Outer measure question (disjoint set st $\lambda^*(A \cup B) < \lambda^*(A) + \lambda^*(B)$)

Question

Let $\lambda^*$ be the outer measure on $\mathcal{P}(\mathbb{R})$. Is it true that there exist $A,B \in \mathcal{P}(\mathbb{R})$ such that $$A \cap B = \emptyset$$ and $$\lambda^*(A \cup B) < \lambda^*(A) + \lambda^*(B)?$$ Do you have any idea if such $A$ and $B$ sets exist? Thanks in advice!

Answer 1

This statement is equivalent to the existence of non-Lebesgue-measurable sets.

According to Caratheodoy's Criterion (see also here), a set $E\subset\Bbb R$ is measurable iff $$\lambda^*(A)=\lambda^*(A\cap E)+\lambda^*(A\cap E^c)$$ for every $A\subset\Bbb R$.

If you assume the Axiom of Choice, then you can construct the Vitali Set, wich is the classical example of non-Lebesgue-measurable set on $\Bbb R$.

Now, if you denote $\Bbb V$ this non-measurable set, by Caratheodory's Criterion, there is some set $A_0\subset \Bbb R$ such that $$\lambda^*(A_0)\ne \lambda^*(A_0\cap \Bbb V)+\lambda^*(A_0\cap \Bbb V^c), $$ wich (thanks to subadditivity of $\lambda^*$) is equivalent to $$\lambda^*(A_0)< \lambda^*(A_0\cap \Bbb V)+\lambda^*(A_0\cap \Bbb V^c).$$

And here are your counterexample: $A:=A_0\cap \Bbb V$ and $B:=A_0\cap\Bbb V^c$.