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When $d(A,B)>0$, we have $$m^*(A)+m^*(B)=m^*(A\cup B)$$I want to examine whether the identity holds if we only have $A\cap B=\varnothing$.

My attempt:

The proof of the origin theorem requires a $\delta$-distance to separate these 2 sets. So the only worse situation may only happen if they have strange boundaries. However, I am unable to construct a counterexample so far.

Isomorphism
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  • If the sets are measurable then the above must hold. So,... – copper.hat Jun 20 '20 at 01:01
  • Can you share by which source you are study? – zkutch Jun 20 '20 at 01:02
  • Are you familiar with Caratheodory's criterion for Lebesgue measurability? As @copper.hat says, you need to look at non-measurable sets. – paul blart math cop Jun 20 '20 at 01:26
  • @zkutch It is a self-made materials from one of our professors. – Isomorphism Jun 20 '20 at 01:50
  • @paulblartmathcop I just know the criterion and constructiong a non-measurable set is not that easy. – Isomorphism Jun 20 '20 at 01:52
  • @dilong A famous example of a non-measurable set is a Vitali set. You can find a proof of its non-measurability on the wikipedia page, so I'll just provide the definition here. Let $\sim$ on $\mathbb R$ be defined by $x \sim y$ if $x - y \in \mathbb Q$. A vitali set is a system of representatives $S$ of this equivalence relation (which exists by the axiom of choice) such that $S \subseteq [0, 1]$. This is non-measurable. This, along with Carathéodory's criterion will solve your question. – paul blart math cop Jun 20 '20 at 02:02
  • I know this example. And the outer measure of a Vitali set is 1. Do you mean that we can choose 2 disjoint Vitali sets (if so, my question is solved)? I am not familiar with AC and do not know whether is it reasonable. – Isomorphism Jun 20 '20 at 02:18

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