What is the integral of the indicator function of VITALI?

Question

For a positive function $f$, the Lebesgue integral is the supremum of the integral of all simple functions below $f$. So if $f$ is the indicator function for the VITALI set, the lebesgue integral for it must exist. But in general, the integral of an indicator function of a set is the measure of that set. But VITALI is unmeasurable!?

Answer 1

Any measurable subset $U$ of Vitali's set $V$ has measure zero. Therefore, if you take the supremum of the integrals of simple functions not larger than the indicator of vitali's set, their integrals are zero, and so is the supremum.

If $U\subset V$ is measurable, then for an enumeration $q_i$ of the rationals we have $U+q_i\pmod{1}$ disjoint for different $i$'s, measurable and of the same measure. Therefore $\sum_{i=0}^{\infty}\nu(U)=\sum_{i=0}^{\infty}\nu(U+q_i\pmod{1})=\nu(\bigcup_{i=0}^{\infty}(U+q_i\pmod{1}))\leq\nu([0,1])=1$. Therefore $\nu(U)=0$.