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 $S=0$.


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 $\int_S f^+ d\lambda =\int_S f^- d\lambda = \infty$

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  • $\begingroup$ Got it! I have one more question: S is not Riemann integrable, so "$\int_0^1 \int_0^1 1/(x-y)=0$" does not make any sense? I got this from online calculator, but don't know what happens. $\endgroup$ – Math Boy Jan 29 '16 at 7:06
  • $\begingroup$ It makes sense, but only in limit. The problem happens along the line $y=x$ where if $x \uparrow y$, then $f(x,y)\downarrow -\infty,$ but if $x\downarrow y$ then $f(x,y) \uparrow \infty$ $\endgroup$ – Christian Chapman Jan 29 '16 at 7:08
  • $\begingroup$ Is this an improper Riemann integral? I thought only integrals like $\int_a^{\infty} f$ is improper Riemann integral, forgot the Cal stuff. $\endgroup$ – Math Boy Jan 29 '16 at 7:11
  • $\begingroup$ You can define it that way, but in alternative definitions, you can say it's Riemann integrable in limit over a set $A$ if you can find a sequence of growing subsets $(S_i),\ S_i\subseteq S_{i+1} \subseteq A$ where $A\backslash (\cup_{i=1}^\infty S_i)$ has volume $0$, and $\lim_{i\rightarrow \infty}\int_{S_i} f(x) dx $ exists. But it's cumbersome and at that point you might as well start talking about principal values and measure. $\endgroup$ – Christian Chapman Jan 29 '16 at 7:21
  • $\begingroup$ Got it, thanks! $\endgroup$ – Math Boy Jan 29 '16 at 7:25

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

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