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Lebesgue vs Riemann integral

  1. Sep 25, 2007 #1

    quasar987

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    In the course I'm taking, we are already done with Lebesgue integration on R, and while we have proven that for continuous fonctions, the Riemann integral and the Lebesgue integral give the same output, we have not investigated further the correspondance btw the two. So I have some questions...

    i. Is the class of (absolutely) Riemann integrable function a subset of Lebesgue integrable ones? Or are there functions that are (absolutely) Riemann-integrable but whole Lebesgue integral diverge?

    ii. For those Riemann-integrable functions that are also Lebesgue integrable, are the Riemann and Lebesgue integral equal?


    I would also be interested to know what were the historical motivations for creating this new theory of integration, and also, what are its use today?
     
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  3. Sep 25, 2007 #2

    morphism

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    Every Riemann integrable function is Lebesgue integrable and their integrals are equal.

    The Lebesgue integral is really an extension of the Riemann integral, in the sense that it allows for a larger class of functions to be integrable, and it does not succumb to the shortcomings of the latter (e.g. interchanging limits and integrals behaves better under the Lebesgue integral).

    The Wikipedia page on Lebesgue integration is nice.
     
  4. Nov 15, 2007 #3
    Does this count even for unproper Riemann integrals?
     
  5. Nov 15, 2007 #4
    Do you have an example of an interesting function over some domain that is not Riemann integrable, but Lebesgue integrable, and compute its integral? I took a course in measure theory, and it seemed that all we did was prove existence theorems and did nothing computationally.
     
  6. Nov 15, 2007 #5
    An example may be the function defined as 1 for rational x, and 0 elsewhere, but of course, this depends on what you consider interesting.
     
  7. Nov 15, 2007 #6

    mathwonk

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    the reason for extending the definition is to have the limit of the integrals equal the integral of the limit more often.

    a sequence of riemann integrable functions can have convergent integrals without the limit function being riemann integrable. lebesgue integrals repair this flaw.
     
  8. Nov 15, 2007 #7

    morphism

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    Sadly, no. There are many examples; a couple of typical ones are (1/x)sin(1/x) on [0,1], or sin(x)/x on R.

    As for computational things, this assignment seems to have some interesting ones.
     
  9. Nov 16, 2007 #8
    It is impossible to construct a non-Lebesaure measurable set which does not involve the axiom of choice. That is a nice advantage.
     
  10. Nov 17, 2007 #9

    mathwonk

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    i recommend wendell fleming, calculus of several variables. i am not an expert but will recite what i just read there for you.

    i consider only functions with bounded domains.

    a step function is a function with a finite number of values, whose level sets are "measurable sets", i.e. sets to which a length can be asigned. for example intervals are measurable. but also much more complicated sets are measurable by approximating them by intervals and taking limits.

    the integral of a step function with values a1,...an on the sets S1,...,Sn is the linear combination a1 m(S1)+...an m(Sn), where m(S) is the length (measure) of S.


    now a function f is integrable if for every e>0, f admits both larger and smaller step functions, whose integrals differ by less than e.

    a function is riemann integrable if the step functions can be chosen to be constant on intervals, i.e. if the sets Si can be chosen as intervals.

    thus the integral is seen to be a direct generalization of the riemann integral, where the level sets of the approximating step functions are allowed to be more general than intervals.

    a function is continuous if and only if the inverse image of each open interval is open. it is called measurable if the inverse image of each interval is measurable.

    then a bounded function is integrable if and only if it is measurable. since open sets are measurable, in particular all continuous functions are integrable.

    i hope this is correct.
     
    Last edited: Nov 17, 2007
  11. Nov 18, 2007 #10

    rbj

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    from an engineer's POV, here is a cute quote from Richard Hamming:

    when i first learned about the Lebesgue integral, i noticed right away that it pointed to problems of interpretation and usage of the Dirac delta function. engineers think of it differently than mathematicians. we treat [itex]\delta(x)[/itex] as a function that is zero almost everywhere, yet with an integral of 1, which is contrary to what Lebesgue integration would say. Lebesgue integrations says that if f(x) agrees with g(x) almost everywhere, and if f(x) is integrable, so is g(x) and its integral is the same. in the physical sciences, i think that is a perfectly good way to look at it (and indeed there are undergraduate Electrical Engineering texts written as such), but the mathematicians do not like it.

    anyway, from a point-of-view of describing real physical processes and quantities, i hadn't worried much about the difference betweent the two.
     
  12. Nov 18, 2007 #11

    mathwonk

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    what a philistine, head in sand, attitude. if it doesn't immediately build a better bridge, who cares?
     
  13. Nov 18, 2007 #12

    rbj

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    it's not about immediately building anything, it's just that it is very convenient in Linear System Theory (the whole sub-discipline where we learn about signals, spectra, linear and time-invariant systems, impulse responses, convolution, etc.) to simply treat the dirac delta as:

    [tex] \delta(t) = \lim_{\Delta \rightarrow 0} \delta_{\Delta}(t) [/tex]

    where [tex] \delta_{\Delta}(t) [/tex] is any of a bunch of "nascent delta" functions all having unit area in them. the simplest example is:

    [tex] \delta_{\Delta}(t) = \frac{1}{\Delta} \mathrm{rect} \left( \frac{t}{\Delta} \right) [/tex]

    and rect() is the unit rectangular function (i can't remember how to express it nicely in LaTeX).

    anyway, it's far better to use this representation with undergrad EEs than it is to try to explain to them the difference between Lebesgue and Riemann or how the Dirac delta is a "distribution" and not a function. also, Richard Hamming was pretty well respected among academics when he was alive (the "Hamming window" is his creation), so i dunno who is the Philistine you're referring to. perhaps me, and that's fine, but all of us EEs are Philistines, then. the differentiation of the simple limit of functions definition and the more proper distribution definition of the Dirac delta has never been shown to be useful in the descriptions of physical systems all of which have some decent continuity in their properties.
     
  14. Nov 18, 2007 #13

    HallsofIvy

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    Actually, the difference between Riemann and Lebesque integrals should be of great importance to engineers! When Fourier, who was himself an engineer rather than a mathematician, developed Fourier series to solve differential equations, he made two claims: first, that any integrable, periodic, function could be expanded in a Fourier series and second, that any (convergent) Fourier series gave an integrable, periodic series. The first statement was obviously true. The second just as obviously false using the Riemann integral. The Lebesque integral was developed specifically to make that second statement true- since Fourier's technique obviously worked!
     
  15. Nov 18, 2007 #14

    matt grime

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    rbj - since the dirac delta isn't a function from R to R, I don't really get your point. It is a functional on a some suitable space, not a function from R to R.
     
  16. Nov 18, 2007 #15

    HallsofIvy

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    matt, rbj made exactly that point- that it isn't a function. In fact, it is possible to treat distributions (or generalized functions) as equivalence classes of sequences of functions. That, in effect, is what rbj is doing.
     
  17. Nov 18, 2007 #16

    matt grime

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    No, I meant 'why did rjb say that lebesgue integration implies problems for...' It doesn't imply any such thing. Whilst some may treat it as if it were a function, it isn't a function, and everyone knows that, including engineers.

    Indeed one doesn't even need to invoke Lebesgue integration. If f=g at all but a point, then the Riemann integrals of f and g are equal if they exist.
     
  18. Nov 20, 2007 #17

    rbj

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    i don't think you understand what it's like in a 4-year undergrad Electrical Engineering program. now it might be worse in Math and/or Physics (i think it's conceptually deeper in Math/Physics than EE), but in a 4-year EE program, each and every semester is packed with required courses and "required electives", if you drop even one, you must make it up in the summer or you will not graduate in 4 years. anyway, in this tight schedule, the only math courses in the Math department you can expect will be 3 semesters of Calculus and 2 semesters of Diff Eq. (and similar) following. in the Diff Eq. the EE student will see Laplace Transforms and will see the Dirac Delta presented much the same as i had mentioned (mostly so we can learn that [tex] \mathcal{L}\{ \delta(t) \} = 1 [/tex]), and that is it. these students have not had "Real Analysis" or "Advanced Calculus" or whatever it is when you get to the real anal definitions of continuity, limits, and the derivative/integral. no "countably infinite" vs. "uncountably infinite". no Lebesgue measure or Lebesgue anything (only Riemann integral/summation). very little "given any epsilon>0, find a delta so that [such-and-such] is less than that epsilon." certainly no functional analysis or "generalized functions" or "distributions" (but they might use the word "distribution" in a course about probability).

    so no. the fact is that nearly no engineer knows that the Dirac delta function is not a limit of functions as was presented to him/her in undergraduate courses. almost every electrical engineer, if they mess with the Dirac delta at all, treats it simply as a function that is zero at all non-zero values of the argument (it is zero "almost everywhere") yet it has an integral of 1, if the limits straddle 0. and from a physical POV, that's okay. not much physical difference between a "real" Dirac delta and a rectangular approximation that is, say, a femto-second wide. and if a femto-second is too wide, make the width a Planck time. i don't think physical reality will know the difference or care.

    i even asked about this a few years back on the sci.physics.research site and there were physicists who admitted to using and thinking about the Dirac delta in the same way that EEs do.

    while i do not challenge the mathematical legitimacy of the pure mathematical treatment of the Dirac delta (you would have to define Lebesgue integration differently to say something other than "if f(x)=g(x) almost everywhere, and if f(x) is integrable, then g(x) must also be and the integral is the same as f(x) over the same region." I know that. but nonetheless, in Linear System Theory, in Control Systems Theory, Communications Systems, Digital (or analog) Signal Processing, any of this to any level of advancement, it is just easier (and it doesn't break anything) to just think of the Dirac Delta function as a function that agrees with f(x)=0 at all points except x=0, and yet does not have an integral of zero. that is the state of things outside the ivory tower.

    not in the limits. the Riemann integral of the "nascent delta functions" are all 1, yet, in the limit, all of the nascent delta functions will equal zero at every value of x except x=0 where it is undefined. the only questionable thing is we treat the Dirac delta as one of those nascent delta functions that is thin enough for our purposes.
     
    Last edited: Nov 20, 2007
  19. Nov 20, 2007 #18

    matt grime

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    "not in the limits. the Riemann integral of the "nascent delta functions" are all 1, yet, in the limit, all of the nascent delta functions will equal zero at every value of x except x=0 where it is undefined. the only questionable thing is we treat the Dirac delta as one of those nascent delta functions that is thin enough for our purposes."

    That doesn't make any sense. I have lost what it is you're attempting to say. As I am not taking limits, and only dealing with functions, just like the same statement when one replaces lebesgue with Riemann, then what you say has no bearing.

    All of the applied mathematics I know (from before doing any rigorous analysis) says that the dirac delta is *like* a function. Not that it is a function from R to R. Clearly it isn't, as any engineer would be happy to agree with, as it is undefined at 0. It is a functional, though it is unnecessary to explain this in order to use it.

    I will reiterate one point. One does not need to invoke Lebesgue integration at all to come up with the "problem" you mention:

    "If f=g at all but a point, then the Riemann integrals of f and g are equal if they exist".
     
  20. Nov 21, 2007 #19
    The delta function cannot be used for a conterexample about theorems about intergration. Because it is not a "function" eventhough it is called a function, rather it is a Schwartz distribution so the known results about integration might not apply to it.
     
  21. Nov 22, 2007 #20

    rbj

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    [tex] \delta(x) = \lim_{a\to 0} \delta_a(x) [/tex]

    where

    [tex] \delta_a(x) = \frac{1}{a \sqrt{\pi}} \mathrm{e}^{-x^2/a^2} [/tex]

    or

    [tex] \delta_a(x)= \frac{\mathrm{rect}(x/a)}{a} [/tex]

    [tex] \mathrm{rect}(x) = \begin{cases}
    0 & \mbox{if } |x| > \frac{1}{2} \\[3pt]
    \frac{1}{2} & \mbox{if } |x| = \frac{1}{2} \\[3pt]
    1 & \mbox{if } |x| < \frac{1}{2}.
    \end{cases} [/tex]

    or a variety of other nascent delta prototypes. in all these cases, if you think of or treat [itex] \delta(x) [/itex] as a function, if [itex] x \ne 0 [/itex] then [itex] \delta(x) = 0 [/itex] and for any [itex] a > 0 [/itex], then

    [tex] \int_{-\infty}^{\infty}\delta_a(x) dx = 1 [/tex]

    so we Neanderthal engineers just ask "why not for [itex] a = 0 [/itex] in the limit? that is if

    [tex] \lim_{a\to 0} \int_{-\infty}^{\infty}\delta_a(x) dx = 1 [/tex]

    why not

    [tex] \int_{-\infty}^{\infty} \lim_{a\to 0} \delta_a(x) dx = \int_{-\infty}^{\infty} \delta(x) dx = 1 [/tex] ?

    but i am.

    i know that it is not, since the mapping from 0 is undefined.

    [tex] f(x) = 0 [/itex]

    [tex] g(x) = \lim_{a\to 0} \frac{1}{a \sqrt{\pi}} \mathrm{e}^{-x^2/a^2} [/tex]

    except for x=0, g(x) is defined everywhere and it is 0. what it is at 0 is undefined. yet for any [itex] a > 0 [/itex], any of the limit functions for g(x) has an integral of 1.

    yes, you are repeating the same thing.

    this issue is pedagogical. (you are not going to be teaching a very large portion of undergraduate engineering students anything about "generalized functions" or "Schwartz distributions" or the fine distinctions between them and functions that behave as an extreme limit of functions that we all agree are functions.)

    the issue is also practical. there is not a system in the Universe that will know the difference between a legit Dirac delta function (if it could exist somewhere in physical reality) with argument t and any nascent delta function where the width is the finite, but very small 10-44 second. wasting our time with the language of "generalized functions" or "Schwartz distributions" has never been useful in the engineering classroom (and it's never done) nor in any practical system where we are trying to model an impulse of some sort.

    Electrical Engineering texts in basic circuits, signals and systems (a.k.a. linear system theory), communications systems, control systems, filter theory, analog or digital signal processing, simply do not use the formal mathematical language of the Dirac delta as a "distribution, not a function". and we don't break anything by not using it.
     
    Last edited: Nov 22, 2007
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