Is $1/x$ integrable over the Cantor set? What about other not Lebesgue integrable functions and sets with Lebesgue measure $0$

Question

I don't know how to integrate functions over fractal sets, so I would appreciate some pointers or references on the topic.

What I want to know - are there cases when a function which is not integrable over $(0,1)$ (for example) is nevertheless integrable over some meausre $0$ set, and has a non-zero integral over this set?

Actually, we can consider a more simple (or more complicated?) case of integrating $1/x$ over the rational numbers on $(0,1)$

I am not well versed in measure theory, so I would appreciate a direct reference (not 'consult with your measure theory textbook').

Answer 1

Yes, and no. Yes, ALL functions are integrable when restricted to a subset of measure zero. No, none of these functions will have a NON-ZERO integral. The integral will always be zero. Remember the DEFINITION of the integral (for the positive part of $f$, say) as the least upper bound of "integrals" (really, sums) for step functions. Since you already restricted to a subset of measure zero, all these integrals/sums for all step functions will be zero, which proves at once that the integral of the positive part of $f$ exists AND is equal to zero. Same for the negative part.