The following is an exercise from Bruckner's Real Analysis:
Let $E$ be a measurable set of positive Lebesgue measure. Show that E can be written as the disjoint union of two sets $E = E_1 ∪ E_2$ so that $μ(E)=μ^∗(E_1)=μ^∗(E_2)$.
I have no idea how to construct a single example let alone proving for any measurable E. A useful hint also would be great, thanks!
Let $F,G$ be two disjoint bernstein sets. Define $E_1:=E\cap F$, $E_2:=E-E_1\supseteq E\cap G$. We claim that the two sets satisfy the hypotesis. To see that, it suffices to show that $\mu^*(E\cap F)=\mu^*(E\cap G)=\mu(E)$, since then $\mu(E)\ge\mu^*(E_2)\ge \mu^*(E\cap G)=\mu(E)$ and so the equality must hold. By simmetry, it suffices to prove the result for $F$. Its clear that $\mu^*(E\cap F)\le \mu(E)$, it remains to prove $\ge$: suppose that this is not true. Then there exists a covering of $E\cap F$ made of open sets (let us write the covering as $\cup_{i\in\mathbb{N}} A_i$), such that $\mu(E-\cup A_i)>0$. Since $E-\cup A_i$ has positive measure, it must contain a closed subset of positive measure $C$ which has to be uncountable. By definition of bernstein set, this implies $F\cap C\neq\emptyset$, a contradiction.