# Thread: Lebesgue integrability of a function

1. ## 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) = $\int_x^b {\frac{f(t)}{t} dt}$
pour 0$\leq$x$\leq$b.
Prove that g is integrable on [0,b] and that
$\int_0^b {g(x) dx}$ = $\int_0^b {f(t) dt}$.

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 :
$\int_0^b {g(x) dx}$ = $\int_0^b {f(t) dt}$.

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:
$\int_0^1 {\int_0^1 {g(x,y) dxdy}}$.

Any ideas ?

2. Originally Posted by matt33
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 $0\leqslant x\leqslant t\leqslant b$.)

Originally Posted by matt33
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].