Lebesgue outer measure of the union is strictly less than the sum of the measures

Question

Show that it exists $A , B \subset \mathbb{R}^n$ such that $A \cap B = \emptyset $ and \begin{align*} \lambda^{*} (A \cup B) < \lambda^{*} (A) + \lambda^{*} (B) \end{align*}

I tried to use Carathéodory's criterium and Vitali's theorem but without success. Any suggestions?

Thanks in advance!

Answer 1

Are you willing to accept the existence of a non-Lebesgue measurable set $V$?

The very definition of measurable states that there must be a set $E$ with $$\lambda^*(E) < \lambda^*(E \cap V) + \lambda^*(E \setminus V).$$

Let $A = E \cap V$ and $B = E \setminus V$.