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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?
Thanks in advice!

One Answer

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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