Editing Fubini's theorem (section)

===Failure of Fubini's theorem for non-integrable functions===
Fubini's theorem tells us that (for measurable functions on a product of σ-finite measure spaces) if the integral of the absolute value is finite, then the order of integration does not matter; if we integrate first with respect to ''x'' and then with respect to ''y'', we get the same result as if we integrate first with respect to ''y'' and then with respect to ''x''. The assumption that the integral of the absolute value is finite is "[[Lebesgue integral|Lebesgue integrability]]", and without it the two repeated integrals can have different values.

A simple example to show that the repeated integrals can be different in general is to take the two measure spaces to be the positive integers, and to take the function ''f''(''x'',''y'') to be 1 if ''x'' = ''y'', −1 if ''x''&nbsp;=&nbsp;''y''&nbsp;+&nbsp;1, and 0 otherwise. Then the two repeated integrals have different values 0 and 1.

Another example is as follows for the function
<math display="block">\frac{x^2-y^2}{(x^2+y^2)^2} = -\frac{\partial^2}{\partial x\,\partial y} \arctan(y/x).</math>
The [[double integral|iterated integral]]s

<math display="block">\int_{x=0}^1\left(\int_{y=0}^1\frac{x^2-y^2}{(x^2+y^2)^2}\,\text{d}y\right)\,\text{d}x = \frac{\pi}{4}</math>
and
<math display="block">\int_{y=0}^1\left(\int_{x=0}^1\frac{x^2-y^2}{(x^2+y^2)^2}\,\text{d}x\right)\,\text{d}y=-\frac{\pi}{4}</math>
have different values. The corresponding double integral does not [[absolute convergence|converge absolutely]] (in other words the integral of the [[absolute value]] is not finite):
<math display="block">\int_0^1\int_0^1 \left|\frac{x^2-y^2}{\left(x^2 + y^2\right)^2}\right|\,\text{d}y\,\text{d}x=\infty.</math>