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