Nonlinearity in general of the Lebesgue integral for nonmeasurable functions reduces in some sense to inner and outer measure of nonmeasurable sets: $$\int_{\Omega}\chi_A+\chi_{\Omega\setminus A}\mathrm{d}\mu=\mu(\Omega)\neq\mu(\Omega)+\sup_{E\subseteq A}\mu(E)-\inf_{A\subseteq F}\mu(F)=\int_{\Omega}\chi_A\mathrm{d}\mu+\int_{\Omega}\chi_{\Omega\setminus A}\mathrm{d}\mu$$ The proof of this statement is the goal of this thread!


Given the measure space $[0,1]$ with Lebesgue measure $\lambda$ and consider a Vitali set $V$.

(Take the classical construction of a Vitali set.)

Every measurable subset has Lebesgue mass zero: $$\lambda(E)=0\quad(E\subseteq V)$$ (The proof imitates the one for nonmeasurability of a Vitali set.)

Moreover, every measurable superset has nonvanishing Lebesgue mass: $$\lambda(F)>0\quad(V\subseteq F)$$ (By nonmeasurability this seems reasonable and indeed follows by completeness.)

However this is not enough yet, so...

Does the Vitali set admit a bound: $$0<C_0\leq\lambda(F)\quad(V\subseteq F)$$ (In other words, is the outer measure necessarily greater zero for any Vitali set?)


While writing this thread I found the answer unfortunately :) - I will post it anyway.

Hope this is not a duplicate of another thread - if so I'm sorry!


Since a Vitali set is nonmeasurable its outer measure cannot vanish.


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