Let $E\subset \mathbb{R}$ be a nonempty set. Show that, $$m^{*}(E)\leq \delta(E),$$ where, $$m^{*}(E)=\inf\left\{\sum_{n=1}^{\infty}\ell(I_n):E\subset \bigcup_{n=1}^{\infty} I_n\right\},$$ is the outer Lebesgue measure of $E$ and, $$\delta(E)=\sup\big\{|x-y|:x,y\in E\big\},$$ is its diameter.
My attempt:
First, I consider $E\subset \mathbb{R}$ to be a bounded set. So, $\delta(E)<\infty$. Then, I consider $x_0\in E$ and I define $I_0=(x_0-\delta(E),x_0+\delta(E))$. Notice that, $E\subset I_0$. Then, I obtain that, $$m^{*}(E)\leq \ell(I_0)=2\delta(E).$$ However, this is not the desired result. Any suggestion?
Best Answer
If $E$ is unbounded, there nothing to prove. Suppose that $E$ is bounded. Let $\alpha=\inf E$ and $\beta=\sup E$. For any $\delta>0$, the intervals $I_1=(\alpha-\delta/4,\alpha+\delta/4)$, $I_2=(\alpha,\beta)$ and $I_3=(\beta-\delta/4, \beta+\delta/4)$ cover $E$ and $|I_1|+|I_2|+|I_3|=\operatorname{diam}(E)+\delta$. Hence $m^*(E)\leq\operatorname{diam}(E)+\delta$ for all $\delta>0$.