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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}-\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 $\delta^{s-1}-\delta^{s}$.

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

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}-\delta^s. \end{align*} I think the key is to express the infimum in terms of $\delta$ and such a cover is a little bit larger than $\delta^{s-1}-\delta^{s}$. I feel like I need to make as few intervals of length $\delta$ as possible.

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}-\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 $\delta^{s-1}-\delta^{s}$.

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}-\delta^s. \end{align*} I think the key is to express the infimum in terms of $\delta$ and such a cover is a little bit larger than $\delta^{s-1}-\delta^{s}$. I feel like I need to make $\delta$ as few as possible.

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}-\delta^s. \end{align*} I think the key is to express the infimum in terms of $\delta$ and such a cover is a little bit larger than $\delta^{s-1}-\delta^{s}$. I feel like I need to make as few intervals of length $\delta$ as possible.