Let $S=[0,1]\times[0,1]$, and $\frak m$ be the Lebesgue measure on $\Bbb R^2$. Define $f(x,y)=(y-1/2)(x-1/2)^{-3}$ if $|y-1/2|<|x-1/2|$. Otherwise $f=0$. What is $\int _S f \,{\rm d}{\frak m}$? Is it defined?
Since we can not compute the integral in orders, then how should we evaluate that integral?
Hint: Consider both diagonals of the square. They divide the square into four pieces. Thinking geometrically, we have that ${\rm supp}\,f$ consists of "left" and "right" pieces, say, $L$ and $R$. Once you check that $f$ is Lebesgue-integrable, you can pass to Riemann integrals and compute $$\begin{align}\int_Sf\,{\rm d}{\frak m} &= \int_Lf\,{\rm d}{\frak m} + \int_Rf\,{\rm d}{\frak m} \\ &= \lim_{b\to 1/2^-}\int_0^{b}\int_x^{1-x}\frac{y-1/2}{(x-1/2)^3}\,{\rm d}y\,{\rm d}x + \lim_{a\to 1/2^+}\int_{a}^1\int_{1-x}^x\frac{y-1/2}{(x-1/2)^3}\,{\rm d}y\,{\rm d}x. \end{align}$$