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

Mathematics Asked by Julian Vené on December 7, 2020

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?

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

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

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 $$A_0subset Bbb R$$ such that $$lambda^*(A_0)ne lambda^*(A_0cap Bbb V)+lambda^*(A_0cap Bbb V^c),$$ wich (thanks to subadditivity of $$lambda^*$$) is equivalent to $$lambda^*(A_0)< lambda^*(A_0cap Bbb V)+lambda^*(A_0cap Bbb V^c).$$

And here are your counterexample: $$A:=A_0cap Bbb V$$ and $$B:=A_0capBbb V^c$$.

Answered by Tito Eliatron on December 7, 2020

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