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Math Help - Lebesgue integrability of a function

  1. #1
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    Lebesgue integrability of a function

    Hi !
    Here is my problem:

    (The questions are independent)
    a/
    If f is a function integrable on [0,b] and
    g(x) =
    pour 0xb.
    Prove that g is integrable on [0,b] and that
    = .



    b/
    Let f be a fonction that is measurable, finite-valued on [0,1], and such that |f(x)-f(y)| is integrable on [0,1]x[0,1].
    Prove that f(x) est intégrable sur [0,1].



    For part a/, using Fubini's theorem, I managed to show g is integrable, but cannot get the equality :
    = .


    For part b/, i set g(x,y) = |f(x)-f(y)|, integrable, and tried to use Fubini's, but don't really see what I can do with:
    .

    Any ideas ?
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  2. #2
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    Quote Originally Posted by matt33 View Post
    a/
    If f is a function integrable on [0,b] and
    g(x) = \int_x^b\frac{f(t)}tdt
    pour 0\leqslant x\leqslant b.
    Prove that g is integrable on [0,b] and that
    \int_0^bg(x)\,dx = \int_0^bf(t)\,dt.

    b/
    Let f be a fonction that is measurable, finite-valued on [0,1], and such that |f(x)-f(y)| is integrable on [0,1]x[0,1].
    Prove that f(x) est intégrable sur [0,1].

    For part a/, using Fubini's theorem, I managed to show g is integrable, but cannot get the equality :
    \int_0^bg(x)\,dx = \int_0^bf(t)\,dt.
    If you reverse the order of integration (quoting Fubini appropriately) then

    \int_0^b\!\!\!g(x)\,dx = \int_0^b\!\!\int_x^b\frac{f(t)}t\,dt\,dx = \int_0^b\!\!\int_0^t\frac{f(t)}t\,dx\,dt = \int_0^b\Bigl[\frac{xf(t)}t\Bigr]_{x=0}^tdt = \int_0^b\!\!\!f(t)\,dt.

    (The limits of integration change so as to ensure that both repeated integrals are over the triangular region 0\leqslant x\leqslant t\leqslant b.)

    Quote Originally Posted by matt33 View Post
    For part b/, i set g(x,y) = |f(x)-f(y)|, integrable, and tried to use Fubini's, but don't really see what I can do with:
    \int_0^1\!\!\!\int_0^1g(x,y)\,dxdy.
    You are told that the integral \int_0^1\!\!\!\int_0^1|f(x)-f(y)|\,dx\,dy exists. Also, the function (x,y)\mapsto f(x)-f(y) is measurable. It follows that the integral \int_0^1\!\!\!\int_0^1(f(x)-f(y))\,dx\,dy exists.

    By Fubini's theorem, \int_0^1\!\!(f(x)-f(y))dx exists p.p.(y). But this integral is \int_0^1\!\!\!f(x)\,dx - f(y). So choosing a value of y for which the integral exists, we see that \int_0^1\!\!\!f(x)\,dx exists. In other words, f is integrable on [0,1].
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