This is from Section 2A, exercise 7 in Axler's MIRA book.
Suppose $a,b,c,d$ are real numbers with $a<b$ and $c<d$. Prove that$$|(a,b) \cup (c,d)| = (b-a) + (d-c) \iff (a,b) \cap (c,d) = \emptyset$$
I am having trouble coming with the right argument. Visually it makes a lot of sense that the intersection is empty, but I am having trouble using this intuition right now. For example, if I assume the intersection isn't empty, then how does that change the equality? In particular, how would the right side $(b-a) + (d-c)$ be affected as a result?
For reference, $||$ denotes the outer measure, so $|(b-a)| = b-a$ says that the outer measure of the open interval (a,b) is equal to b-a.
Updated with proof: Let $A_1 = (a,b)$ and $A_2 = (c,d)$. Take the union $A = A_1 \cup A_2$, where $A_1$ and $A_2$ are not necessarily disjoint. Then from elementary set theory, we can express $A$ as the union of disjoint sets $A = A_1 \cup A_2 \bigcup A_1 \cap A_2$. Then we have $$|A_1 \cup A_2 \bigcup A_1 \cap A_2 | = |A_1| + |A_2| + |A_1 \cap A_2|.$$ Then the original equality assumption implies that $A_1 \cap A_2 = \emptyset$ as desired.