2
$\begingroup$

Let $f:(0,1)\rightarrow \mathbb{R}$ be Lebesgue integrable. Prove that $g(x)=\int_{(x,1)} \frac{f(y)}{y} d\lambda$ is measurable, integrable and that $\int_{(0,1)}g d\lambda=\int_{(0,1)}f d\lambda$

I know that since f is a Lebesgue integrable function it needs to be measurable, the function $f(y)/y$ for $y$ in $(x,1)$ is perfectly well defined and does not come into trouble since $x>0$, but besides that, I don't see how may we engage the fact of f being lebesgue integrable in order to continue with the problem.

I also have that:

$\int_{(0,1)}g(x)dx=\int_{(0,1)}\int_{(x,1)}\frac{f(y)}{y}dydx$, but don't feel certain of whether a limit here would be of any use in order to get to the right hand side of the equality.

Thanks in advance for any useful tip on how this problem may be approached.

$\endgroup$
  • $\begingroup$ Can we apply Fubini's theorem here? $\endgroup$ – Berci Aug 9 '18 at 21:34
  • $\begingroup$ You can write $f$ as $f^{+}-f^{-}$ so the proof actually reduces to the case $f$ non-negative. So Tonelli's Theorem can be applied. $\endgroup$ – Kavi Rama Murthy Aug 9 '18 at 23:44
2
$\begingroup$

Define $h: [0,1]\times [0,1]\rightarrow \mathbb{R}$, $h(x,y)= \cases{\frac{f(y)}{y} \,\, \text{if}\,\, y>x\\ 0 \,\, \text{otherwise}}$.

Then $\int\limits_{[0,1]} g \, d\lambda = \int\limits_{0}^{1} \int\limits_{x}^{1} \frac{f(y)}{y} \, dy \, dx = \int\limits_{0}^{1} \int\limits_{0}^{1} h(x,y) \, dy \, dx $, which by Fubini's theorem is

$\int\limits_{0}^{1} \int\limits_{0}^{1} h(x,y) \, dx \, dy = \int\limits_{0}^{1} \int\limits_{0}^{y} \frac{f(y)}{y} \, dx \, dy = \int\limits_{0}^{1} y\cdot \frac{f(y)}{y} \, dy = \int\limits_{0}^{1} f(y) \, dy = \int\limits_{[0,1]} f \, d\lambda$.

$\endgroup$
  • 1
    $\begingroup$ The question is: why can you use Fubini here? $\endgroup$ – amsmath Aug 9 '18 at 21:51
  • 1
    $\begingroup$ I don't think that is a serious problem. As $f$ is Lebesgue integrable and both measures are finite, there shouldn't be any trouble. (The horizontal segments are step functions, and the vertical segments are clearly integrable, so everything is nice and the calculation makes sense.) $\endgroup$ – A. Pongrácz Aug 9 '18 at 22:01
  • 2
    $\begingroup$ Of course, the calculation makes sense, and it is correct. But "shouldn't be any trouble" is not a rigorous reasoning. I propose the following: First show that $|h|$ is integrable. Here, you can use Fubini since $|h|\ge 0$. Hence, you can use Fubini on $h$ itself. $\endgroup$ – amsmath Aug 9 '18 at 22:14
  • 1
    $\begingroup$ Sure, this is the rigorous argument. But this is fairly straightforward, I think. That is what I meant by "shouldn't be any trouble": simple. Anyway, you are right, this is how it is done. So good point. $\endgroup$ – A. Pongrácz Aug 9 '18 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.