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Math Help - Lebesgue integration

  1. #1
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    Lebesgue integration

    My question is from Folland "Real Analysis" 2nd ed. p.77 Chapter 2, #56.

    If f is lebesgue integrable on (0,a) and
    g(x) = integral (from x to a) [t^-1 f(t)dt]

    Prove then that:

    1. g is integrable on (0,a) and
    2. integral (from 0 to a) [g(x)dx] = integral (from 0 to a) [f(x)dx]
    Last edited by Fubini; November 16th 2005 at 08:48 PM.
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  2. #2
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    Quote Originally Posted by Fubini
    My question is from Folland "Real Analysis" 2nd ed. p.77 Chapter 2, #56.

    If f is lebesgue integrable on (0,a) and
    g(x) = integral (from 0 to a) [t^-1 f(t)dt]

    Prove then that:

    1. g is integrable on (0,a) and
    2. integral (from 0 to a) [g(x)dx] = integral (from 0 to a) [f(x)dx]
    Are you sure you have copied this correctly?

    g(x)\ =\ \int_0^a\ \frac{f(t)}{t} dt

    when it exists is a constant.

    RonL
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  3. #3
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    Re: Lebesgue integration

    You're absolutely right it is constant, it should be the integral from x to a. Sorry.
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  4. #4
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    Quote Originally Posted by Fubini
    My question is from Folland "Real Analysis" 2nd ed. p.77 Chapter 2, #56.

    If f is lebesgue integrable on (0,a) and
    g(x) = integral (from x to a) [t^-1 f(t)dt]

    Prove then that:

    1. g is integrable on (0,a) and
    2. integral (from 0 to a) [g(x)dx] = integral (from 0 to a) [f(x)dx]

    I'm still a bit confused by this

    My Lebesgue theory is a bit rusty, but if this were true it should
    also hold for a continuous bounded Riemann integrable function.

    In fact should it not hold for f(x)\ =\ x^n, n > 1\
    when a < \infty ?

    RonL
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  5. #5
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    solution

    I was able to get this by following the advice of the book. First apply Tonelli's theorem then, Fubini's theorem, from the previous section on p.68.

    int (0 to a) [ |g(x)| dx] is non-negative so we can switch the order of integration by Tonelli's theorem.

    int (o to a) [ |g(x)|] = int (0 to a) int (x to a) [ |f(t)| t^-1 dt dx]
    =int (0 to a int (0 to t) [ |f(t)| t ^-1 dx dt]

    noting the regions (x<t<a, 0<x<a)
    and (0<x<t, 0<t<a)
    are identical.

    evaluating the inside integral gives:

    int (0 to a) [ {x |f(t)| t^-1 (x=t to x=0)} dt]
    =int (0 to a) [ |f(t)| dt ]
    < infinity (by assumtion, it is given f if Lebesgue
    integrable on (0,a) ).

    Hence g(x) is Lebesgue integrable on (0,a).

    Now the equality is clear as the methodology on how to achieve it is given above (switching integration order) and apply Fubini's theorem for integrable functions.
    Last edited by Fubini; November 28th 2005 at 04:32 PM.
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  6. #6
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    It certainly works for f(t) = t^2 (a bounded Riemann integrable function)

    0<a<infinity

    one gets 1/3 a^3 either way.

    Any Riemann integrable function is Lebesgue integrable on a bounded interval.
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