Let $A, B\subset \mathbb{R}$ be two sets such that $A$ is Lebesgue measurable and $B$ is not Lebesgue measurable. Let $m^*$ denote the Lebesgue outer measure on $\mathbb{R}$.
I know that $m^*(A\cup B)=m^*(A)+m^*(B)$ whenever $A\cap B=\emptyset$ (that is, the intersection of $A$ and $B$ is empty).
But if we have $A\cap B\neq\emptyset$, is it true that $m^*(A\cup B)=m^*(A)+m^*(B)-m^*(A\cap B)$?
I think that this is not true, but I could not find a counterexample. What would be a counterexample, if any? Thanks in advance.
It is true. You need Caratheodory's criterion: Since $A$ is measurable it satisfies for any set $S\subset\mathbb{R}$ $$m^*(S)=m^*(S\setminus A)+m^*(S\cap A).$$ If you put $S=B$ and $S=A\cup B$ you obtain two equations: $$m^*(B)=m^*(B\setminus A)+m^*(B\cap A)$$ $$m^*(A\cup B)=m^*((A\cup B)\setminus A)+m^*((A\cup B)\cap A)=m^*(B\setminus A)+m^*(A).$$ If you add these two you obtain $$m^*(A\cup B) +m^*(B\setminus A)+m^*(B\cap A)=m^*(B)+m^*(B\setminus A)+m^*(A),$$ which is our conjecture.