# Outer measure of intervals: $|(a,b)|=|[a,b)|=|(a,b]|=b-a$

I have proved the following statement and I would like to know if I have made any mistake, thanks.

"Prove that if $$a,b\in\mathbb{R}, a, then $$|(a,b)|=|[a,b)|=|(a,b]|=b-a$$

NOTE: $$|\cdot|$$ refers to outer measure, i.e. for $$A\subset\mathbb{R},\ |A|:=\inf\{\sum_{k=1}^{\infty}l(I_k): I_1,I_2,\dots\text{ are open intervals such that }A\subset\bigcup_{k=1}^{\infty}I_k\}$$; the length of an open interval $$I\subset\mathbb{R}$$ is defined as

$$\ell(I):=\begin{cases} b-a & \text{if }I=(a,b),\ a,b\in\mathbb{R}, a

I already know: $$(1)$$ countable subadditivity of outer measure, $$(2)\$$countable sets have measure $$0$$, $$(3)$$ outer measure preserves order, $$(4)\ |[a,b]|=b-a$$ outer measure of a closed interval

(I) $$(a,b)\subset [a,b]\overset{(3)}{\Rightarrow} |(a,b)|\leq |[a,b]|$$

(II) $$|[a,b]|=|\{a\}\cup (a,b)\cup \{b\}|\overset{(1)}{\leq} |\{a\}|+ |(a,b)|+ \{b\}\overset{(2)}{=} |(a,b)|$$

(I), (II) $$\Rightarrow$$ (III) $$\fbox{|(a,b)|=|[a,b]|\overset{(4)}{=}b-a}$$

$$(a,b)\subset [a,b)\subset [a,b]\overset{(3)}{\Rightarrow}|(a,b)|\leq |[a,b)|\leq |[a,b]|\overset{(III)}{\Rightarrow}\fbox{|[a,b)|=b-a}$$

$$(a,b)\subset (a,b]\subset [a,b]\overset{(3)}{\Rightarrow}|(a,b)|\leq |(a,b]|\leq |[a,b]|\overset{(III)}{\Rightarrow}\fbox{|(a,b]|=b-a}$$

Putting all together we have $$\fbox{|[a,b)|=|(a,b]|=|(a,b)|=|[a,b]|=b-a}$$, as desired.

Your proof is correct. It can be simplified a bit. If you prove $$|(a,b)|=|[a,b]|= b-a$$ first then the remaining equalities follow from $$(a, b) \subset (a, b] \subset [a, b]$$ and $$(a, b) \subset [a, b) \subset [a, b]$$ because of the order preserving property.