never reach the discussion. We're not trying to imitate Rudin. Revolver 02:06, 7 Feb 2004 (UTC)

If the definitions are incomprehensible, the discussion is irrelevant (and likewise incomprehensible). The most you can say to someone who doesn't understand the definition is "All definitions of the integral are the same for easy cases, but the Lebesgue integral handles some more difficult cases."
Well, before the Lebesgue integral was developed, many mathematicians discussed why they thought a new type of integral was necessary, what problems it should attack, what specific examples motivated it. The idea that it's impossible to "discuss" a mathematical subject without going through the details of the definitions doesn't seem right to me -- again, if for no other reason than that the mathematicians did it themselves for years before arriving the modern definition (think of groups, topological spaces, non-Euclidean geometries, etc.) I think it's quite possible to explain the specific issues that eventually led to the creation of the Lebesgue integral, without defining the Lebesgue integral itself. And yes, I'll try to do this and put it on the talk page someday...after I finish my thesis, find a job, move, etc.,...;-) Revolver 22:56, 8 Feb 2004 (UTC)

I think the (short) section just before the definition should say that. A secondary problem is that the discussion, as it stands, is needlessly verbose; it needs a lot more focus. Wile E. Heresiarch 09:20, 7 Feb 2004 (UTC)

What should ideally be done, is that the "discussion" and the "formal definition" should become merged into a single thread of narration, not separated into two pieces. This of course, would take a good deal of thought and effort to do it right. Revolver 02:08, 7 Feb 2004 (UTC)

I'm not convinced that is desirable. Maybe you can put a version of this kind of presentation on your talk page and have people look at it. Just a thought. Wile E. Heresiarch 09:20, 7 Feb 2004 (UTC)

Wile, see below, there's a guy who claims this article is too advanced. The text now under "Discussion" was my attempt at answering that gripe.

Loisel 21:13, 6 Feb 2004 (UTC)

Hello Loisel, thanks for bringing up this topic. I moved the discussion sections below the definitions, etc., because the discussion is quite long winded and goes off on several tangents. That's fine, really, since the Lebesgue integral is important and there's a lot to be said about it. But the article works better as a reference if we get right to the point.
Maybe to address the concern that the article is too abstract, we can make the "Introduction" section (just before the defn) better (not necessarily longer). Without getting technical, one can say the Lebesgue integral is more general than some other defns, and also the it's defined as the limit of a sequence of simpler integrals. Perhaps an example of a sequence of simple functions will make a good companion article. Wile E. Heresiarch 01:49, 7 Feb 2004 (UTC)

Michael, why did you remove the previous article and leave a corpse in its stead? I've now restored the article.

Loisel 08:04, 3 Feb 2004 (UTC)


I've made a correction to the Technical Difficulties regarding improper Reimann Integrals on this page. What was defined was not the Improper Reimann Integral but the improper Cauchy Pricipal Value. The improper Reimann Integral does not exist in this example.

-joshua

Look - I'm sure that this is a well-writen article, but could someone (knowledgeable) put in a 2-3 line explanation of what it actually about, and what it is used for? The opening sentence "Let m be ..." isn't really au fait in a generalist encyclopedia. I'm really glad we have this sort of advanced stuff, but in this case I don't even understand what I don't understand. - MMGB

Shouldn't it be at some point made clear that m is the Lebesgue measure? Otherwise, there's no guarantee the integral will correspond to Riemann integration - if m is the counting measure, it will correspond to summation. But I'm not sure where precisely to work that in.

The link to monotone convergence theorem is to a different theorem.

Charles Matthews 14:00, 6 Sep 2003 (UTC)

Fixed. Loisel 06:38, 7 Sep 2003 (UTC)

[edit] Too advanced!

This encyclopedia article is too advanced; the only people who will be able to get the meaning are those who have already learned Lebesgue theory thoroughly. The modern, extremely compact, definitions of mathematics are simply a shorthand for people already familiar with the underlying concepts and some illustrative examples; without including the latter, the former is unintelligible.

In fairness, an article on the Lebesgue integral is certainly only of interest to a specialized audience. And it would perhaps be appropriate to include the existing article as an advanced appendix as a minimal, extremely formal, refresher, less formal explanation is needed for anyone who doesn't remember or hasn't learned what ideas the formal definition corresponds to.

For the main body of the article, it would be nice to follow history somewhat: (1) explanation of the problems encountered with the Riemann integral including specific functions or applications (2) elementary definition (NOT a formal argument: emphasis on underlying IDEAS, not on formal definitions) and explanation of how the problems of Riemann are resolved by Lebesgue (that is, examples) (3) perhaps a little modern history on the growth of analysis since Lebesgue. The formalities could be a conclusion of sorts. I would write such an article myself, but I don't know Lebesgue theory sufficiently well (hence my need to refer to an encyclopedia article); perhaps someone else might? (Anon.)

A page on improper integrals in the Riemann theory would be a help. But no one-page intro to the Lebesgue integral is going to be easy.

Charles Matthews 16:10, 23 Oct 2003 (UTC)


This is a valid point, however I'm not very good at history. I could attempt to vulgarize the concepts discussed in the "formal construction" section, but I couldn't color the text with historical anecdotes about integration theory: I don't know any.

Would that help?

Loisel 16:47, 24 Oct 2003 (UTC)

Is this better? Loisel 18:15, 24 Oct 2003 (UTC)

I'm sorry, this topic is advanced pdenapo Wed Jan 14 03:09:34 UTC 2004

[edit] Well done

With the addition of the new introductory material, this article takes the place of a good professor - at first giving a general indication of the questions and problems at hand, and once the ideas involved are explained a little further (with some nice examples) providing the formal framework which enables those ideas to be put into rigorous use. Great job.

--

Couple of points where I can carp (I do concur with the post above: the work on this page was very worthwhile).

  • The Lebesgue approach is not the most elementary area-based integration theory; that distinction goes to the Riemann integral.

Don't think that's actually true (cf. Dieudonne's Treatise on Analysis ona sub-Riemann theory).

  • Uniform convergence of Fourier series.

Rare? A couple of derivatives will do.

Charles Matthews 07:24, 18 Dec 2003 (UTC)

About the uniform convergence: it is true that twice differentiable periodic functions have uniformly convergent Fourier series. This is a very thin set in L^2. That is one of the meanings of "rare." Feel free to adjust the wording if you think you can improve it, but the goal of that passage (to show that there are many common examples where the uniform convergence theorem is insufficient) should remain, I think.

About "elementary." I meant to recognize the order in which this is usually taught. I don't know the sub-Riemann theory you mention, but I should've thought there would be a gazillion variants that claim to be more elementary than the Riemann integral. Again, feel free to adjust the wording if you think you can improve it. If you're changing it, I think it should probably say that Riemann is "more elementary" than Lebesgue, but I doubt it should refer to some obscure theory that's not generally taught.

Anyway, have a ball. Loisel 02:04, 8 Jan 2004 (UTC)

Charles Matthews refers here to what Dieudonne calls regulated functions in the English translation (satisfying a condition something like: both one sided limits exist at every point = uniform norm closure of linear span of indicator functions of intervals ). It is more elementary in the sense that all its properties follow by continuity. Dieudonne disparages theRiemann integral (which in fact is the prevalent attitude among the mathematicians I know)

I just removed the following text:

Correction: The improper Reimann integral does not exist for f or g since the improper Reimann Integral is defined as a double limit and you cannot subtract infinities, i.e. the improper Reimann integral is defined as limlim∫abf(x) dx where the limits are taken as a goes to infinity and b goes to infinity. What is true is that the Improper Cauchy Principal value (about zero) for f does in fact exist and it is PV∫-∞∞f(x)=0.

I will modify the text to say something like (sometimes called the Improper Cauchy Principal value about zero). Please note Wikipedia:Integrate_changes, which is part of the Wikipedia policy, requests that you integrate your changes so that they form a seamless part of the article. Loisel 02:14, 8 Jan 2004 (UTC)

I think that maybe this whole paragraph should be junked, since with the proper definition with 2 limits, there is no preferred point and no translation noninvariance. Thus this is not a deficiency of the Riemann integral.MarSch 15:30, 11 Mar 2005 (UTC)

I've rewriten this article since the old version focused only on the technical difficulties of Riemman integral, rather than defining the concept of Lebesgue integral. I've included the examples in the old version, though pdenapo Wed Jan 14 03:09:34 UTC 2004

The new introduction is nice, but not consitent with the definition that appears below. [User:pdenapo|pdenapo]]Wed Jan 14 03:09:34 UTC 2004

I think the rewrite has destroyed a large quantity of useful information and has generally decreased the quality of the article. I'm tempted to revert.

Loisel 08:05, 1 Feb 2004 (UTC)

[edit] Equivalent Formulations

I think this page is very good. Hello, Do you have references for the equivalent formulations of the integral? (unique continuous extension of a linear functional...) Thanks Sergei Vieira srgvie2000@yahoo.com.br

That's elementary functional analysis. See either Rudin's Real and Complex Analysis or Rudin's Functional Analysis, for instance. If f is a continuous functional defined on X, a dense subset of the Banach space Y, and if g and h are continuous functionals that agree with f when restricted to X, then g=h. To see this, let x be arbitrary in Y and let x_k be a Cauchy sequence in X converging to x. Then g(x_k) converges to g(x) and h(x_k) converges to h(x). But g(x_k)=h(x_k)=f(x_k), and f(x_k) is Cauchy. Hence f(x_k) has a unique limit point, and g(x)=h(x). You can get existence out of (say) the Hahn-Banach theorem (but you can also get it in an elementary way.)

The statement that L^1 is the completion of C_c in the norm given by the Riemann integral is just a rehash of the uniqueness of the completion of a metric space. We know that in the Lebesgue construction, C_c is dense in L^1 (and so L^1 is a concrete completion of C_c). If we define a norm on C_c using the Riemann integral, we know that it agrees with the norm you would get out of restricting the L^1 norm on C_c. Hence, C_c with the Riemann integral L^1 norm, is the same as C_c with the Lebesgue integral L^1 norm. C_c with the Riemann integral L^1 norm has an abstract completion, as per metric space theory, but the Lebesgue theory is just an explicit version of this completion.

Loisel 19:44, 27 Jun 2004 (UTC)

[edit] Indicators

Does anyone agree with me that there should be more consistency with the notation here? On the pages indicator function and Riemann integral, I(your set here) is used. I definitely prefer this, or a notation with χs, to the current one on this page. Best, mat_x 15:59, 18 Aug 2004 (UTC)

[edit] Mountains

The Riemann integral of the mountain requires lower sums and upper sums, in each vertical slice find the highest and lowest points. The integral beng described is more like the Cauchy integral. CSTAR 15:42, 19 Dec 2004 (UTC)

[edit] Towards a better integration theory

Is there any objection to deleting this section? CSTAR 17:42, 22 Dec 2004 (UTC)

I have no objections. You are doing a good job on the page, but I wonder whether you could write slightly more on the definition of measure besides refering the reader to the measure (mathematics) article. I fear that many readers may not get pass the sentence "Let μ be a (non-negative) measure on a sigma-algebra X of subsets of E." -- Jitse Niesen 21:28, 22 Dec 2004 (UTC)
Yes I do plan to fix that soon.CSTAR 21:35, 22 Dec 2004 (UTC)

[edit] Edits

Remarks on measurability condition.. MEasurability of a function as defined in the article is a property of the underlying sigma-algebras of source and target space. The user who made the latest edits, confused this with measurability of functions with respect to the sigma-algebra of the completion in the source space. This is not necessary and is nowehere used in the article.. Of course this means that if f = g a.e. and is measurable g may not be. There are a number of other changes which I don't see add anything to the article. For instance why add vector-valued functions here? If there is no objection I am going to revert. The only improvement was the use oof \liminf and \limsup whic I propose to keep. CSTAR

[edit] Integral as area under curve

Intuitively, continuity is not necessary: Two disjoint rectangles with vertical sides and of different heights have an area and aren't representedby continuous functions. CSTAR 15:29, 11 Mar 2005 (UTC)

The first sentence of the article talk about an area being bounded. If the fuction is discontinuous then no area is bounded. Of course I should also have made changes to say that the area is bounded from below by 0. I realize that even then you will have difficulty on the sides, so I will be bold and change it to something which mentions area below a (positive) function. MarSch 15:40, 11 Mar 2005 (UTC)
Huh? Discontinuity implies unboundedness? What about the indicator function of the interval [0, 1]? That's a bounded function with a discontinuity. Michael Hardy 00:09, 14 Mar 2005 (UTC)
Perhaps MarSch means with "unbounded" that the graph does not run along the boundary. For instance, with your indicator function, the graph looks like this
              ------



      --------      ------
and the vertical sides of the square — running from (0,1) to (0,0) and from (1,1) to (1,0) — are not bounded by the graph. MarSch's change from "bounded" to "contained" solves the problem. -- Jitse Niesen 14:12, 14 Mar 2005 (UTC)
Referring to the region under the curve and above the x-axis is about as unambiguous as you can get in natural language. And the anti 2D bias isn't very informative in my view. CSTAR 14:26, 14 Mar 2005 (UTC)
I think we should try to keep the first sentence as easy as possible and I edited the article accordingly (bringing back the "2D bias"). -- Jitse Niesen 15:10, 14 Mar 2005 (UTC)
As far as I've seen there are also articles about integration and Riemann integration. Your sentence would be perfect for integration. For those reading this article it is needlessly simplistic and ugly. -MarSch 14:40, 4 Apr 2005 (UTC)

[edit] Generalization

This article defines the integral for functions f: R -> R. From what I can tell from reading it, it's trivial to extend this definition to cover any function f: S -> R where S is any measurable set - should the article cover this? I'm particularly thinking that it's easy to define an integral on f: R x R -> R — ciphergoth 22:30, 2005 Apr 28 (UTC)

From what I see, the integral is defined exactly as you want, on a general space. The first several paragraphs deal with this. Oleg Alexandrov 23:08, 28 Apr 2005 (UTC)

[edit] Limitations of the Riemann integral

You show that the two functions f(x)=-1+2[x>0] and g(x)=-1+2[x>1] are not riemann integrable, and that the cauchy values of the integrals differ. Sadly. But it is not clear to me if the lebesgue integral improves the situation ? Bo Jacoby 13:57, 22 September 2005 (UTC)

[edit] Reversion

Hi CSTAR. You reverted my simplification. Your argument is incorrect. The function |f|, mapping x to |f(x)|, allows any real function to be split into a positive and negative part: f=(f+|f|)/2+(f-|f|)/2. The ad hoc notations f+ and f- are not necessary. Bo Jacoby 08:23, 27 September 2005 (UTC)

Reply.

  1. Your definition of the integral was expressed as a difference of two values both of which could be +∞ You must assume that at least one of the values is finite.
  2. Re your comment: The ad hoc notations f+ and f- are not necessary and should be avoided. On the contrary, this a very common notation and is hardly ad-hoc. See
  • P. Halmos, Measure Theory, or
  • W. Rudin, Real and Complex Analysis.
Why do you believe that this notation should be avoided?
--CSTAR 15:57, 27 September 2005 (UTC)
  1. Thank you for your reply.
  2. I agree that the finity condition should not be postponed.
  3. Using the better-known absolute value function |x| should make it easier to read. x+ = (x+|x|)/2 and x- = -(x-|x|)/2. You may assume that the reader knows |x|, but as you do not assume that the reader knows x+ and x-, you state that 'we need a few more definitions' and then you define x+ and x- in the text. Well, we don't need these definitions. We can use standard mathematical expressions instead.
  4. Please compare my version with your reverted version. You state that the integral is defined for real functions, but you actually define it for complex functions too. Why not state that it is defined for complex functions? You call the set X when you mean E. It's merely a typo. Your definition made in the 'indicator function' section was not used in the 'simple function' section , so either the former could be omitted or the latter be changed to build on the former. I made many small improvements.
  5. I wonder why the classical notation f(x)dx is abandoned. How do you express the integrand ax^2dx as opposed to ax^2da when you write ax^2d\mu? Only the repeated dummy variable tells what the integration is about.

Bo Jacoby 11:53, 28 September 2005 (UTC)

i second CTARS's comments above. these notations are standard, not "ad hoc". Mct mht 18:01, 16 August 2006 (UTC)

[edit] Silly quote

I removed a silly quote about the Lebesgue integral not being useful "physically". I don't know what this means. Is there a physical usefulness to having the irrationals? The reals are to the rationals as Lebesgue is to Riemann. And, should we add a ton of pro-Lebesgue quotes? -cj67

Re: And, should we add a ton of pro-Lebesgue quotes? I don't care one way or the other.
Re The reals are to the rationals as Lebesgue is to Riemann. I don't know what this means. More specifiuically, the reals are the metric completion of the rationals; the extension of Riemann to Lebesgue is more subtle than an extension by continuity to a completion.--CSTAR 17:16, 26 May 2006 (UTC)
Nope. Lebesgue integrable functions are the completion of the continuous Riemann integrable functions w.r.t. the L^1 norm (which can be defined using the Riemann integral, since you are only considering continuos functions).(Cj67 20:05, 30 May 2006 (UTC))
Nope. The elements have to be represented as functions. As abstract Banach spaces, L^1 is the completion of the continuous Riemann integrable functions w.r.t. the L^1 norm, as you point out. But representing an element of this Banach space as an equivalence class of measurable functions is another matter, because pointwise evaluation is not continuous as a functional on L^1. --CSTAR 20:24, 30 May 2006 (UTC)
L^1 is a representation of the completion of Riemann integrable functions, just as the reals are a representation of the completion of the rationals. I think the analogy is clear. Of course, there is more involved, since L^1 does not equal the reals. But I think my point about the physical "usefulness" of the irrationals is clear. (Cj67 22:15, 30 May 2006 (UTC))
the analogy looks fine to me. completion of metric spaces are by construction equivalence classes. the Lebesgue-a.e. condition gives an explicit identification. Mct mht 18:08, 16 August 2006 (UTC)
My point in saying that "the extension of Riemann to Lebesgue is more subtle than an extension by continuity to a completion" was that the "pointwise" properties of L^1 do not follow by continuity, since f_k converges to f in L^1 does not imply f_k converges to f almost a.e. (though it is true that some subsequence of f_k converges to f ae.) . ANyway, this is ome of those WP talk page discussions I'm sorry I ever got into.--CSTAR 18:22, 16 August 2006 (UTC)
  • A curious thing: ain't integration by parts allowed when performing a lebesgue integral under certain measure (i.e Gaussian measure), i don't know if this is omitted in the article or it simply can't be a Lebesgue integration by parts under a given measure.

[edit] extended real number line.

Article quote:

Let f be a non-negative measurable function on E which we allow to attain the value +∞, in other words, f takes values in the extended real number line.

"in other words" is not quite true, as the extended real number line also contains negative real numbers and minus infinity.

Bo Jacoby 13:46, 7 February 2007 (UTC).

[edit] The 'intuitive' interpretation

Hi. The 'intuitive' interpretation for the Lebesgue integral is one I have seen before, but doesn't seem to correspond to what is actually going on. According to the mountains diagram for the Lebesgue integral we are multiplying increments in the value of f by the measure of the set of all x which map to equal or higher values of f. This is incorrect, in reality we multiply f (not it's increments) by the measure of all x which map to that value of f. I realize that the result is the same, but the process is different, and so the 'intuition' is a bit confusing when trying to understand the actual correct definition. For a more correct mountains diagram you could imagine a curve with upright rectangles filling it from below (like the dark blue rectangles for the Riemann integral) but where rectangles of different heights have different colours, and those of the same height have the same colour and correspond to one term of the summation for the integral of a simple function.

Benjaminveal 11:59, 3 June 2007 (UTC)

Benjaminveal is right; that section could use some reworking. also, in comparing with the Riemann integral, the following can be pointed out: the Riemann sum is obtained by partitioning the domain of a function. therefore the limit may not be well-behaved if the function exhibits violent local behavior on its domain, e.g discontinuous. this is rectified in the Lebesgue approach by instead partitioning the range of a function and consider the preimages, which necessitates the notion of a measure of a fairly arbitrary set, the family of which is called the Lebesgue-measurable sets. an indication of the effectiveness of the Lebesgue approach is that there's no constructive example of a non-Lebesgue measurable set.

also, perhaps i missed it but the article doesn't seem to explicitly mention that Lebesgue integral extends the Riemann integral and that the Riemann integral functions are classified via the Lebesgue measure. Mct mht 14:19, 4 June 2007 (UTC)

I have rephrased this section because I didn't think it was all that well written. I have also added a reference for those of you who are detractors. Loisel 21:59, 18 June 2007 (UTC)

to build the simple function corresponding to a given partition of the y axis, one passes back to the preimages and thus ends up with vertical-looking rectangles. if we insist on having a picture, wouldn't it be better, more precise, to reflect this? Mct mht 02:35, 19 June 2007 (UTC)


[edit] Figure in red and blue

I think that the figure in red does not match with lebesgue integral definition (although blue one does)

--Nicolaennio 21:23, 9 September 2007 (UTC)

[edit] Faulty discussion of integration on unbounded intervals

The discussion of the faults of the improper Riemann integral seems faulty to me. For some reason, the only definition of improper integral considered is the symmetric limit (a.k.a. the Cauchy principal value), with the justification that it is the "simplest possible" definition. This definition is too simple, for the reasons described in the example: it fails translation invariance (or, as I try to explain to my calculus students, making a u-substitution can change the answer). But this is a strange argument: on the one hand, for these reasons this CPV-style limit is not the definition of an improperly Riemann integrable function given (even) in calculus books: rather, one picks a real number a, breaks up into two improper integrals and requires that both integrals converge. (There are of course more elegant ways to express the same thing, e.g. by convergence of a net based on the collection of all closed subintervals of the real line ordered under containment.) With the standard definition, the given function is not improperly Riemann integrable. Moreover, it is not Lebesgue integrable either, so this example is quite misleading. Indeed, one of the features of the Lebesgue integral (whether or not it is a drawback depends on your perspective) is that it does not include all improperly Riemann integrable functions.

I will give the writers of this otherwise very nice article a chance to correct this before I remove it. 70.155.76.162 (talk) 02:15, 27 November 2007 (UTC)Plclark

[edit] Mix between the general Lebesgue integral and the specific one that extends the Riemann integral.

At some places in the article there is a mixture between the two meanings of Lebesgue integration. To quote from the article: The term "Lebesgue integration" may refer either to the general theory of integration of a function with respect to a general measure, as introduced by Lebesgue, or to the specific case of integration of a function defined on a sub-domain of the real line with respect to Lebesgue measure.

It is not a bad thing to use the second meaning troughout the article as a guidline and/or example but it should be clear in every statement which meaning is used. I would do it myself but I don't feel proficient enough (yet!) with Lebesgue integration at this point.

Stigin (talk) 16:04, 16 January 2008 (UTC)

Lebesgue integration is almost universally understood to mean the general theory of integration with respect to a countably additive measure. You're right though that this made need clarification in the article.--CSTAR (talk) 16:19, 16 January 2008 (UTC)