Necessity of absolute value in The Fubini–Tonelli theorem? In The Fubini–Tonelli theorem(reference in wiki):
What's the point of taking absolute value in $\int_{X\times Y}|f(x,y)|\,d(x,y)$? Isn't $f$ integrible in $X\times Y$ automatically implies $|f|$ integrable in $X\times Y$? So we only need $f$ integrable to apply  Fubini–Tonelli theorem?
 A: To say that $f$ is integrable means (in this context) that the integral of $|f|$ is finite. The following statements both say the same thing:


*

*$f$ is integrable.

*The integral of $|f|$ is finite.


There are cases in which the two iterated integrals not equal to each other, but that can happen only when the integral of $f$ over that part of the domain in which $f\ge0$ is $+\infty$ and over that part of the domain in which $f\le 0$ is $-\infty.$ An example is
$$
\int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy \right) \, dx \ne \int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dx \right) \, dy.
$$
(One of these is $+\pi/2$ and the other is $-\pi/2.$) Thus Fubini's theorem tells us that
$$
\iint_{[0,1]^2} \left| \frac{x^2-y^2}{(x^2+y^2)^2} \right| \,d(x,y) = +\infty.
$$
Tonelli's theorem is applicable even when the integral is infinite. It assumes $f\ge0$ everywhere.
A: No .The theorem holds for all non-negative measurable functions; you compute the repeated iterated integral of the absolute value of of f and if it comes out finite you CONCLUDE the $f$ is integrable so you may then apply Fubini to $f$ to calculate the integral of $f$ by iterated integrals -very useful result   
