Why the following integral is Riemann integrable but not Lebesgue integrable?

Question

I know Riemann integrable implies Lebesgue integrable, but why the following integral is Riemann integrable but not Lebesgue integrable?

S=$\int_E {1\over{x-y}}dm$, where $E=[0,1]\times[0,1] $.

I think $S$ is Riemann integral because $S=-S$ by symmetric property of E and $S=S$ implies $S=0$.

Answer 1

The function you have posted is not Riemann integrable in a strict sense because of what happens along the diagonal. For any partition you give me, I can always find a refinement of that partition whose upper and lower Riemann sums are arbitrarily far apart. (Consider taking a refinement only in the part of the domain where the function is very positive.)

It is, however, Riemann integrable in limit. If you cut out a diagonal strip from the domain: $\{(x,y): |x-y|<\varepsilon\}$ then no matter how small $\varepsilon$ is, the integral over the 'cut' domain is 0.

However, Riemann integrability in limit does not imply Lebesgue integrability. (You have given an exmaple here, and I am sure that by now you have seen the example involving $\mathsf{sinc}(x)$)

The function is not Lebesgue integrable because $\int_S f^+ d\lambda =\int_S f^- d\lambda = \infty$

Answer 2

I don't have enough reputation to comment, but it might be helpful considering what's happening along the diagonal of $E$.