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For $f:(0,1)\rightarrow \mathbb{R}$ Lebesgue integrable define:

$g: [0,1]\times [0,1]\rightarrow \mathbb{R}$, such that

$$g(x,y)= \cases{\frac{f(y)}{y} \,\, \text{if}\,\, y>x \\ \\ 0 \,\, \text{otherwise}}$$

Then $$ \int\limits_{0}^{1} \int\limits_{x}^{1} \frac{f(y)}{y} \, dy \, dx = \int\limits_{0}^{1} \int\limits_{0}^{1} g(x,y) \, dy \, dx = \ ..$$

I need to use Fubini to complete my proof, but I don't know how to show that $|g|$ is integrable to be able to do so?

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    $\begingroup$ Tonelli's theorem says that $\int_{[0,1]^2} \lvert g(x,y)\rvert \, dx\otimes dy=\int_0^1dx\int_0^1 \lvert g(x,y)\rvert\,dy$ holds unconditionally (given measurability), so push comes to shove you can do all the manipulations you need on $\int\lvert g\rvert$ to check if it's integrable or not. $\endgroup$ Mar 18, 2023 at 11:38

1 Answer 1

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Use Tonelli's Theorem.

$\int_0^{1}\int_0^{1}|g(x,y)|dxdy=\int_0^{1}\int_0^{y}dx \frac {|f(y)|} ydy=\int_0^{1}|f(y)|dy<\infty$

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