Let $E_1,E_2$ be some sets, I'd like to prove that if $d(E_!,E_2)>0$ then $m^*(E_1\cup E_2) = m^*(E_1)+m^*(E_2)$, where $m^*$ is outer Lebesgue measure.
this statement is used often, and although it is very intuitive I can't seems to go anywhere.
p.s. we already know that $m^*(E_1\cup E_2) \leq m^*(E_1)+m^*(E_2)$
thanks ahead
Take $d(E_1,E_2)>\delta >0$. Let $\varepsilon >0$. There are $\{Q_j\}$ cubes s.t. $E\subset \bigcup_{j=1}^\infty Q_j$ and $$ \sum_{i=1}^\infty |Q_j|\leq m^*(E)+\varepsilon.$$We can suppose WLOG that the cubes has dimater smaller than $\delta $. Therefore, each $Q_j$ can intersect at most one set $E_i$. So let $$J_i=\{j\mid E_i\cap Q_j\neq\emptyset\},\quad i=1,2$$ So, $E_i\subset \bigcup_{j\in J_i}Q_j.$ Now, $$m^*(E_1)+m^*(E_2)\leq \sum_{j\in J_1}|Q_j|+\sum_{j\in J_2}|Q_j|\leq \sum_{j=1}^\infty |Q_j|\leq m^*(E_1\cup E_2)+\varepsilon ,$$ what prove the claim.