I have proved the following statement and I would like to know if I have made any mistakes or if there is something I could improve. Thank you.
Suppose $a,b,c,d\in\mathbb{R}$ with $a<b$ and $c<d$. Prove that $|(a,b)\cup (c,d)|=(b-a)+(d-c)\Leftrightarrow (a,b)\cap (c,d)=\emptyset$.
My proof:
$\Rightarrow$| Suppose $(a,b)\cap (c,d)\neq\emptyset$: then either
- $c<a<b<d\Rightarrow |(a,b)\cup (c,d)|=|(c,d)|=d-c\neq (b-a)+(d-c)$, contradiction;
- $a<c<d<b\Rightarrow |(a,b)\cup (c,d)|=|(a,b)|=b-a\neq (b-a)+(d-c)$, contradiction;
- $a<c<b<d\Rightarrow |(a,b)\cup (c,d)|=|(a,d)|=d-a\neq (b-a)+(d-c)$, contradiction;
- $c<a<d<b\Rightarrow |(a,b)\cup (c,d)|=|(c,b)|=b-c\neq (b-a)+(d-c)$, contradiction;
and since in all possible cases we get a contradiction it must be $(a,b)\cap (c,d)=\emptyset$.
$\Leftarrow$| From countable subadditivity of outer measure we know that $|(a,b)\cup (c,d)|\leq |(a,b)|+|(c,d)|$. It remans to show that $|(a,b)|+|(c,d)|\leq |(a,b)\cup(c,d)|$; so, let $I_1,I_2,\dots$ be a sequence of open intervals whose union contains $(a,b)\cup (c,d)$. For $n\geq 1$ let $J_n:=I_n\cap (-\infty,a),\ K_n:=I_n\cap (a,b),\ L_n:=I_n\cap (b,\infty)$. $K_1, K_2,\dots$ is a sequence of open intervals whose union contains $(a,b)$ and $J_1,L_1,J_2,L_2,\dots$ is a sequence of open intervals whose union contains $(c,d)$. Thus $\sum_{n=1}^{\infty}l(I_n)=\sum_{n=1}^{\infty}(l(J_n)+l(L_n))+\sum_{n=1}^{\infty}l(K_n)\geq |(c,d)|+|(a,b)|$ and taking the $\inf$ on both sides we have $|(a,b)\cup (c,d)|\geq |(a,b)|+|(c,d)|$ and this concludes the proof.