What is the proof of riemann integral?

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The Riemann integral is based on the concept of approximating the area under a curve using rectangles, where the area can be determined by taking the limit of upper and lower sums. A function is Riemann integrable if the upper and lower sums converge to the same limit, which defines the area under the curve. However, not all functions are Riemann integrable; for example, functions with discontinuities can fail this criterion. The discussion highlights that the existence of a Riemann integral is related to the measure of discontinuities, with functions having a set of discontinuities of measure zero being integrable. More advanced integration theories, like the Lebesgue integral, extend the concept to a broader class of functions.
  • #31
Oh I just saw MasterWonk's post, yea we are more or less talking about the same thing.

Rather than talking from a text not everyone has (eg Rudin), I just googled for a few seconds and found a course paper with some of the relevant definitions
http://www-finmath.uchicago.edu/Courses/Stochastic2.pdf

Look on page 6, they use absolute convergence (not absolute continuity, I never said anything about that) in the definition of what is or is not integrable.

(Incidentally,being a theoretical physicist, we use the word pole somewhat informally and sloppily - mea culpa. Regardless in the writeup of the HK integral that function that Fourier Jr suggested is mentioned in several places, it was a motivating concept behind their work. Moreover, please keep in mind the f(0) = 0 condition)

-edited to remove potential ambiguites in decorum =)
 
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  • #32
Here's a problem from Royden, third edition, page 93, 10b:

The improper Riemann integral of a funciton may exist without the function being integrable (in the sense of Lesbegue), e.g., if f(x) = (sin x)/x on [0, ∞]. ...
 
  • #33
haelfix,

i apologize for accusing you of the error involving absolute continuity. that one was in Fourier jr's post.

perhaps i was too hard on you for your error calling a nice differentiable point of a function a "strong pole". to me a pole is a point where the value of an analytic function is infinity (look for example in rudin where you say you learned the subject), whereas this function had value zero at this point. I fail to see how there could be any difference of opinion about such a matter.

question: when i give you useful and correct historical information from a standard famous source, why do you fault me for using a source "not everyone has"?

when i reveal to you that ideas you think are "cutting edge" mathematical research are 90 years old, why are you unhappy to learn this?

this is not a debate, this is an attempt to arrive at understanding. some of us here are actually mathematical scholars, and we use appropriate source material. if this were merely a contest to see who could top the other person we might limit our sources sportingly to the american mathematical monthly.

if we were discussing theoretical physics which you say is your specialty, i would be glad to learn from you about it.

have a nice day.
 
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  • #34
Haelfix said:
Look on page 6, they use absolute convergence (not absolute continuity, I never said anything about that) in the definition of what is or is not integrable.

They use absolute convergence when referring to the integrability of a simple function. Which makes some sense, since the Lebesgue integral of a simple function is defined with a summation, and you want that summation to converge absolutely.

But saying that integrability requires absolute convergence makes no sense when referring to an arbitrary function. Absolute convergence is a property of a series, not of a function.
 
  • #35
let me try to say this this as simply as possible (limited by my knowledge and understanding).

MY impression is the following:

if you think "integral" should mean area under the graph, then there is no definition more general than lebesgues.

i.e. to integrate a positive function, there is no better or newer approach than that of lebesgue, all later approaches agree with his, and none work on a larger class of functions.

to integrate a function with both positive and negative values, there are various definitions yielding different results.

they all agree when both positive and negative parts have finite integrals, but when both positive and negative parts have infinite integrals, then there are various different ways to decide how to combine these integrals.

1) lebesgue's method apparently says to look separately at the positive and negative parts, integrate them separately and subtract.

in particular if f has finite lebesgue integral, then so does |f|.
this is the most restrictive convention.

2) other methods such as improper riemann and HK integration allow one to cancel the positive and negative parts against each other so that some functions whose positive and negative parts both have infinite integrals, yield when combined, a finite integral.

If this interests you, you may be interested in the HK integral.

nonetheless, whenever |f| has a finite HK intergal, then it already has a finite lebesgue integral, so nothing is gained from the HK approach, as proved in the paper i cited, which is available on the web.

If this is in error, please correct me.
 
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  • #36
my apologies haelfix, I've been acting like an A**H***.

Some of the things i said may nonetheless be correct however. (in addition to this.)
 
  • #37
I think we've all been a tad hasty and misunderstanding one another here, again my apologies. I was referring to Rudin (my book) when I said 'rather than use sources others don't necessarily have' and pulled out the weblink, I personally have no problems whatsoever and welcome sources outside the internet.

I really don't know much about the so called Gauge integral, I appreciate your comments, they are much more informed than what I know of the subject. I also think we got a little hasty on the terminology and started talking by one another. Physicists often use the word 'pole' to mean a place where something goes wrong, not necessarily the complex analysis point of view, again my fault since I am in the wrong forum. Other than that, I agree with everything in your last posts, all of that makes perfect sense to me.

I'd like to point out that IMO from a historical point of view, I think the Lebesgue integral's primary draw is not just that it can make sense of integration outside of the reals, but also that it has various nice convergence properties that the Riemann integral does not have. Rudin talks about some of them.
 
  • #38
thank you. i agree with your comment on convergence properties. that is certainly to me a key virtue of lebesgues method.

i am not sure what use you wish to make of new integration theories but there may be some out there that would interest you.

i am myself not up on these but there is a very new one (to me) in algebraic geometry called "motivic integration", and of course we are still challenged to make sense of feynman integrals.

there is a lot of work on this area, but i sense that it tries to get round the difficulty of making sense of them, by various means. here i am told the difficulty is that of integrating over path spaces of infinite dimensions.
 
  • #39
re convergence properties, i have even found a source where the lebesgue integral is defined by this property.

i.e. a function is lebesgue integrable if and only if it is almost everywhere the pointwise limit of a sequence of (measurable) step functions, whose integrals converge to some number.

then the inetgeral is proved to be independent of thes equence of step functions and to depend only on the a.e. pointwise limit, and we're off.

have a good day!
 
  • #40
Hrm, that definition sounds like it would apply to (sin x)/x over [0, ∞)... just have the n-th approximation be zero on [n, ∞).
 
  • #41
mathwonk said:
i.e. a function is lebesgue integrable if and only if it is almost everywhere the pointwise limit of a sequence of (measurable) step functions, whose integrals converge to some number.

I don't think that's correct. I've seen a definition very close to that one, but it required the integrals of the absolute values of the step functions to converge to a number. Without that stronger condition, the "if" doesn't hold.
 
  • #42
thank you master coda. i was trying to simplify the statement so as not to use technical language and i vastly oversimplified it.

(I try to take it as a compliment when I am corrected in an error, since at least it means I was understood correctly!)

the correct statement seems to be that f is integrable if and only if the sequence of step maps is not only pointwise convergent to f almost everywhere, but also "Cauchy in the L^1 norm", which as you know, means that the integrals of the absolute values of the differences |fn-fm| of the step functions, converge to zero as both n,m go to infinity.

This mdefinition makes sense even for Banach space valued functions, and may be called the bochner (version of lebesgues) integral.

It is not I think appropriate only to assume the integrals of the absolute values of the step functions converge, as that would not imply the step functions are L^1 Cauchy.

for instance we would like to be able to compute the integral of the limit from the limit of the integrals of the step functions. but suppose the nth step function were equal to 1 on the interval [n,n+1] and zero elsewhere. then all the integrals of all the absolute values would equal the integrals of the functions themselves and would all be 1. the constant sequence 1,1,1,1... certainly converges, but the pointwise limit function of these step functioins is the zero function, whose integral is not 1, i.e. is not the limit of the integrals of the approximating step functions.

so one needs also the step functions to be Cauchy in the L^1 sense, it seems.

what do you think of this, hurkyl?

the point of course is to make the lebesgue integrable functions the completion of the step functions in the L^1 norm. thus the convergence properties are "built in".

now what about cos(x)/x?
 
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  • #43
mathwonk said:
the correct statement seems to be that f is integrable if and only if the sequence of step maps is not only pointwise convergent to f almost everywhere, but also "Cauchy in the L^1 norm", which as you know, means that the integrals of the absolute values of the differences |fn-fm| of the step functions, converge to zero as both n,m go to infinity.

I'm not sure if that's strictly necessary, although you're probably right. I thought that you only needed stronger convergence if you actually wanted the integral of the limit to equal the limit of the integrals (which is obviously much more useful, but just not what I was thinking about).
 
  • #44
master coda, are you perhaps thinking of convergence of series of functions? as opposed to sequences?
 
  • #45
mathwonk said:
master coda, are you perhaps thinking of convergence of series of functions? as opposed to sequences?

I was thinking of sequences, but series does make a lot more sense. The combination of your definition and the vague memory of another definition seems to be confusing me.
 

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