Measurable set in real numbers with arbitrary lebesgue density at some point I'm not sure if this is easy or not, but i can't see the solution (or that it is wrong!) 

Suppose that $\alpha \in (0,1)$ is given, Can you find a Lebesgue measurable set in $\mathbb{R}$, such that at point $0$, it has Lebesgue density $\alpha$?
Let $\alpha _n =\alpha \{\frac 1 n -\frac 1 {n+1}\}$ and choose a subset $E_n$ of $ (\frac 1 {n+1},\frac 1 n) \cup (-\frac 1 n,-\frac 1 {n+1})$ whose measure is $\alpha _n$. Let E be the union of the sets $E_n$. Then E has density $\alpha$.
Let $\alpha _n =\alpha \{\frac 1 n -\frac 1 {n+1}\}$ and choose a subset $E_n$ of $ (\frac 1 {n+1},\frac 1 n) \cup (-\frac 1 n,-\frac 1 {n+1})$ whose measure is $\alpha _n$. Let E be the union of the sets $E_n$. Then E has density $\alpha$.