We begin with the case in which we have a closed bounded interval, say $[a,b]$ . Since the open interval $(a-\varepsilon,b+\varepsilon)$ contains $[a,b]$ for each positive number $\varepsilon$ , we have $m^*[a,b]\le b-a+2\varepsilon$ . But since this is true for each positive $\varepsilon$ , we must have $m^*[a,b]\le b-a$ . Thus we only have to show that $m^*[a,b]\ge b-a$ ; for this it suffices to show that if $\{I_n\}$ is a countable open cover by intervals of $[a,b]$ , then \begin{equation*} \sum l(I_n)\ge b-a. \end{equation*}By the Heine-Borel theorem, any collection of open intervals covering $[a,b]$ contains a finite subcollection that also cover $[a,b]$ and since the sum of the lengths of the finite subcollection is no greater than the sum of the original one, it suffices to prove the inequality for finite collections $\{I_n\}$ that cover $[a,b]$ . Since $a$ is contained in $\bigcup I_n$ , there must be one of the $I_n$ 's that contains $a$ . Let this be the interval $(a_1,b_1)$ . We then have $a_1<a<b_1$ . If $b_1\le b$ , then $b_1\in [a,b]$ , and since $b_1\not\in (a_1,b_1)$ , there must be an interval $(a_2,b_2)$ in the collection $\{I_n\}$ such that $b_1\in (a_2,b_2)$ , that is $a_2<b_1<b_2$ . Continuing in this fashion, we obtain a sequence $(a_1,b_1),\dots,(a_k,b_k)$ from the collection $\{I_n\}$ such that $a_i<b_{i-1}<b_i$ . Since $\{I_n\}$ is a finite collection our process must terminate with some interval $(a_k,b_k)$ . But it terminates only if $b\in (a_k,b_k)$ , that is if $a_k<b<b_k$ . Thus

If $I$ is any finite interval, then given $\varepsilon>0$ , there is a closed interval$ J\subset I$ such that $l(J)>l(I)-\varepsilon$ . Hence $$ l(I)-\varepsilon<l(J)=m^*J\le m^*I\le m^*\overline I=l(\overline I)=l(I), $$ where by $\overline I$ we mean the topological closure of $I$ . Thus for each $\varepsilon>0$ , we have $l(I)-\varepsilon<m^*I\le l(I)$ , and so $m^*I=l(I)$ .

If now $I$ is an unbounded interval, then given any real number $\Delta$ , there is a closed interval $J\subset I$ with $l(J)=\Delta$ . Hence $m^*I\ge m^*J=l(J)=\Delta$ . Since $m^*I\ge\Delta$ for each $\Delta$ , it follows $m^*I=\infty=l(I)$ .

Bibliography

Royden, H. L. Real analysis. Third edition. Macmillan Publishing Company, New York, 1988.