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Thread: 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
    $\displaystyle g(x) = \int_x^b\frac{f(t)}tdt$
    pour $\displaystyle 0\leqslant x\leqslant b$.
    Prove that g is integrable on [0,b] and that
    $\displaystyle \int_0^bg(x)\,dx = \int_0^bf(t)\,dt$.

    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 :
    $\displaystyle \int_0^bg(x)\,dx = \int_0^bf(t)\,dt$.
    If you reverse the order of integration (quoting Fubini appropriately) then

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

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

    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:
    $\displaystyle \int_0^1\!\!\!\int_0^1g(x,y)\,dxdy$.
    You are told that the integral $\displaystyle \int_0^1\!\!\!\int_0^1|f(x)-f(y)|\,dx\,dy$ exists. Also, the function $\displaystyle (x,y)\mapsto f(x)-f(y)$ is measurable. It follows that the integral $\displaystyle \int_0^1\!\!\!\int_0^1(f(x)-f(y))\,dx\,dy$ exists.

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