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


*

*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.


*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.

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.
