Editing Iterated integral (section)

===Lack of commutativity===
The order in which the integrals are computed is important in iterated integrals, particularly when the integrand is not continuous on the domain of integration. Examples in which the different orders lead to different results are usually for complicated functions as the one that follows.

Define the sequence <math>a_0=0<a_1<a_2<\cdots</math> such that <math>a_n\to1</math>. Let <math>g_n</math> be a sequence of continuous functions not vanishing in the interval <math>(a_n,a_{n+1})</math> and zero elsewhere, such that <math display="inline">\int_0^1 g_n=1</math> for every <math>n</math>. Define
:<math>f(x,y)=\sum_{n=0}^\infty \left( g_n(x)-g_{n+1}(x)\right)g_n(y).</math>
In the previous sum, at each specific <math>(x,y)</math>, at most one term is different from zero.
For this function it happens that<ref>Rudin, W., ''Real and complex analysis'', 1970</ref>
:<math>\int_0^1 \left(\int_0^1 f(x,y) \,dy\right)\,dx =\int_0^{a_1}\left(\int_0^{a_1}g_0(x)g_0(y)\,dy\right)\,dx= 1\neq0 = \int_0^1 0\,dy = \int_0^1 \left(\int_0^1 f(x,y)\, dx\right)\,dy</math>