Let $s, \delta \in (0,1)$. Consider the outer measure on $\mathbb{R}$, $\mu^s\_{\delta}$, defined by
\begin{align\*}
 \mu^s\_{\delta}(E):=\inf \left\{\sum\_{j}\lvert I\_{j}\rvert^s: E \subset \bigcup\_{j} I\_{j}: I\_{j} \text { closed intervals, } \lvert I\_j\rvert\leq\delta\right\}.
\end{align\*}
For an interval $I \subset \mathbb{R}$, $\lvert I\rvert$ denotes the length of $I$. I want to prove that if $E$ is an interval and $\delta< \lvert E\rvert$, then
\begin{align\*}
 \mu^s\_{\delta}(E) \geq \delta^{s-1}\left|E\right|-\delta^s.
\end{align\*}
I think that by definition, $\exists \epsilon>0$ such that $\mu^s\_{\delta}(E)+\epsilon\geq \sum\_{j}\left|I\_{j}\right|^s\geq \left(\sum\_{j}\left|I\_{j}\right|\right)^s\geq \left|E\right|^s>\delta^s.$ I wonder how to get $\delta^{s-1}$. I feel like I need to show that the delta cover is the smallest one among all the best covers, thankswhich is achieved through infimum. Such a delta cover is a little bit larger than $\delta^{s-1}\left|E\right|-\delta^{s}$.



An outer measure defined on $\mathbb {R}$
$\newcommand\de\delta\newcommand\ol\overline$Your goal cannot be attained in general. Indeed, suppose that $\de\in(0,1/2)$. Take any interval $E$ of length $\de\_1:=|E|\in(\de,2\de)$. Then for the closure $\ol E$ of $E$ and some closed intervals $I\_1$ and $I\_2$ of lengths $|I\_1|=\de$ and $|I\_2|=\de\_1-\de\le\de$ we have $E=I\_1\cup I\_2$, so that
$$\mu\_\de^s(E)\le|I\_1|^s+|I\_2|^s
=\de^s+(\de\_1-\de)^s\underset{\de\_1\downarrow\de}\longrightarrow
\de^s<\de^{s-1}-\de^s,$$
so that for some interval $E$ with $\de<|E|$ we have
$$\mu\_\de^s(E)\not\ge\de^{s-1}-\de^s.$$


(It is actually easy to see that for any interval $E$ of length $t$ we have $\mu\_\de^s(E)=k\de^s+(t-k\de)^s$, where $k:=\lfloor t/\de\rfloor$.)


