If $(a,b)\cap (c,d) \neq \emptyset$ then, by considering cases one can see that the outer measure of $(a,b)\cup (c,d)$ is not the same as $b-a + d-c$.For the other side, assume they are disjoint, moreover assume that $b\leq c$, then by countable subadditivity, $\lvert (a,b)\cup (c,d)\rvert\leq b-a+d-c$. On the other hand $(a,d)=(a,b)\cup [b,c] \cup (c,d)$, so again by countable subadditivity, $d-a\leq \lvert (a,b) \cup (c,d)\rvert + (c-b)$. Taking $c-b$ to the other side we obtain the result.
↧