A measurable set $E\subset \Bbb R$ such that $\limsup_{h\to 0^+} \frac{1}{2h} m(E\cap [-h,h]) = \alpha$ for some $\alpha \in (0, \frac 12)$

Question

Construct a measurable set $E\subset \Bbb R$ containing $0$ such that
$$\limsup_{h\to 0^+} \frac{1}{2h} m(E\cap [-h,h]) = \alpha$$ for some $\alpha \in (0, \frac 12)$ where $m$ denotes the Lebesgue measure on $\Bbb R$.
Hint: Consider $E:= \{0\} \cup \bigcup_{j=1}^\infty [2^{-j}, 2^{-j} \beta]$ for some appropriate $\beta \in (1,2)$.

I understand that this might possibly be a duplicate, but I'd like feedback on my work below, and other suggestions for constructing the required set $E$.

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Let $E:= \{0\} \cup \bigcup_{j=1}^\infty [2^{-j}, 2^{-j} \beta]$ for some $1 < \beta < 2$ (to be determined later.) Let $0 < h < \frac12$. Then, there exists $k \in \mathbb N$, $k \ge 1$ such that $2^{-k-1} \le h < 2^{-k}$. Note that $k \to \infty$ as $h\to 0$. We have $$\frac{1}{2h} m(E\cap [-h,h]) = \frac{1}{2h} m(E\cap [0,h]) = \frac{\beta - 1}{2h}\sum_{j\ge k+1} 2^{-j} = \frac{\beta - 1}{2h} \cdot 2^{-k} = \frac{2^{-k-1}(\beta - 1)}{h}$$ Since $2^{-k-1} \le h < 2^{-k}$, we have the bounds $$\frac{\beta-1}{2} < \frac{1}{2h} m(E\cap [-h,h]) \le \beta - 1$$ giving $$\frac{\beta-1}{2} < \limsup_{h\to 0^+}\frac{1}{2h} m(E\cap [-h,h]) \le \beta - 1$$ We show that $\limsup_{h\to 0^+}\frac{1}{2h} m(E\cap [-h,h]) = \beta - 1$. For any $0 < \epsilon < \frac 1 2$, we can find $k \in \mathbb N$ such that $2^{-k} < \epsilon$. Then, let $h \in [2^{-k-1}, 2^{-k})$. If $h = 2^{-k-1}$, then $$\frac{1}{2h} m(E\cap [-h,h]) = \frac{2^{-k-1}(\beta - 1)}{h} = \beta - 1$$ As $\epsilon \in (0, \frac{1}{2})$ is arbitrary, we have $$\limsup_{h\to 0^+}\frac{1}{2h} m(E\cap [-h,h]) = \lim_{\epsilon \to 0} \sup\left\{\frac{1}{2h} m(E\cap [-h,h]): h \in (0,\epsilon)\right\} = \beta - 1$$ Finally, choosing $\beta = 1 + \alpha$ gives the desired equality.

I'd like to know if my work is correct, and suggestions for alternative constructions of the set $E$ (i.e., different from the hint.) Thank you!

Answer 1

Here is a general construction of a set $\subset[0,\infty)$ whose density at $0$ is $\alpha$:

Let $a_n>0$, $n\in\mathbb{Z}_+$, be a strictly decreasing sequence converging to $0$. For $0<\alpha<1$ define $E$ as a symmetric set about $0$ so that \begin{align*} E\cap(0,\infty)=\bigcup_n[a_{n+1},(1-\alpha)a_{n+1}+\alpha a_n] \end{align*}

Given $0<r<a_1$, there is $n\in\mathbb{Z}_+$ such that $a_{n+1}\leq r<a_n$. Then \begin{align*} \frac{\alpha a_{n+1}}{a_n}\leq \frac{\lambda(E\cap[0,a_{n+1}])}{a_n}\leq \frac{\lambda(E\cap[0,r])}{r}\leq\frac{\lambda(E\cap[0,a_n])}{a_{n+1}}\leq \frac{\alpha a_n}{a_{n+1}} \end{align*} This shows that the density of $E$ at $0$ is $\alpha$. In particular, $0$ is not a Lebesgue point of $E$. Choosing $a_n$ so that $\lim_m\frac{a_{n+1}}{a_n}=1$ gives another set $E$ satisfying the conditions of the OP.

Edit:

The set proposed by the OP is of the form $$E=\bigcup_n\big[\frac{1}{2^{n+1}}, \frac{1}{2^{n+1}}+(\beta-1)\big(\frac{1}{2^n}-\frac{1}{2^{n+1}}\big)\big]$$ In which case, \begin{align} \liminf_{r\rightarrow0+}\frac{|E\cap[0,r]|}{r}&=\beta-1\\\ \limsup_{r\rightarrow0+}\frac{|E\cap[0,r]|}{r}&=\frac{2(\beta-1)}{\beta} \end{align} See for example here.