An outer measure defined on $\mathbb {R}$

Question

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 $\delta^{s-1}\left|E\right|-\delta^{s}$.

Answer 1

$\newcommand\de\delta\newcommand\ol\overline$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 $k:=\lfloor t/\de\rfloor$.)