Suppose that our underlying space is $\Bbb R^2$ and $f:\Bbb R^2\to \Bbb R$, for concreteness. It is not hard to artificially construct such a function $f$ such that $$ \int_Y\int_X f(x,y)dxdy\ne \int_X\int_Y f(x,y)dydx $$ ,i.e. the interchanging of the order of integration in not applicable. Note that Fubini's Theorem implies that any $f$ with the above property is not absolutely integrable on $\Bbb R^2$.
Although we, the mathematicians, are quite cautious about this matter, it seems to me that many physicists don't really pay much attention to the problem at all. I was trying to convinced my physicist friend that this is a serious matter but failed. The function I gave as an example seem too ad hoc for him. I want a more real-life example.
Is there any $f:\Bbb R^2\to \Bbb R$ that arises naturally from a physical situation such that the interchanging the order of integration fail? I want the function $f$ to be in a classical sense i.e. not a distribution or any generalized function.
Let $f(x,y)=\frac{x-y}{(x+y)^3}$ for $(x,y)\in[0,1]^2$. Then \begin{align} \int_0^1\int_0^1 f(x,y)\ \mathsf dy\ \mathsf dx &= \int_0^1\int_0^1 \frac{x-y}{(x+y)^3}\ \mathsf dy\ \mathsf dx\\ &= \int_0^1 \int_0^1\left(\frac{2 x}{(x+y)^3}-\frac{1}{(x+y)^2}\right) \mathsf dy\ \mathsf dx\\ &= \int_0^1 \frac1{(1+x)^2}\ \mathsf dx\\ &= \frac12, \end{align} but \begin{align} \int_0^1\int_0^1 f(x,y)\ \mathsf dx\ \mathsf dy &= \int_0^1\int_0^1 \frac{x-y}{(x+y)^3}\ \mathsf dx\ \mathsf dy\\ &= \int_0^1-\frac1{(1+y)^2} \ \mathsf dy\\ &= -\frac12. \end{align} Here $f$ does not satisfy the assumptions of Fubini's theorem since $|f|$ is not integrable.