Talk:Lebesgue integration

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.

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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)

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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∫_a^bf(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)

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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)

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

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

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      --------      ------

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)

Almost everywhere

The definition of "almost everywhere" in https://en.wikipedia.org/wiki/Lebesgue_integration#Basic_theorems_of_the_Lebesgue_integral seems to be different than the one in the article https://en.wikipedia.org/wiki/Almost_everywhere . The difference is that here the set where two functions differ has to have measure zero, but in the dedicated page it is enough that it is contained in a set of measure zero. In other words, the set where the functions differ does not have to be measurable in the second definition. Can someone confirm this? Thank you for a nice article otherwise. 129.194.183.160 (talk) 17:47, 24 September 2020 (UTC)

Excellent point. This article is incorrect. StrokeOfMidnight (talk) 00:54, 25 September 2020 (UTC)

The diagram with length-wise rectangles seems wrong

The diagram in the section titled "Intuitive Interpretation" seems misleading (these are the red rectangles underneath the blue ones). The approximation used by Lebesgue integration is not by length-wise rectangles. In particular, nothing in the approximations of the Lebesgue integral have a height that corresponds to the heights of these rectangles. I know intuitions are, by definition, not perfectly rigorous or accurate. But this seems to be far enough from the reality of the Lebesgue integral, that it would actually cause more misunderstanding than understanding. Also, the diagram right below it seems to accurately represent the Lebesgue integral. So I vote we take down diagram with the misrepresentation. Addemf (talk) July 2021

Agreed. The red diagram is flat out wrong, not just misleading. It doesn't reflect the fact that, for simple non-negative functions,

$\int _{S}f\,d\mu \,{\stackrel {\text{def}}{=}}\,\sum _{i}c_{i}\mu {\Bigl (}\{x\mid f(x)=c_{i}\}\cap S{\Bigr )}.$

And yes, the diagram underneath the red one makes a lot more sense. StrokeOfMidnight (talk) 20:22, 13 August 2021 (UTC)

Please read the section [[1]]. The Lebesgue integral is defined by partitioning the range and adding up the contributions from the horizontal slabs of the partition. Apart from being horizontal, the main difference with the Riemann approach is that the slabs are of the form S×[t,t+dt] where S is a measurable set rather than an interval. I have put this section back in, since it addresses this very point. 164.52.242.130 (talk) 11:41, 23 September 2021 (UTC)

Regarding your example of a simple function, it can be verified as follows (although drawing a picture such as that in the relevant section of the article helps to make the required summation by parts a little clearer). To get the integral of a simple function $\sum _{n=1}^{N}c_{n}\chi _{A_{n}}$, we may assume without loss of generality that the $c_{n}$ are distinct, positive, and arranged in (strictly) increasing order. By convention, set $c_{0}=0$. Then consider, for $n=0,\dots ,N$, the "slab" $S_{n}=\mu \{x|f(x)>c_{n}\}\Delta y_{n}$, where $\Delta y_{n}=c_{n+1}-c_{n}$. Note that we may write $\mu (A_{n})=\mu \{x|f(x)>c_{n-1}\}-\mu \{x|f(x)>c_{n}\}$

so that summation by parts gives:

$\sum _{n=0}^{N}S_{n}=\sum _{n}(\mu \{x|f(x)>c_{n}\}-\mu \{x|f(x)>c_{n+1}\})c_{n+1}=\sum _{n}\mu (A_{n})c_{n},$

as required. 164.52.242.130 (talk) 12:30, 23 September 2021 (UTC)

Please read the edit summary. I did not say the claims in this section are wrong, only that they are too heavy for an introduction. This section gives an alternative definition of Lebesgue integral, ... while preparing for the main one below? Interesting logic, isn't it? The point is: a supposedly informal intro cannot be as heavy or heavier than the main subject. Therefore I have moved the section past the main definition. As a side note, the article already has a section named "Intuitive interpretation". StrokeOfMidnight (talk) 13:32, 24 September 2021 (UTC)

I've tried to extrapolate the informal part for the intuition section. The point is that the Lebesgue integral is really not that far removed from the intuitive explanation. I think the use of simple functions as the first approach seems to obscure that. 164.52.242.130 (talk) 14:14, 24 September 2021 (UTC)

I'm not sure what the advantage is in bringing in the Lebesgue-Vitali theorem. Monotone functions are Riemann integrable, easily checked from the definition. Existence and interpretation of the improper Riemann integral in the section poses no difficulty. 74.111.98.245 (talk) 00:25, 26 September 2021 (UTC)

Please stop disruptive editing:

• as I previously stated, the condition $\mu (E)<\infty$ you insist on is too restrictive and eliminates a huge number of important cases;
• the improper Riemann integral does not always exist in this context, and that needs to be acknowledged;
• the Riemann integral can be equivalently defined via $\mu (\{x\mid f(x)\geq t\})$ and $\mu (\{x\mid f(x)>t\}).$ This duality needs to be stated and explained;
• the information in the edits you reverted was both relevant and correct. Therefore reverting the edits entirely, while insisting on something completely wrong (see above), constitutes disruptive behavior. StrokeOfMidnight (talk) 02:17, 26 September 2021 (UTC)

Please do not accuse editors acting in good faith (and discussing on the talk page!) of acting disruptively. That is not constructive. If you feel that I am disruptive, WP:ANI is thataway. The added discussion of a.e. equality of $\mu (f(x)>t)$ and $\mu (f(x)\geq t)$ does not seem important here since it is not required to define the integral. In particular it is not present in the cited source. They define the integral as an improper Riemann integral without further elaboration, it being obvious that this can be defined without difficulty (to me at least). In case readers find the idea of an improper Riemann integral too vexing, I have given a definition at length in a footnote. 74.111.98.245 (talk) 11:58, 26 September 2021 (UTC)

Disruptive editing is frequently done is good faith, and I have a very good reason (see above) as to why you were being disruptive. StrokeOfMidnight (talk) 14:29, 26 September 2021 (UTC)

oh, what is your "very good reason" for accusing another editor of disruption? I'll note that you're the one who changed the status quo here by section-blanking. 74.111.98.245 (talk) 15:05, 26 September 2021 (UTC)

The fact that $\mu (\{x\in E\mid f(x)>t\})=\mu (\{x\in E\mid f(x)\geq t\})$ (a.e.) needs to be mentioned. Otherwise the reader will wonder what makes ">" special and "≥" not. The answer, of course (and this is not immediately obvious), is that "≥" is just as special as ">", and that's where the a.e.-equality becomes relevant. StrokeOfMidnight (talk) 14:54, 26 September 2021 (UTC)

I'm ok with that in a footnote, which seems like a reasonable compromise. But this insistence that one must do something special to interpret the improper integral is silly. If the distribution is infinite at some t>0, the "proper" Riemann/Darboux integral is infinity, directly from the definition! There is no need to do any further analysis of that case. But since you seem to be uncomfortable with that, I included in a footnote a standard way of defining the improper integral in the presence of singularities which handles the special cases you seem to think requires some further analysis. 74.111.98.245 (talk) 15:20, 26 September 2021 (UTC)

You say, "If the distribution is infinite at some t>0, the "proper" Riemann/Darboux integral is infinity, directly from the definition!". Well, if $\mu (\{t\in E\mid f(t)>0\})=\infty ,$ for some $t>0,$ then $\mu (\{t\in E\mid f(t)>0\})$ is not even Riemann-integrable. There is no improper Riemann integral in this case either. StrokeOfMidnight (talk) 15:38, 26 September 2021 (UTC)

I think you mean μ(f(x)>t)=∞ for some t>0. The upper and lower Darboux sums are both infinite in that case. So, by definition, so is the Riemann integral (at least according to the Darboux approach). But I think the more important point here is that the cited source didn't identify the issues with the definition that you seem to have. It's hair-splitting: the integral is clearly well-defined, and attempting to pin it down too much here runs the risk of original research (e.g., are Darboux sums kosher for unbounded integrands?) For proof you need look no further than the fact that you, a humble mathematician, were able to supply such a definition (by a non-standard and ad hoc approach). I have given a different approach that is sourced to secondary literature. I consider the matter settled. 74.111.98.245 (talk) 15:57, 26 September 2021 (UTC)

• The way you phrased it, you don't need an alternative definition of improper Riemann integral because the ordinary one fits perfectly;
• you say "I think you mean μ(f(x)>t)=∞ for some t>0. The upper and lower Darboux sums are both infinite in that case. So, by definition, so is the Riemann integral (at least according to the Darboux approach)." Sorry, yes, I did mean that notationwise. And no. The (proper) Riemann integral is defined for bounded functions on a finite interval, and those functions only. This integral was not created with infinity in mind. That's what the improper one is for. So, if $\mu \{x\in E\mid f(x)>t_{0}\}=\infty$ for some $t_{0}>0,$ then $\mu \{x\in E\mid f(x)>t\}=\infty$ for every $t\in [0,t_{0}],$ and the concept of (proper) Riemann integral is inapplicable. Therefore there is no improper one either. StrokeOfMidnight (talk) 20:07, 26 September 2021 (UTC)

I'd agree that in general the Riemann integral doesn't like unbounded functions. It's just not that interesting in those cases, because the integral always fails to exist then. However, here the "proper" Riemann/Darboux integral diverges to infinity, which is not a problem. Anyway, that's not important, since there is a standard definition of the improper integral in the literature that completely resolves this issue. 74.111.98.245 (talk) 21:04, 26 September 2021 (UTC)

There is a well-defined limit according to at least one definition of the improper integral. 74.111.98.245 (talk) 20:37, 26 September 2021 (UTC)

So glad that we're in agreement. I've gone ahead and put back in the version that has a citation. If you would like to add your method of defining the integral, go ahead and restore it with a source. Thanks, 74.111.98.245 (talk) 20:44, 26 September 2021 (UTC)

What is the title, and what is the page number in this source? StrokeOfMidnight (talk) 20:51, 26 September 2021 (UTC)

It's cited in the article. 74.111.98.245 (talk) 20:58, 26 September 2021 (UTC)

Only p.269 is cited. There is no title. StrokeOfMidnight (talk) 21:01, 26 September 2021 (UTC)

If it's "Elementary classical analysis" (1974), then p.269 does not have that definition. StrokeOfMidnight (talk) 21:05, 26 September 2021 (UTC)

It might be a different page in your edition. Have you considered looking for "Improper integral" in either the index or the table of contents? 74.111.98.245 (talk) 21:09, 26 September 2021 (UTC)

p462 in the second edition, with Michael Hoffman (via Google books). 74.111.98.245 (talk) 22:05, 26 September 2021 (UTC)

• I've looked at both. It's not there. There is a relevant definition on both of these pages, but it's completely different and has nothing to do with what you have. Thus, you don't have a relevant source.
• You said "If you would like to add your method of defining the integral, go ahead and restore it with a source." There is no "my" method here. I use the standard definition. I can link to a relevant WP article if you like, and that will be my source. StrokeOfMidnight (talk) 00:33, 27 September 2021 (UTC)

The definition in the cited source says that the (improper) integral of a positive function on a set A is obtained by cutting off the function above, integrating the bounded function, and then taking the limit as the cutoff tends to infinity. That is precisely what is happening in the footnote. Perhaps you would like to clarify your understanding of the footnote and the definition in the cited source. I am guessing (without much evidence based on this interaction) that you have the competence required, although if you like we can enlist the help of someone at the math project. In the mean time have reverted your revert to restore the relevant, cited definition of the improper integral that applies to this case. Obviously another source of similar quality that defines the imroper integral so that the expression $\int _{0}^{\infty }f^{*}(t)dt$ makes sense would be welcome. But you seem to have reached the conclusion that it doesn't make sense, and therefore needs to be redefined in an ad hoc way to reach the bad cases. That requires a source!

Alternatively, we can agree to disagree. I think the expression is well-defined (without further elaboration!). (Although I am familiar with how to define improper integrals of functions singular at interior points, and many readers - and certain editors - appear not to be.) Since that seems to be the attitude of the original source as well, we can remove all explanations of its meaning here. 2600:387:F:4917:0:0:0:6 (talk) 11:17, 27 September 2021 (UTC)

• Stop disruptive editing. You are demonstrably wrong: there is a clear discrepancy between the source and your usage of it. The function

$f_{M}(x)={\begin{cases}f(x)&{\text{if}}\ x\leq M\\0&{\text{if}}\ x>M\end{cases}}$

in the source cuts $f$ down to zero when $x>M.$ In your version, however, $\min(A,f^{*}(t))$ merely freezes $f^{*}$ at $A,$ once $f^{*}>A.$

• One can only overlook this difference if one chooses to.
• The other reason this source is unsuitable here (this time on terminological grounds) is the part of it that says "if this limit is finite, and in that case we say that $f$ is integrable", in reference to $\textstyle \lim _{M\to \infty }\int _{A}f_{M}.$ We need to allow the integral to be both finite and infinite. StrokeOfMidnight (talk) 12:36, 27 September 2021 (UTC)

Ah you're right about that source. 2600:387:F:4917:0:0:0:6 (talk) 13:17, 27 September 2021 (UTC)

Thanks! StrokeOfMidnight (talk) 13:53, 27 September 2021 (UTC)

Wow, apparently this started outright warfare. Sorry about that, I just thought that the depiction of length-wise rectangles deceptively makes it look like the width of the rectangles is due to the width of the cells of the partition on the vertical axis. But that's not true, the width of the rectangles is determined by the distance all the way to the horizontal axis. I just worry that a picture suggesting a falsehood like this would mislead a person not yet familiar with the subject, in a way that's unnecessary. Addemf (talk) 02:45, 30 October 2021 (UTC)

@Addemf: Yes, I'd been wondering the same thing. I had difficulty with the length-wise rectangles in the diagram labelled "Lebesgue integration (in red)". Funnily enough, I think the edit war you're worried this may've started was probably quite constructive: the article seems better comparing versions a bit before and after.

Having re-read the article and the accompanying discussion, I can see some attractions in comparing the Lebesgue method to a contour map, ie. describing it as in effect partitioning the range, rather than the domain.

• The approach seems to be mathematically equivalent to the construction via simple functions.
• The diagram in red is easy to understand and contrast with the Riemann approach.
• It ties in with several key characteristics of Lebesgue's approach, eg. that the domain just needs to be a measure space (rather than requiring all the properties of the real line).
• If one wants to, it's probably fairly easy to see how an approximation as horizontal slabs can be converted into one expressed as simple functions, either geometrically by dropping some verticals onto the x-axis or perhaps algebraically by the summation by parts described above. This might be needed to reconcile the diagram with the preceding quotation from Lebesgue that "I order the bills and coins according to identical values and then I pay the several heaps one after the other".

I would keep it, perhaps with a pointer to how it will be interpreted.

I had even more difficulty with the next diagram, containing the small rectangles of height dt. I'm may be too strongly affected by having originally learnt the construction of the Lebesgue integral via simple functions, but it might be more readable to take the shortest route towards this solution and move the final paragraphs of the Intuitive interpretation and its accompanying diagram into the Definition section Via improper Riemann integral. However neat the derivation, we shouldn't assume that a reader will know about Riemann integrals (they already have to get their heads around measure theory!). NeilOnWiki (talk) 14:07, 19 January 2022 (UTC)

I write this having not lately looked at the changes or conversation in a while. But I would say that the problem isn’t the idea of partitioning the range, that is totally valid. Diagrams which display this are totally fine with me. This is indeed equivalent to the Lebesgue integral (I’ve proved it in my spare time a while ago, so I’m fairly confident in this.).

The issue is then diagramming the integral as horizontal rectangles with lateral sides determined by the partition, which I believe is not equivalent to the Lebesgue integral. To diagram it faithfully, the lateral sides of the rectangle should still go from some point on the curve, down to the x-axis. I see you mention that it is easy to correct the given picture, along these lines. That may be, although : I really do worry that a diagram which so fundamentally mis-matches the formalism will cause rather than resolve confusion. Perhaps if the diagram displays a prominent warning about this mis-match then it might be ok. But I’m not in control of the page, so I’ll let everyone else do what they will do.

I totally agree that I like diagrams which resemble the familiar Riemann picture but show the slight difference with the Lebesgue integral—-but I believe the important difference between them doesn’t easily show up in a static picture. The difference between them is the order in which points are chosen. In the Riemann integral we partition the x-axis and determine corresponding heights. In the Lebesgue integral we partition the y-axis and determine corresponding regions in the domain. So I believe a helpful diagram will somehow have to demonstrate a kind of “order of choices”. It would be especially useful when applied (1) to a continuous function, where there is obviously no difference between the Lebesgue integral and the Riemann integral, and (2) to the Dirichlet function, where you can more easily see the two different order of choices result in different integrals (especially how, by partitioning the codomain first, the preimage is not even a partition of the domain and therefore the quantities involved are clearly not “reachable” by the Riemann order of choices).

As for rearranging the intuitive explanation part, I’m fairly neutral—-I only joined in editing that part because I thought the intuitive explanation that I encountered wasn’t intuitive enough. But if anyone wants to reorganize or change that for the better, I am all in favor of it.Addemf (talk) 19:25, 19 January 2022 (UTC)

For what it's worth, some authors actually do define the Lebesgue integral of a non-negative measurable function as the limit:

$$\int f=\lim _{n\to \infty }\sum _{k=0}^{n2^{n}}k2^{-n}\mu \{x|k2^{-n}<f(x)\leq (k+1)2^{-n}\}+n\mu \{x|f(x)>n\}$$

which is in a rather obvious manner the sum of slabs obtained by partitioning the range. (This definition is proven equivalent in green Rudin, but I recently saw this as the primary definition given in Liptser and Shiryaev.) I think probably a source of confusion is that in the Lebesgue approach, the slabs are not rectangular. Anyway, the Lebesgue integral can be written as an equivalent (improper) Riemann integral, integrating up the horizontal (non-rectangular) slabs $\{t|f(t)<x\}dx$. Tito Omburo (talk) 11:00, 30 January 2024 (UTC)