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Let $f:[0,1]^2\rightarrow \Bbb R$ be defined by

$$f(x,y)=1/y^2 \text{ if }0\leq x <y<1$$ $$f(x,y)=-1/x^2\text{ if }0\leq y <x \leq 1$$ $$f(x,y)=0 \text{ otherwise}$$ If can show that $$\int_{0}^{1}\int_{0}^{1}f(x,y)dxdy \neq \int_{0}^{1}\int_{0}^{1}f(x,y)dydx$$ In my course, it is written that the last equation does not contradict fubini's theorem since $\mid f\mid$ is not integrable on $[0,1]^2$. I don't know why and I can't compute the integral of $\mid f \mid$ on $[0,1]^2$. Any help would be appreciated

Dicordi
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By Tonelli's Theorem $\int |f| \geq \int_0^{1}\int_x^{1}\frac 1 {y^{2}} dydx=\int_0^{1}(\frac 1 x-1) dx =\infty$.

Kavi Rama Murthy
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  • why do you invoque tonelli's theorem ? can you detail a bit more your answer please ? – Dicordi Jul 27 '20 at 10:35
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    You can evaluate the double integral of $|f|$ using iterated integrals. This is becasue $|f|$ is a non-negative measurable function and you don't need integrability to write the double integral of $|f|$ using iterated integrals in any order you like. This is exactly what Tonelli's Theorem tells us. @Dicordi – Kavi Rama Murthy Jul 27 '20 at 11:40