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Find a measurable set $A\subseteq [0,1]$ such that $$\lim\limits_{\varepsilon\to0}\frac{m(A\cap[0,\varepsilon])}{\varepsilon}=\frac{1}{2}$$


I'm also interested in a set $B\subseteq [0,1]$ such that $$\forall\alpha\in(0,1)\quad \frac{m(B\cap[0,\alpha])}{\alpha}=\frac{1}{2}$$


My Attempt

I thought to try constructing:

  1. Fat Cantor set

This does not answer the second question, and I'm not sure if $0$ is a density point for fat cantor set, or not.

  1. Splitting $[0,1]$ by the binary expansion of every element.

Maybe something like:

$\forall x\in[0,1]$ define $i(x)$ to the the first index in the binary expansion of $x$ in which $1$ appear.

Then, $$x\in A\iff \sum\limits_{j=i(x)}^{2i(x)} x_j = 1 \mod 2$$

Intuitively, It seems that $m(A)=m([0,1]\setminus A)$, but I'm not sure how to prove it formally.

Thanks in advance for any help.

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A hint for the first part: let $\{x_k\}\downarrow 0$ such that $x_0=1$ and $x_k>x_{k+1}$ for all $k\in \Bbb N_{\geqslant 0} $, also choose some $r\in [0,1]$. Now set $m_k:=r\ell _k$ for $\ell _k:=x_k-x_{k+1}$, also we set $E_k:=[x_{k+1},x_{k+1}+m_k)$. Then $|E_k|=m_k$, and setting $E:=\bigcup_{k\geqslant 0}E_k$ we found that $|E\cap [0,x_k)|/x_k=r$ for all $k$. Pick $r=1/2$ in your case and try to see how to choose $(x_k)$ to make the density of $E$ at zero equal to $1/2$.

A hint for the second part: use the Lebesgue differentiation theorem to show that it cannot be possible.

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  • $\begingroup$ Thanks for the answer! I didn't understand the sentence "try to see how to choose $(x_k)$ to make the density of $E$ at zero equal to $ 1/2$". I thought you proved that $$|E\cap [0,x_k)|/x_k=r\text{ for all $k$}$$ regardless of the choice of $(x_k)$ as long as $\{x_k\}\downarrow 0$ $\endgroup$ Feb 9, 2020 at 11:58
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    $\begingroup$ @TheHolyJoker because you need to show that the limit exists. Just a sequence is not enough to show that the limit exists. You need to see what values have $|E\cap [0,t)|/t$ for any chosen $t$ in between of the sequence to figure out how this sequence must be to ensure that the limit exists $\endgroup$
    – Masacroso
    Feb 9, 2020 at 12:00

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