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Something a little odd - Lebesgue vs Riemann integral

  1. Dec 18, 2007 #1

    quasar987

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    In the Riemann theory for a function f defined on all of R, we define its improper integral over R as the sum of two limits:

    [tex]\int_{-\infty}^{+\infty}f(x)dx = \lim_{a\rightarrow -\infty}\int_{a}^0f(x)dx+\lim_{b\rightarrow +\infty}\int_{0}^b f(x)dx[/tex]

    and in general, this is not equal to

    [tex]\lim_{L\rightarrow +\infty}\int_{-L}^L f(x)dx[/tex]

    For instance, for f(x)=sin(x), neither of the two limits above exists, but the limit below is 0.

    But what does Lebesgue says? We know that given f:R-->R integrable, the function [tex]\nu:\mathfrak{L}_{\mathbb{R}}\rightarrow \mathbb{R}[/tex] defined by

    [tex]\nu(E)=\int_Ef(x)dx[/tex]

    is a (signed) measure. Signed measures enjoy the same continuity properties as finite and positive measures. That is, if E_n is an increasing sequence of measurable sets, then

    [tex]\nu\left(\bigcup_{n=1}^{+\infty}E_n\right)=\lim_{n\rightarrow +\infty}\nu(E_n)[/tex]

    In particular, for E_n = ]-n,n[ and the measure defined above,

    [tex]\int_{-\infty}^{+\infty}f(x)dx=\lim_{n\rightarrow +\infty}\int_{-n}^nf(x)dx[/tex]

    (Since U E_n=R)

    What's up with that? (as in : any comments welcome)
     
    Last edited: Dec 18, 2007
  2. jcsd
  3. Dec 18, 2007 #2

    HallsofIvy

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    With that choice of E_n, you are doing, effectively,
    [tex]\lim_{n\rightarrow \infty}\int_{-n}^{n} f(x) dx[/itex]
    the "Cauchy Principal Value", not the general Lebesque integral.
     
  4. Dec 18, 2007 #3

    quasar987

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    Explain please!

    What would the general Lebesgue integral be?
     
  5. Dec 18, 2007 #4

    mathwonk

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    i seem to recall the lebesgue integral subtracts the negative part of the integral from the positive one (i.e. does the integrals of max(f,0) and min(f,0) separately) so you would simply get a non integrable function for sin since both integrals are infinite. but i have not looked at this in 40 years.
     
  6. Dec 18, 2007 #5

    HallsofIvy

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    It would be using a sequence of measurable sets that was NOT symmetric as you have it.

    Just after I read your question, I then asked myself the question you are probably thinking of: in order to be integrable, it shouldn't MATTER which measurable sets we choose! That's true but it is also true of Riemann integrable functions (in terms of intervals rather than measurable sets).

    Note that
    [tex]\lim_{L\rightarrow \infty}\int_{-L}^L f(x)dx= \lim_{a\rightarrow -\infty}\int_a^0 f(x)dx \lim_{b\rightarrow\infty}\int_0^L f(x)dx[/tex]
    is true if all the integrals exist.

    In terms of the Lebesque integral, just as for the Riemann integral, even if
    [tex]\int_{-\infty}^{+\infty}f(x)dx=\lim_{n\rightarrow +\infty}\int_{-n}^nf(x)dx[/tex]
    exists the "more general" integral,, using other measurable sets may not exist.
     
  7. Dec 18, 2007 #6

    quasar987

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    mathwonk spotted my error: sin(x) is not Lebesgue integrable because |sinx| isn't.

    It would be nice then, if we could show that if f 's improper Riemann integral diverges, then it isn't Lebesgue integrable. This way, we then never encounter the humbling situation

    [tex]\int_{-\infty}^{+\infty}f(x)dx \ \ \ \mbox{(Riemann)} \neq\lim_{L\rightarrow +\infty}\int_{-L}^L f(x)dx \ \ \ \mbox{(Riemann)} = \int_{-\infty}^{+\infty}f(x)dx \ \ \ \mbox{(Lebesgue)}[/tex]

    Bed.
     
  8. Dec 19, 2007 #7

    Hurkyl

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    I don't understand your dilemma. If a function is Riemann integrable, then the Riemann and Lebesgue integrals agree on any bounded interval, right? And isn't it a straightforward proof that

    [tex]\int_{-\infty}^{+\infty} f(x) \, dx = \lim_{
    \substack{a \rightarrow -\infty \\ b \rightarrow +\infty }}
    \int_a^b f(x) \, dx[/tex]

    whenever either side exists? (for both kinds of integrals)
     
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