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Given the function $f: \Bbb R^2 \rightarrow \Bbb R$,

$f(x, y) = \begin{cases}1/y^2, & \text{$0 < x < y < 1$} \\ -1/x^2, & \text{$0 < y < x < 1$} \end{cases}$

I would like to check whether the premises of the theorem of Fubini are given. We already got the solutions for this, but I don't understand them. The solution begins like this:

We know that $f$ is measurable and we take a look at $f_+$ with the help of the theorem of Tonelli:

$\int_{\Bbb R^2} f_+ (x, y) d\lambda^2(x, y) = \int_{[0, 1]} \int_{[0, y]} 1/y^2 \ d\lambda(x) \ d\lambda(y) = \ ... \ $

I don't understand why he chooses to work in the intervals $[0, 1]$ and $[0, y]$, plus, I don't see why he puts $1/y^2$ into the inner integral. Shouldn't we assume something like $x < y$ first in order to do so?

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"Tonelli's theorem (named after Leonida Tonelli) is a successor of Fubini's theorem. The conclusion of Tonelli's theorem is identical to that of Fubini's theorem, but the assumption that ${\displaystyle |f|}$ has a finite integral is replaced by the assumption that ${\displaystyle f}$ is non-negative." Wikipedia

So the suggested solution begins by considering the function $f_+$, which i guess it is defined as follows: $$f_+(x,y)=max\{0,f(x,y)\} $$

Thus, we have $$f_+(x,y)=\left\{\begin{array}[cr] \ \frac{1}{y^2}&\ if\quad 0

Then, the choise made in the solution concerning the integration intervals $[0,1]\times[0,y]$ can be chosen otherwise.

for example: $[0,1]\times[x,1]$, with permutation of the integrals.