Find a Borel subset satisfying the condition Let $\alpha \in (0,1)$. Find a fixed Borel subset $E$ of $[-1,1]$ such that 
$$
\lim_{r \rightarrow 0^+} \frac{m(E \cap [-r,r])}{2r} = \alpha
$$

I think it is the trickiest problem for studying Lebesgue measure Differentiability. Firstly, I came up with setting $E = [-\alpha r, \alpha r]$, but it does not meet the condition of ''fixed'' subset. It is so tricky that even my TA said this problem is "hard as hell". 
Anybody can suggest any idea? 
Thanks in advance.
 A: Here is a try...
For every $ n \in \mathbb N$ define 
$$ C_n= \left[ -\frac{1}{n} , - \frac{1}{n+1} \right)  \cup \left( \frac{1}{n+1} , \frac{1}{n} \right]$$
We will use the following claim:
Claim: If $\displaystyle{ A \subset \mathbb R}$ is measurable with $\displaystyle{ 0 < m(A) < \infty}$ and $\alpha \in (0,1) $ then  we can choose a subset $ B \subset A$ such that $ m(B) = \alpha m(A) $.
Proof of claim: Define the function $\displaystyle{ f: \mathbb R \to \mathbb R}$ with
$$ f(x) = m(A \cap (- \infty , x]$$
We can easily see that $f$ is contninuous (moreover is $1-$Lipschitz).
Now, we see that
$$ \lim_{n \to \infty} f(n) = \lim_{n \to \infty} m(A \cap (- \infty, n]) = m(A)$$
and 
$$ \lim_{n \to \infty} f(-n) = \lim_{n \to \infty} m(A \cap (- \infty, -n]) = m(\emptyset)=0 $$
Thus,
$$ 0= \lim_{n \to \infty} f(-n) < \alpha m(A)  < \lim_{n \to \infty} f(n) m(A) $$
since $f$ is contninous, by the intermediate value property there exist $x \in \mathbb B$ such that $\displaystyle{ f(x) = m(A \cap (-\infty,x]) = \alpha m(A)}$.
So, it is enough to take $\displaystyle{ B = A \cap (- \infty ,x].  \quad \square}$
Using the claim we can choose $\displaystyle{ E_n \subset C_n }$ such that $\displaystyle { m(E_n) = \alpha m(C_n)}$
Define now 
$$ E= \cup_{n=1}^{\infty} E_n$$
If $\displaystyle{ \frac{1}{n+1} < r \leq \frac{1}{n}}$  then,
$ \displaystyle{ \frac{m(E \cap [-r,r]}{2r} \leq \frac{m(E \cap [ - 1/n ,1/n]}{2/(n+1)} = \frac{2\alpha /n}{2/(n+1)} = \alpha \frac{n+1}{n} \leq \alpha (1+2r)}$
and
$\displaystyle{ \frac{m(E \cap [-r,r]}{2r}  \geq \frac{ m(E \cap (-1/(n+1) , 1/(n+1) ) }{2/n} = \frac{2 \alpha /(n+1)}{2/n} = \alpha \frac{n}{n+1} \geq \alpha (1-2r)}$
Let now $r \to 0 ^+ $ in the above two inequalities to get that 
$$ \lim_{r \to 0^+ } \frac{ m(E \cap [-r,r])}{2r} =\alpha $$
