Constructing a particular Borel set I'm working on a question that requests the following:
Given $\alpha \in (0,1)$, Construct a Borel set $E_\alpha \subseteq [-1,1]$ such that $\lim_{r\to 0^+} \dfrac{m(E_\alpha \cap [-r,r])}{2r} = \alpha$.
I'm totally stumped.  This question is in the same chapter that discusses antiderivatives, but I can't figure out how to use that to my advantage here.
 A: Hint: Break up $(0,1]$ (and similarly $[-1,0)$) into an infinite sequence intervals $I_n$ approaching $0$, and for each $n$ choose a subset $J_n\subset I_n$ which has an $\alpha$ proportion of its measure.  Then if $E_\alpha$ is the union of the $J_n$, $E_\alpha$ will have exactly an $\alpha$ proportion of the measure of $[0,r]$ whenever $r$ is an endpoint an $I_n$.  If you choose the $I_n$ appropriately, you can make it so the error when $r$ is in the middle of an $I_n$ is small enough that $E_\alpha$ satisfies the desired condition.
A: For $n\in \Bbb N$ let $A_a(n)=(\frac {1}{n+1}, \frac {1}{n+1}+\frac {a}{n^2+n})$, which is a subset  of $(\frac {1}{n+1},\frac {1}{n}),$ and let $B_a(n)=A_a(n)\cup \{-x:x\in A_a(n)\}.$ Observe that $$\frac {m(A_a(n))}{   \frac {1}{n}-\frac {1}{n+1}   }=a.$$ Let $E_a=\cup_{n\in \Bbb N}B_a(n).$
Observe that for $n\in \Bbb N$ we have $$\frac {m(E_a\cap [-\frac {1}{n},\frac {1}{n}])}{ \frac {2}{n}}=a.$$ For $n\in \Bbb N$ and $\frac {1}{n+1}<x<\frac {1}{n}$ we have $$\frac {2a}{n+1}=m(E_a \cap [-1/(n+1),1/(n+1)])\leq$$ $$\leq m(E_a\cap [-x,x])\leq$$ $$\leq  m(E_a\cap [-1/n,1/n])=\frac {2a}{n}$$  and we have    $$\frac {n}{2} <\frac {1}{2x}<\frac {n+1}{2}$$ so, therefore,       $$\frac {2a}{n+1}\cdot \frac {n}{2}<\frac { m(E_a\cap [-x,x])}{2x}<\frac {2a}{n}\cdot \frac {n+1}{2}.$$
