If $I$ is an interval, then

$$m^{\ast }(I)=\ell (I)$$

Suppose $I=[a,b]$. Then $I\subset (a-ϵ,b+ϵ)$ for all $ϵ>0$. But then we have

$$m^{\ast }(I)\le \ell (a-ϵ,b+ϵ)=b-a+2ϵ$$

Thus we have $m^{\ast }(I)\le b-a$.

Now we show $b-a\le m^{\ast }(I)$. Let $\{I_n{\}}_n$ be a collection of open intervals such that $[a,b]\subset \bigcup _n{I}_n$.

Now, since $[a,b]$ is compact (by the Heine-Borel theorem), there exists a finite subcover $\{J_k{\}}_{k=0}^m\subset \{I_n{\}}_n$ such that $[a,b]\subset \bigcup _{k=0}^m{J}_k$

Since $a\in \bigcup _{k=0}^m{J}_k$, there exists $k_1$ such that $a\in J_{k_1}$. Rearranging intervals, I let $k_1=1$ and $a\in J_1=(a_1,b_1)$. If $b_1<b$, then $b_1\in [a,b]$. So ther eis some $k_2$ such that $b_1\in J_{k_2}$. By rearranging again, I assume $k_2=2$. So $b_1\in J_2=(a_2,b_2)$.

Continuing in the same manner, we conclude that there exists a $K,1\le K\le m$ such that $b<b_K$. And for all $k\in \{1,2,\dots ,K-1\}$ we have $b_k\le b$ and $a_{k+1}\le b_k<b_{k+1}$.

But then we have

$$\begin{array}{rl}\sum _n\ell (I_n)& \ge \sum _{k=1}^m\ell (J_k)\\ & \ge \sum _{k=1}^K\ell (J_k)\\ & =(b_K-a_K)+(b_{K-1}-a_{K-1})+\cdots +(b_1-a_1)\\ & =b_K+(b_{K-1}-a_K)+(b_{K-2}-a_{K-1})+\cdots +(b_1-a_2)-a_1\\ & \ge b_K-a_1\\ & \ge b-a\\ & =\ell (I)\end{array}$$

Thus we have $m^{\ast }(I)\ge \ell (I)=b-a$

Thus we conclude $m^{\ast }(I)=\ell (I)$

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