Talk:Riemann integral

Animated image

Hello. Riemann integral is a very nice article. I wonder about the animated image. I agree that it's useful for viewing in a browser, but I don't think it would be as useful on a printed page. I don't see a good reason to rule out printed copies; this article could easily be something a student would print out for reference, for example. Can we convey the same information by displaying each frame of the animation separately? -- Sequences of graphs are going to be pretty common in the math articles, and so I think it's important to set a good precedent here. I look forward to your comments. Wile E. Heresiarch 16:38, 21 Jun 2004 (UTC)

Personally I think the animated image is great. It really gets accross the idea well and takes advantage of what the web format can offer. I do see your point that in print it would not work. So the answer would not be to remove the animation from the web version, but to make sure it translated to a series of images perhaps in a future print version. - Taxman 20:38, Jun 21, 2004 (UTC)

Older discussion

~~Maybe needs paragraph about Stieltjes generalization. Or perhaps link to separate article about same.~~

Done

Do you mean totally bounded functions and all that stuff? Loisel 07:30 30 Jun 2003 (UTC)

Image

Isn't it far TOO BIG?

Merge?

I think the Riemann sum and the Riemann integral have too much in common. I suggest merging information from them into Riemann integral, and make Riemann sum a redirect. —Preceding unsigned comment added by Igny (talk • contribs)

I would not be opposed to it. But I think you would need to also write your proposal on talk:Riemann sum and maybe even contact User:Icedemon who had put a lot of effort into that one. Oleg Alexandrov (talk) 18:19, 5 December 2005 (UTC)[reply]

I was not aware that a Riemann integral site existed. I agree that hte two should be merged, perhaps entering the work I did on Right/Left/Middle/Trapezoidal into the Examples section of the Riemann Integral page. Icedemon 03:29, 7 December 2005 (UTC)[reply]

Well, in this case I would not be for the move. The article Riemann integral is rather well-written and self-contained, while the article Riemann sum has a lot of wording which I find reduntant. All those sections on left Riemann sum, right Riemann sum, middle Riemann sum could be described just in a couple of sentences instead of several screenfuls of text. In fact, I already did that, see the ==Defintion== section in Riemann sum. So I would agree to move from Riemann sum to Riemann integral just the picture and a paragraph or two; most of the work there is reduntant. Otherwise, I would suggest leaving thing the way they are. Oleg Alexandrov (talk) 04:26, 7 December 2005 (UTC)[reply]

I agree that Left, Right, Middle, and Trapezoidal sums can be explained in a few sentences, but not to anybody that is not a mathematician. Please remember that the goal of Wikipedia is not only to be succint, but also to explain the subject in a manner that is understandable by everybody. I have attempted to do so in Riemann sum, and you still refer to my work as "redundant". Redundancy is necessary, in this case, in order to properly explain the material to anybody that does not have a degree in mathematics. As for hte merge, I maintain that either the Riemann Sum Left,Middle,Right, and Trapezoidal be moved in their entirety to Riemann integral or the pages remain separate. Icedemon 00:32, 9 December 2005 (UTC)[reply]

That's what I am saing too. The material at Riemann sum is way too detailed to be included in Riemann integral. So then the two articles should be separate. Oleg Alexandrov (talk) 01:29, 9 December 2005 (UTC)[reply]

LaTeX formatting

Who tagged this page as needing LaTeX formatting, and what do you need done? –Ryan McDaniel 15:02, 15 March 2006 (UTC)[reply]

You can tell who put the tag from the article history. The work to be done is converting html formulas to TeX formulas, which I think is not urgent at all, or maybe not even necessary. I'd suggest the template be removed from this talk page. Oleg Alexandrov (talk) 02:29, 16 March 2006 (UTC)[reply]

direct proof of f *g is Riemann integrable?

Given that both f and g are real functions, Riemann integrable on [a,b], then how to prove DIRECTLY that f * g (f times g) is also Riemann - integrable on [a,b]? —The preceding unsigned comment was added by 142.104.2.101 (talk) 18:50, 16 March 2007 (UTC).[reply]

Original definition

This article is really well-written and interesting. Does this article give the original definition of Riemann?

In most textbooks, e.g. Cohn: Measure theory, the equivalent definition by Darboux is given and they still call it the Riemann integral.

The Darboux definition is easier for a student to understand. Perhaps the main article should give the easier definition by Darboux, and then another article should give the original definition by Riemann. Pierreback 00:56, 25 March 2007 (UTC)[reply]

Link to Java program

I included the following link:

[[1]] Interactive computer program about the Darboux integral (in German)

I think it is a really nice java program which gives a feeling about the definition. Unfortunately it is in German and has now been removed (by another user) from the article. In my opinion it is good enough to be included. What is the general opinion? 83.253.10.236 23:55, 25 March 2007 (UTC)[reply]

Merger proposal

Rectangle method and Riemann sum should be merged into Riemann integral. Rectangle method is about the method of finding a Riemann sum; it is not a long article and could be its own section in Riemann integral. Riemann sum is the same thing as Riemann integral, with a simpler name. (See above for earlier discussion). Mazin07^C₪_T 15:49, 7 April 2007 (UTC)[reply]

I don't think that all those facts about left Riemann sum, right sum, etc, and all those pictures from Riemann sum belong at Riemann integral. That's too much detail. In Riemann integration all that matters is that the Riemann sum converges regardless of how the points inside the intervals are chosen.

So if a merger is done, I think very little from Riemann sum should make it in Riemann integral.

My preferred solution would be to keep things the way they are, and refer to the Riemann sum article in the "Riemann sum" section on Riemann integral, for people who want extended knowledge of the sums. Oleg Alexandrov (talk) 16:13, 7 April 2007 (UTC)[reply]

I'd agree that Riemann sum goes into too much detail for this article, and there is even more to say about Riemann sums if you want to. It does make sense to merge Rectangle method and Riemann sum, but I'd keep them separate from Riemann integral. -- Jitse Niesen (talk) 04:16, 11 April 2007 (UTC)[reply]

Rectangle method would be much better merged with midpoint method than Riemann integral. 69.231.241.75 06:12, 29 June 2007 (UTC)[reply]

Remember the Reader

I looked at the article on Riemman sum, and it's done up in advanced formal notation. While there are, of course similarities, Rectangle and Riemman are in in completely different leagues; not to mention the significant historical and pedagogical differences. To give an other example, playdoh and Riemman's contributions to differential geometry are related, and deformable solids such as playdoh (or some unspecified dough) are often used to convey the idea of smooth continuous deformation; however, it really just wouldn't be appropriate to lump (no pun intended) playdoh and non-euclidian geometry together in the same article; instead you have links pointing to each. And likewise, the Riemman integral could point to and be pointed to by the Rectangle article.

A merger could hold back Riemman (properly in the realm of advanced mathematics), and would certainly add luggage to the Rectangle article (a famous person, familiar with non-euclidean geometry once said: 'as simple as possible, just not too simple' [paraphrased from the original german])--ElectronicsEnthusiast 00:28, 22 July 2007 (UTC)[reply]

Unencylopedic tone and scope

This piece is very nicely written, but it has a didactic quality which is not appropriate for an encylcopedia article. It reads like a set of lecture notes, often saying "we see" and giving first provisional definitions and then after some discussion fixing them up to give the correct definition.

The article is also sorely lacking in historical information. References to Riemann's original papers are required. As another example, the claim that the Riemann integral is the first rigorous definition of an integral needs to be supported and explained by a comparison to the concept of integral used in the 200 years between the founding of calculus and the Riemann integral. Plclark 04:07, 30 September 2007 (UTC)Plclark[reply]

If you don't like it, be bold and fix it: Write out all the "we see"s and find references to Riemann's papers. I agree that the article sounds didactic, but when I wrote it (I'm responsible for most of the text) I figured this was the most comprehensible way to describe the Riemann integral to non-specialists.

I don't think the claim that the Riemann integral is the first rigorous definition is correct, though I did hear it made somewhere; what is more likely true is that it is the first definition of an integral that is independent of the notion of the derivative. My understanding is that before Riemann, the Fundamental Theorem of Calculus was taken as the definition of a definite integral. However, I'm no historian and I could be wrong (even completely wrong) about this. 141.211.62.20 00:43, 4 October 2007 (UTC)[reply]

Continuity

I want to point out that the discussion on continuity requirements for Riemann integration refer to measure theory. This seems unfortunate as one reason for using Riemann's approach is to avoid measure theory! Is there a concise and precise way to describe the requirements that does not use measure theory? Acolombi 17:07, 15 October 2007 (UTC)[reply]

Yes and no. The condition, precisely stated, is that the set of discontinuities has measure zero. But the concept of measure zero (with respect to Lebesgue measure) can be defined independently of the whole framework of measure theory: A measure zero set is simply one which can be covered by a set of balls B(x_i, ε_i) such that the sum of all ε_i (2ε_i if you're pedantic) can be made arbitrarily small. Perhaps that should be added to the article on almost everywhere. 141.211.62.20 01:11, 18 October 2007 (UTC)[reply]

What the damn?

what the damn does half this stuff mean? I am currently doing senior mathematics in high school and I have no idea what this means can someone explain it in layman's terms if need be? what i know is 'elongated "s" (integral sign) of a (lower limit) and b (upper limit) of f(x) dx is the integral. Can someone please explain it to me without the x sub i's and weighted sums and stuff like that what i mean is is in the form of ....(b-a/2 + f(a) + (a+b/6),etc...) something like that so that I can understand what the **** they are talking about. —Preceding unsigned comment added by 61.68.208.134 (talk) 08:32, 29 April 2009 (UTC)[reply]

You may find Darboux integral a little more approachable. In general, however, I would advise you to study hard and think carefully about what the article is saying. The subject is intrinsically difficult; there is no royal road. You will understand integrals better the more you think about them. Ozob (talk) 17:02, 29 April 2009 (UTC)[reply]

If you're in high school, you've probably gotten the same relatively shallow mathematics education that almost everyone else gets in high school. The topic of this article is a fair bit deeper than that and if you want to understand it all you'll probably have to get some more background. In other words it's not likely that someone can explain all that much more simply than the article already does. Perhaps try reading the article again and focus on the overview parts, then dig into the notation and specifics as you gain experience. The topic of this article is commonly covered in textbooks with titles like Advanced Calculus. Try picking up a few used or new texts like that until you find one that seems the most approachable, and consider getting some books that lead up to it. Most likely you will need to become comfortable with the x sub i notation (known as subscript notation). Also try looking for open textbooks online. There are many good sources of free math texts, though most are at a high level. If you look enough you can find good ones. - Taxman ^Talk 17:10, 3 May 2010 (UTC)[reply]

Cauchy's definition of the integral

I've been reading a little about different definitions of the integral, and a couple of books mention the "Cauchy Integral" which was formulated before the Riemann Integral and is in fact a special case of the latter where the "tag" of each interval in the partition is chosen to be the left endpoint of the interval. I notice that Wikipedia (and seemingly most other online sources from a quick google) doesn't mention it at all, and Riemann Integral even goes as far to say "the Riemann integral, created by Bernhard Riemann, was the first rigorous definition of the integral of a function on an interval." when it seems that Riemann merely generalized Cauchy's integral. Cauchy integral is also a redirect to Cauchy's integral theorem.

I feel like it should be included for completeness, at least from a historical perspective if nothing else. I'm no expert by any means, but I might have a go at making an article. It seems fairly odd that it is not referred to on WP at all so I'm not sure whether to just plow ahead and make an article (and change the Integral and Riemann Integral accordingly). It seems it would be fairly straightforward to include since the definitions are so similar to the Riemann.

So basically I'm just wondering if anyone objects to including this integral in WP, or knows anything about it. (also posting this at Talk:Integral) slimeknight (talk) —Preceding undated comment added 02:38, 25 November 2009 (UTC).[reply]

If Cauchy really did have a working rigorous definition of an integral, that would be news to me. I don't know where I first heard the claim that it was Riemann who first formulated a correct definition of an integral—it may have been this very article. If you have a reference that says otherwise and explains what Cauchy did, then adding it would be great! Ozob (talk) 20:59, 25 November 2009 (UTC)[reply]

Slimeknight's comment above appears to be quite correct. When I read the statement "the Riemann integral, created by Bernhard Riemann, was the first rigorous definition of the integral of a function on an interval", my immediate reaction was that this is dead wrong. I have seen this integral referred to as the Cauchy-Riemann integral. The catch is that Cauchy appears to have applied this definition only to continuous functions. Riemann seems to deserve credit for having made the first examination of Cauchy's definition applied to all functions (or, at least, all bounded ones) and to have asked the question "What functions are integrable?". You can read all about it here: ^[1] Look at Chapter 1, Section 1.2 (Catgod119 (talk) 00:10, 18 November 2016 (UTC))[reply]

References

^ "Lebesgue's Theory of Integration: It's Origins and Development" by Thomas W. Hawkins Jr., Ams Chelsea Publishing Series, Amer Mathematical Society.

Riemann integral "unworkable"?

The final text about improper integrals, from "This is also unacceptable..." seem to me a little harsh and misleading. First, I think that the final alternative of a improper integral definition (over the entire real line), with the double limit, has some undesirable properties but it is nonetheless "workable", and is the commonly accepted definition of that improper integral in the Riemmann integral context. This should be stressed. Second, the example of a uniformly convergent function that -in that definition- does not conmute with the integral is interesting, but misleading: it seems to imply that the example can be dealt by the Lebesgue integral, but (AFAIK) it cannot, nor could it: the integral of the succession fuctions are obviously 1, the function to which they converge is identically zero, so no integral (Riemman, Lebesgue or whatever) would permit us to interchange limits and integral in that case. Finally, to state that "This makes the Riemann integral unworkable in applications." is way too much. Riemman integral is just less useful for some theoretical work, that's all. —Preceding unsigned comment added by Leonbloy

Riemann integral equivalence to Darboux Integral

I have a question. Is the Riemann Integral really equivalent to the Darboux Integral?. For example, consider the function f(x)=sin(x)/x. I want to know if it is integrable between -1 and 1. The problem arises at x=0. Now, this function is clearly Darboux integrable, since you solve the problem of the non-existance of f(0) with the supremum operator. But I don't think it is Riemann integrable, since you can not evaluate f(0), and the Riemann definition clearly states that the sum must exist fon ANY election of evaluation numbers. If you defined f(0) as 1, you would not be solving the problem, because you would be integrating a different function,not f(x), since you are changing the domain of the function. I think it would be better to say that the Riemann and Darboux integral are equivalent iff f(x) is well-defined for every single point of the interval. Please correct me if I am wrong.I will deeply appreciate any replies. (DanteEspinoza1989 (talk) 06:20, 1 May 2010 (UTC))[reply]

I think the problem is that your f(x) is not integrable between −1 and 1. It's not defined everywhere in that interval, so it can't be integrable. After all, you can't integrate something that's not a function, and since f doesn't assign a value to zero, it's not a function.

A perhaps confusing aspect of this is that we can still integrate partially defined "functions" like f. We can change the value of a function at any single point (or more generally, on a set of measure zero) without changing the integral. What this means is that we can set f(0) = 1 and integrate, and we get the same thing as if we set f(0) = -999 and integrate. Or you could set f(0) to 37, or to 245.20485254, or whatever. So in a sense f does have an integral: To compute it, you first choose a value for f(0), then you take its (Riemann or Darboux or Lebesgue or ...) integral. That doesn't depend on your choice of f(0), so you can call that the integral of f.

You might appreciate the article Lp space (which is unfortunately a bit technical). We're considering p = 1. Ozob (talk) 20:06, 2 May 2010 (UTC)[reply]

Not all Riemann Integrals are Lebesgue Integrable

A function f is Lebesgue Integrable over a region if and only the function of the absolute value of f is Lebesgue Integrable over the same region. It is possible to construct a function f that is Riemann Integrable over a region but abs(f) is not Riemann Integrable over the same region. Provided that abs(f) were not Lebesgue Integrable over that region, neither would f be. I remember this as an exercise in my Measure Theory course in grad school, to provide such an f, but I don't remember how to provide an example.--Moly 12:41, 10 June 2010 (UTC) —Preceding unsigned comment added by Moly (talk • contribs)

This is not true. It may be true if you consider improper Riemann integrals, because improper integrals do funny things. But choose a and suppose that for all ε > 0 there exists a δ > 0 such that |x - a| < δ implies |f(x) − f(a)| < ε; then also ||f(x)| − |f(a)|| < ε, so |f| is continuous at a. I.e., the set where |f| is continuous contains the set where f is continuous, and hence the set of discontinuities has measure zero. Ozob (talk) 21:46, 10 June 2010 (UTC)[reply]

And even if you consider improper Riemann integrals, you should as well consider improper Lebesgue integrals, and again the latter include the former.—Emil J. 12:16, 11 June 2010 (UTC)[reply]

Cleanup of this Article

It looks as if this article needs some cleanup, as to increase clarity. I did so with the "Partitions of an Interval" section, and I would encourage others to do so too. Try to reduce redundancy of the same equations, and do not use excessive latex in the middle of the paragraph. Fraqtive42 (talk) 00:16, 22 July 2011 (UTC)[reply]

Counter example

I think it would be interesting to show as well what kind of integral is not a Riemann integral. — Preceding unsigned comment added by 189.115.235.6 (talk) 18:31, 4 October 2011 (UTC)[reply]

The article already gives an example of a non-Riemann-integrable function, and it links to Lebesgue integral, an example of another integration theory. Ozob (talk) 21:52, 4 October 2011 (UTC)[reply]

Definition

In the definition part of the article the following equation is found:

\left|\sum _{i=0}^{n-1}f(t_{i})(x_{i+1}-x_{i})-s\right|<\varepsilon .

Shouldn't it be made clear that "s" (the value of the integral) is not part of the summation (by using parenthesis)? — Preceding unsigned comment added by 178.199.110.247 (talk) 16:27, 19 October 2011 (UTC)[reply]

Wrong proof at "Integrability section"

It was noticed by User:186.48.9.230 that the given proof at Integrability of the Vitali-Lebesgue criterion is wrong. The discontinuity points of the function may be dense, so that there is no chance to cut a given (finite) partition of the segment into those that contain discontinuities of the function and the others. What shoud be done is to consider the closed set with measure zero where the oscillation of the function is larger than some ε. I don't have time to do it now, can someone do it? Bdmy (talk) 10:05, 31 May 2013 (UTC) Bdmy (talk) 10:10, 31 May 2013 (UTC)[reply]

I made the needed corrections - actually, the whole proof had to be rewritten. This was tiring... Dan Gluck (talk) 20:14, 23 October 2015 (UTC)[reply]

Generalizations section

I think the assertion of an improper integral not being easily extended is not entirely correct, or at least not the best way to explain it. My textbook defines $\int _{-\infty }^{\infty }f(x)dx=\int _{-\infty }^{a}f(x)dx+\int _{a}^{\infty }f(x)dx$ and further states that the integral on the left exists if and only if both integrals on the right exist, where a is any arbitrary number. Since neither of those two exist in the example given, the integral over the entire real line of that function does not exist.

I alternatively viewed the problem as a multivariate limit $\lim _{(a,b)\to (\infty ,\infty )}\int _{-a}^{b}f(x)dx$ . Then the notation $\lim _{c\to \infty }\int _{-c}^{c}f(x)dx$ is really $\lim _{(c,c)\to (\infty ,\infty )}\int _{-c}^{c}f(x)dx$ , i.e. only one path (the graph of a=b) along which a and b can approach infinity. If the first limit where the path is arbitrary exists, then it is unique, so the last one must also exist too (in other words, if that bivariate limit exists, then we automatically have the Cauchy principal value as the value of the integral, but the converse is false). Likewise, the nested univariate limits simply consider a path consisting of line segments parallel to a=0 and/or b=0, which again can't prove that the bivariate limit along any path exists. Therefore, the example given in this section is overdoing it: the bivariate limit does not exist, so all the talk of the integral of that function over the entire real line is meaningless.

Basically, I'm asserting that this problem with Riemann integrals over the entire real line is not an issue with interchanging limits, "shifting the function", or anything along those lines. It's just a direct consequence of the definition of a bivariate limit. I personally would find this to be a much simpler explanation.--Jasper Deng (talk) 07:18, 7 April 2014 (UTC)[reply]

Philosophically, if you think of an integral as computing a signed area, then you should simply be able to compute the positive area, compute the negative area, and subtract to get the result. When you can't do this, then either you need to worry about symmetry (as with the Cauchy principal value or singular integrals) or your integral is too weak (as with the Riemann integral). For a well-behaved integral the question of taking a limit of the domain should never come up; it should either be unnecessary or should be obviously and always true. What I think the current text is trying to get at is that the problem with improper Riemann integrals is embedded in their very definition as limits. We can't always avoid taking limits of the domain when working with integrals, but usually their presence indicates a problem, and they should never play a fundamental role in the theory. This is the case for the Lebesgue integral, for example: The Lebesgue integral is defined as the positive area minus the negative area, each of those areas is defined by a limiting process with respect to the values of the function, and taking a limit of the domain is unnecessary; and if for some reason we do need to take a limit of the domain, then various theorems (dominated convergence, monotone convergence, etc.) make this easy (at least in principle).

I'm not sure that the article is expressing this well. Yes, it's true that the bivariate limit you described does not exist, and that its failure to exist underlies everything else that the article is saying. But it's also true that a definition is justified by its properties, and the article shows why the definition as an improper integral has the wrong properties. We really want a definition of the integral which takes the right value for, say, compactly supported continuous functions, and which is continuous with respect to an appropriate topology on that function space (and therefore defines and extends to L¹). That might make the article better than details of limits. Ozob (talk) 14:26, 7 April 2014 (UTC)[reply]

Well, I think the explanation of those issues could be made more concise using something like "The Riemann integral of a function over the entire real line can be considered to be the bivariate limit

\lim _{(a,b)\to (\infty ,\infty )}\int _{-a}^{b}f(x)dx

. However, this creates several problems. For example, in the case of

f(x)=1

, this limit does not exist, meaning the Cauchy principal value is not translation-independent." etc. We could elaborate on how this makes defining improper integrals as limits problematic as general, as you say, and thus emphasize this limitation of the Riemann integral. This would replace all of those "tries" at limit definitions that don't work as aformentioned.--Jasper Deng (talk) 07:19, 8 April 2014 (UTC)[reply]

OK, I rewrote that section. What do you think? Ozob (talk) 15:00, 8 April 2014 (UTC)[reply]

Seems pretty good. I made one minor addition.--Jasper Deng (talk) 17:14, 8 April 2014 (UTC)[reply]

Links [8] and [9] at "Notes" aren't in the article

The links [8] and [9] don't make a appeareance anywhere on the article, and in fact when i Click them, nothing occurs. I think someone deleted the links in the article but they remained in the "Notes" section. — Preceding unsigned comment added by Santropedro (talk • contribs) 20:43, 11 July 2015 (UTC)[reply]

These notes appear in the proof between notes [7] and [10], which is displayed only when you click on "show". D.Lazard (talk) 02:45, 12 July 2015 (UTC)[reply]

Assessment comment

The comment(s) below were originally left at Talk:Riemann integral/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Needs some basics historical information, when was it introduced. Some more comparison with Lebesgue integral would compete a nice article. --Salix alba (talk) 16:46, 18 April 2007 (UTC)[reply]

Last edited at 16:46, 18 April 2007 (UTC). Substituted at 02:33, 5 May 2016 (UTC)

Lebesgue's criterion

In the section "Integrability", it is stated A bounded function on a compact interval [a, b] is Riemann integrable if and only if it is continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). Later, in the section on "Generalisations", it is stated Moreover, a function ƒ defined on a bounded interval is Riemann-integrable if and only if it is bounded and the set of points where f is discontinuous has Lebesgue measure zero.

These two statements are close but not identical; in particular, the second statement says something about unbounded functions on a bounded interval (namely, it claims they are not Riemann-integrable), whereas the first statement does not. I believe that the second statement is correct (at least if the 'bounded interval' is replaced with a compact interval); and if this is so, perhaps the first statement should be amended to give this stronger claim. 130.88.123.107 (talk) 14:56, 24 February 2017 (UTC)[reply]

ℛ...?

This page is the destination of the redirect on ℛ. But the ℛ character is never mentioned. Is the redirect correct? Railwayfan2005 (talk) 22:17, 15 October 2018 (UTC)[reply]

I have nominate it for deletion (Wikipedia:Redirects for discussion/Log/2018 October 16#ℛ). D.Lazard (talk) 09:27, 16 October 2018 (UTC)[reply]

Lebesgue-Vitali Theorem

The proof of the Lebesgue-Vitali characterization theorem could maybe easier to read if broken as follows:

A bounded $f:[a,b]\to \mathbb {R}$ is Riemann integrable if and only if its upper and lower Darboux integrals coincide;
For any bounded $f:[a,b]\to \mathbb {R}$ , the difference between the upper and lower Darboux integrals of $f$ is equal to the upper Darboux integral of the point oscillation of $f$ :

{\overline {\int _{a}^{b}}}f(x)dx-{\underline {\int _{a}^{b}}}f(x)dx={\overline {\int _{a}^{b}}}\omega _{f}(x)dx.

The point oscillation is a bounded, upper semicontinuous, non-negative function.
For any bounded, upper semicontinuous non-negative function $g:[a,b]\to \mathbb {R}$ one has

{\overline {\int _{a}^{b}}}g(x)dx=0

if and only if

g(x)=0

a.e.

The point oscillation of $f$ vanishes exactly at the continuity points of $f$ .

Therefore, a bounded function on $[a,b]$ is Riemann integrable if and only if it is continuous at almost any point $x\in [a,b]$ . pm a 21:46, 24 March 2021 (UTC)[reply]

Riemann Darboux Equivalence

In Definition subsection Riemann integral, it's stated that the Darboux Integral article proves equivalence to the second definition given. I'm not convinced this is the case. The Darboux article proves that Darboux integrable implies Definition 2 and that Definition 1 implies Darboux Integrable. But the second proof (under "Details of finding the tags" on the Darboux page) doesn't begin with a tagged partition that is then refined; it just shows that arbitrarily fine partitions can be tagged to be arbitrarily close to the lower Darboux sum.

Maybe there's a subtle reason that I'm missing for why the second proof works when starting from Definition 2. But to me, beginning with that definition gives you that for some $\varepsilon >0$ there is a tagged partition whose refinements are that close to some value, and we want to show that some of those refinements can be made close to the upper and lower Darboux integrals, but we have to deal first with the tags we already have and must keep, which won't necessarily be close to their interval's supremum and infimum and so we would need to enclose them in within a sufficiently small interval, and the rest of the intervals could be assigned tags close to their infimum.

If I'm right and the Darboux article's proof is insufficient, one would need to be added to this section to satisfactorily prove the two definitions' equivalence, at which point it might make sense to make the equivalence arguments a separate section. The page Characterizations of the exponential function would be a good template. I don't think that I have the math or Wikipedia editing experience to make those changes though. Lnkov1 (talk) 00:42, 11 November 2022 (UTC)[reply]

[1] "Lebesgue's Theory of Integration: It's Origins and Development" by Thomas W. Hawkins Jr., Ams Chelsea Publishing Series, Amer Mathematical Society.

[1]