I have proved the following statement and I would like to know if I have made any mistake, thanks.
"Prove that if $a,b\in\mathbb{R}, a<b$, 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<b; \\
0 & \text{if }I=\emptyset; \\
\infty & \text{if } I=(-\infty, a)\text{ or } I=(a,\infty);\\
\infty & \text{ if }I=(-\infty,\infty)
\end{cases} $
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.
Best Answer
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.