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What is the reason that 1/x is not lebesgue integrable?

  1. Dec 1, 2008 #1
    What is the reason that 1/x is not lebesgue integrable where as 1/x^2 is integrable. You can use any theorems: monotone convergence, dominated convergence you want.
     
  2. jcsd
  3. Dec 2, 2008 #2

    morphism

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    Integrable where?
     
  4. Dec 2, 2008 #3
    Probably in some unbounded region not containing zero; if the region contains zero then 1/x2 is not integrable, and if the region is bounded and does not contain zero, both are integrable.
     
  5. Dec 2, 2008 #4
    [0,1]
    Both 1/x and 1/(x^2) are not lebesgue integrable on [0,1]?
    I thought 1/(x^2) is while 1/x isn't
     
  6. Dec 2, 2008 #5
    Let X be a subset of R, let f(x)=x^-1, and g(x)=x^-2. If both f and g are bounded on X, this implies that X does not contain 0, implying that both f and g are continuous on X. Thus f and g are measurable on X. Furthermore, if X is compact and does not contain zero, things get much nicer, as the Riemann integrals are guaranteed to be finite, implying that the function must be Lebesgue integrable.

    In general, let K be a compact set. If f, a positive valued function is Riemann integrable on K then, f belongs to L_1[K] (Lebesgue integrable on K). This is a nice theorem to use when one is trying to prove that a function is Lebesgue integrable( Using the fundamental theorem of calculus). However, if K=[0,1], both x^-1, and x^-2 are non Riemann integrable on the compact set [0,1]. So this does not help much.

    In fact both 1/x and 1/x^2 are non Lebesgue integrable on [0,1]. I will just prove that 1/x is non Lebesgue integrable on [0,1] as the second case is very similar.
    Proof:
    Let A=[a,a+e], where a in [0,1], and e is small enough such that A is included in [0,1]. Now, let X_A be the characteristic function of A. Let us defined phi(x)=((a+e)^-1)X_A. phi is a simple function thus the Lebesgue integral of phi is equal to m(A)((a+e)^-1)=e((a+e)^-1), since m(A)=a+e-a=e. We observe then, that the integral of phi depends on the value of a in [0,1]. Now let a converge to 0, then the Lebesgue integral of phi converges to infinity. Now, we know that both phi and f belongs to L+ which is the set of all the positive measurable functions not necessarily Lebesgue integrable though. Also, we know that for any x, 1/x >= phi(x), thus, int(1/x)>=int(phi(x))= infinity. Hence 1/x is non-Lebesgue integrable.

    Vignon S. Oussa
     
    Last edited: Dec 2, 2008
  7. Dec 2, 2008 #6
    Well, 1/x2 ≥ 1/x on (0, 1].
     
  8. Dec 2, 2008 #7
    Even better, in the extended real number set, 1/x2 ≥ 1/x on [0, 1].
     
  9. Dec 3, 2008 #8
    Thanks for the reply. But on the extended real set, 1/x=1/x^2 at x=0 right. both are equal to +inf
     
  10. Dec 3, 2008 #9
    I think the crucial step in your proof is 1/x >= phi(x)
    where phi(x)=((a+e)^-1)X_A
    May I ask why? This is a myth to me.
     
  11. Dec 3, 2008 #10
    Ok let me clarify then. Take any x in (0,1].

    Case1: if x does not belong to A,
    phi(x) =((a+e)^-1)X_A(x)=((a+e)^-1)0=0
    f(x)=1/x. Thus f(x)> 0 ---> f(x) = phi(x)

    Case2: if x does belong to A
    phi(x) =((a+e)^-1)X_A(x)=((a+e)^-1)1=((a+e)^-1)
    x is in [a,a+e] thus x<=a+e, implying that 1/x >=((a+e)^-1). Thus f(x)>=((a+e)^-1)=phi(x) when x belongs to A

    In conclusion, in either case we must always have f(x)>=phi(x) when x is in (0,1]. I hope I cleared things up for you

    Vignon S. Oussa
     
    Last edited: Dec 3, 2008
  12. Dec 3, 2008 #11
    both statements with " = " or ">= "are equivalent statements. Remember from logic theory that (True OR False) is equivalent to True.

    Vignon S. Oussa
     
  13. Nov 2, 2011 #12
    The lebesgue integral of phi = e/(a+e), no? So letting a -> 0 shows that the integral of phi goes to 1, not infinity.
     
  14. Nov 2, 2011 #13
    It's considerably easier just to take the anti-derivatives: ln |x| and -1/x. These don't behave well as x approaches 0.
     
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