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What is the proof of riemann integral?

  1. Feb 16, 2005 #1
    so riemann integral pretty much says that if you take closer and closer approximations, then you can find the area of whatever(not too precise, I know, but doing the rectangles and stuff).

    I'm looking for a proof of it, but all I can find are more general things, i.e, Riemann integeral is integeral on [a, b], etc etc.

    how was it proven that taking approximations and all that, actually worked??
     
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  3. Feb 16, 2005 #2

    HallsofIvy

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    But it doesn't say that for EVERY function.

    What you can do is this: inside of each interval take M to be the largest value of f(x) and m to be the smallest. for all x in that interval m< f(x)< M so m&delta;x< f(x)&delta;x< M&delta;x. That is: the "true area" inside each interval lies between the
    rectangle area m&delta;x and the rectangle area M&delta;x. The "true area" under the curve lies between the sum of m&delta;x and M&delta;x.

    IF the limit of sum m&delta;x and sum of M&delta;x are the SAME, THEN it is clear that the "true area" must be equal to that joint limit (the "pinching" theorem).

    Of course, THAT is not always true! For example if f(x)= 1 for x irrational, f(x)= 0 for x rational, since every interval contains some rational and some irrational, the "upper sum" (sum of M&delta;x) is always 1 and the "lower sum" (sum of m&delta;x) is always 0. Those are not the same and so that function is NOT "Riemann integrable".
     
  4. Feb 17, 2005 #3

    matt grime

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    It is the definition that IF the approximations using rectangles works THEN it is (Riemann) integrable. It is not something one proves.
     
  5. Feb 17, 2005 #4
    Tom Apostol's Calculus Volume I has pretty good coverage of what I think you're looking for. You might want to check it out of a local library and give pages 57-80 a look.
     
    Last edited: Feb 17, 2005
  6. Feb 17, 2005 #5

    so for example, this link

    http://www.hyper-ad.com/tutoring/math/calculus/Construction of the Riemann Integral.html


    is this pretty much it? to me, it seems like it.


    and reading it, how did they conclude that the upper limit equals teh lower limit?
     
    Last edited: Feb 17, 2005
  7. Feb 17, 2005 #6

    arildno

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    There is an unfortunate representation in the link, that it seems to regard the concept of "area under the curve" as well-defined, IRRESPECTIVE of whether or not the upper and lower partial sums converge to a common number in the limit.
    But, it is precisely when the sums DO converge to a single, common number that it is meaningful to speak of the "area" beneath the curve, in the Riemann sense (that single, common number is then the assigned area value).
    HallsofIvy's example shows a function in which an area in the Riemann sense of the word is impossible to assign to the curve.
     
    Last edited: Feb 17, 2005
  8. Feb 17, 2005 #7

    ah...ok. so basically, it's saying that IF the upper and lower sums are the same, then we can find the area under teh graph? correct? and that is pretty much the definition of reimann integral right?
     
  9. Feb 17, 2005 #8

    arildno

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    Yep, as matt grime said, the definition of being Riemann integrable is that the upper and lower sums converge to a single limit.
    The fancy word "Riemann integrable" says the same as saying it is meaningful to assign an area beneath the curve (in the Riemann sense of "area").
     
  10. Feb 17, 2005 #9

    mathwonk

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    riemann integration means trying to squeeze the graph in between some rectangles which lie below the graph, and some other rectangles which lie above the graph, so that the amount of area caught between the two sets of rectangles can be made as small as desired.

    one acse whne this is true, i.e. when the function does have a riemann integral, is for an increasing function.

    the reason is as follows: subdivide the interval [a,b] say by a point c with a < c < b, and put a rectangle under the part of the graph between a and c, and another one above the graph, and make them fit as close to the graph as possible.

    then move over to the interval [c,b] and do the same thing. notice since the graph is increasing that the lower rectangle fopr the right hand interval will be the same height as the upper rectangle for the elft ahnd interval.

    hence if you consider the "brick" lying between the upper and lowers rectangles, there is a brick over each interval, but the total height of the bricks, is at most the total height of the whole graph between a and b.

    thus if you subdivide into a ,lot of loittle intervals, each of width 1/N, and if the total heoght of the graph is B, then the total area of these bricks, i.e. the difference between the upper and lowers rectangles, wiill be B/N. hence as N gets larger this difference gets smaller.

    so the function is riemann integrable.

    a picture makes this obvious.

    this argument ios due to newton, and here is a site with hnewtons own illustration of it:

    http://www.math.sunysb.edu/~tony/132F02/imagecaption.html
     
    Last edited: Feb 17, 2005
  11. Feb 18, 2005 #10

    mathwonk

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    i have noticed that this site accepts "fazzbazz", but not "oink" as an acceptable post. I find this prejudicial against pigs.

    of course it does take "yellow pig", no doubt in deference to m. spivak.
     
  12. Feb 18, 2005 #11
    Zinnnnnnng :biggrin:
     
  13. Feb 19, 2005 #12

    Haelfix

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    Incidentally, is there a more general theory of integration than integration with say a Lebesgue measure? I'm looking for the cutting edge of mathematical research.
     
  14. Feb 19, 2005 #13

    matt grime

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    Yes, measure theory is used in integration. I doubt that you'd call it cutting edge though. Measure theory is now more of a tool than a research topic. It's useful in probability, calculus (real and complex), hilbert spaces, banach spaces, dynamical systems.....
     
  15. Feb 19, 2005 #14

    mathwonk

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    the lebesgue definition of the integral is over 100 years old, hence certainly not cutting edge research. It is however a staple of all cutting edge research in analysis, i.e. no one doing cutting edge research uses the riemann version of the integral.

    lebesgue's integral is a very simple variant of riemann's, but introduces much mroe technical difficulty.

    for example, suppose you wanted to estimate the average incomes of the readers of a magazine.

    The Riemann method says take all the readers in each state, and pick one of them at random, and use his income as the estimate for that state.

    then do it again, taking only all readers in one county, and again pick one of them at random, and use his income as your estimate for that county.

    This is obviously an incredibly inefficient method, that only works well if all people living near each other have similar incomes. i.e. some requirements of continuity are built into riemann's definition.

    the lebesgue method says instead, pick a range of incomes and ask readers to check off whether their income lies in the range fro 10K to 20K, or 20K to 30K, etc....

    Then you add up all the people who answer 10k-20k, and use say 15k as an estimate for all their incomes.

    this is much better, but requires you to be able to count how many people check each block.


    so for a function, lebesgue says to subdivide, not the x interval, but the y axis, into subintervals. then for each subinterval of values, you measure how "many" points x of the domain interval exist such that f(x) is in the given interval.

    I.e. you try to measure the "size" or "length" of the inverse image under the function, of an interval of the y axis. this is technically quite hard and leads to the theory of measure.

    nonehteless, its superiority leads eventually to a better theory, which assigns in some way an integral to essentially all functions.


    E.g. an example given earlier was the function which equals 0 at all rationals in [0,1] and 1 at all irrationals. riemann's method simply never converges to an answer for this function, because all nearby points are treated alike by his estimates, whereas there exist nearby points with radically different values.

    Under lebesgues method, one says the integral of this function equals 1 times the size of the irrationals, plus zero times the size of the rationals in [0,1].

    now since essentially all the points in [0,1] are irrational the integral is 1.

    I.e. given any interval of length e>0 no matter how small, one can chop it up into an infinite sequence of disjoint pieces, of lengths e/2, e/4, e/8, e/16,.... and still cover up all the rationals in [0,1] using these pieces, since the rationals are countable.

    so the rationals have length zero. "hence" the irrationals in [0,1] must reasonably have length 1.
     
    Last edited: Feb 19, 2005
  16. Feb 19, 2005 #15

    GCT

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    Trying to recall from what I learned last week. Basically the graph of the region in which we wish to take the (riemann) integral has to be continuous over the region.

    You start by observing a partion [tex]P[/tex], represented by the interval [itex] x - x_{i-1} [/itex] . The area of the rectangle at a point [itex]x_i^*[/itex] which is between the interval is represented by [itex] A^n = f(x_i^*)( \Delta x) [/itex] . The function itself as well as the [itex] \Delta x[/itex] can be expressed in terms of [itex]n[/itex] where n basically relates to its position in [itex] \sum_{i=1}^n [/itex] . So the Riemann sum, [itex]R[/itex] represents the sum of all of the rectangles with variable partitions as well as the variability in regard to the position of [itex]x_i^*[/itex] within the intervals. [itex]R= \sum_{i=1}^n f(x_i^*)( \Delta x) [/itex] . Now place all of this in terms of [itex]n[/itex] and take the limit as [itex]n[/itex] infinity. So the definition of the integral in relation to the Riemann sum is [itex] \lim_{x\rightarrow n} R [/itex] as [itex]R[/itex] is defined above.
     
    Last edited: Feb 19, 2005
  17. Feb 19, 2005 #16

    mathwonk

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    the correct theorem on existence of riemanns integrals is the following:

    a function f defined on a closed bounded interval [a,b], has a Riemann integral over that interval if and only if the set of discontinuities of f has measure zero, (in the sense I demonstrated above that the rationals have measure zero).

    In particular you see you cannot even discuss the problem of riemann integrability definitively without the concept of lebesgue measure, or at least of measure zero.

    It has always been a pet peeve of mine, for stduents to come away from a calculus thinking that a function needs to be continuous in order to be Riemann integrable.

    I.e. since the theory of the integral is designed to strengthen the idea of area, one would surely want to be able to integrate a function whose graph was a series of rectangles, and hence had a finite number of jump discontinuities.

    However from the way we teach them, i.e. by saying that all continuous functions are integrable, they get exactly this impression, as perhaps general Chem tutor has done.

    for that reason i always emphasize the result i proved above that all monotone fucntuions are integrable, whether continulus or not.

    also all bounded functions with only a finite set of discontinuities are integrable.
     
  18. Feb 19, 2005 #17
    godel proved that there is no theory of integration which can deal with every possible function, so i guess there's a limit (in a manner of speaking) on how general a theory of integration can be. there's also the daniell integral & the henstock-kurzweil integral (aka generalised riemann integral).
     
  19. Feb 20, 2005 #18

    Haelfix

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    Thats cool, I was not aware of the HK-integral, it looks interesting I'll read up on it.

    My knowledge of integration theory more or less stops at Lebesgue-Stieltjes types (well I have a working knowledge of other notions of integration in different settings, like say using the Haar measure.. I guess that counts) after undergrad analysis.

    Is the HK integral more or less the most general version that you can think off?
     
  20. Feb 20, 2005 #19

    mathwonk

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    i do not believe any of these integrals works on a larger class of real valued functions than lebegue's.

    i think the extra generality is a bit illusory, i.e. one abstracts the properties of an integral in them.

    in fact even to produce a function which is not lebesgue integrable as I recall, you need to use the axiom of choice.

    i am not an expert either though, and am just thinking out loud here from ancient memory.

    haar measure is also not more general then lebesgue integration, or than lebesgue measure. rather the opposite, it is more special. as you surely know, it is merely the name for an ordinary measure which is translation invariant under a given group operation.

    let me clarify my use of the words " more general".

    if you mean a definition of integration which works on a space with less structure than euclidean space, then these other definitions are more general. but if you mean by more general, as i do, a definition which works on a larger class of functions on euclidean space than lebesgue measure does, then they are not more general.

    i.e. in the usual setting these other definitions do not give anything new. they are merely abstractions of the usual definition, which are designed to make sense in a larger class of spaces.

    for example, the unique "haar measure" on eucliden space (if normalized by assigning measure 1 to a unit cube, and for the usual group operation of translation), is lebesgue measure.

    the daniell integral is treated on page 132-3, of the classic text by riesz nagy, published over 50 years ago, indicating again its traditional nature as a theory of integration. (The original papers of daniell and frechet cited there appeared in 1915 and 1917.) the treatment moreover there is only 2 pages, since the arguments are the same as they have given for lebesgue measure.

    briefly, the "daniell integral" consists of defining an integral as a positive linear functional, i.e. something that assigns numbers to functions, so as to be linear, and to take "positive" (in the european sense) functions to positive numbers.

    this is nice, but not a big deal in comparison to lebesgue's innovation.
     
    Last edited: Feb 20, 2005
  21. Feb 20, 2005 #20

    mathwonk

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    my impression from frequenting this skte is that the questioners are mainly people who use the web to do their learning and the people who answer the questions (correctly) are mainly people who read books to learn.

    I.e. I think it is very hard to actually learn anything on the web alone.

    e.g. if you want to know whjat lebesgue meaure means, you might happen upon the following site:

    http://planetmath.org/encyclopedia/Integral2.html

    where you find an almost useless explanation.

    of course it may be that with persistence you will also find somewhere else an explanation of lebesgue measure, the only deep ingredient of this explanation, and the one which is missing here.

    OK, I admit that two links further away I found the definition of the key concept of measurability and the definition of lebesgue outer measure.

    (say...isn't this version of the theory due to caratheodory? who is not credited.)
     
    Last edited: Feb 20, 2005
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