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 $\alpha$?

Here’s full justification for the answer provided by Kavi: 
Suppose first that $t=\frac 1 n$, then $$|(-t,t)|=2\cdot(\sum \alpha\cdot(\frac 1 n-\frac{1}{n+1}))=2\alpha t,$$ hence $\frac{|E\cap (-t,t)|}{2t}=\alpha.$
Now suppose we’re given arbitrary $t\in (\frac 1 n,\frac{1}{n+1})$. Indeed,
$$\frac{2\cdot (\frac{1}{n+1})\alpha}{2\cdot \frac{1}{n}}\leq \frac{|E\cap (-t,t)|}{2t} \leq \frac{2\cdot (\frac{1}{n})\alpha}{2\cdot \frac{1}{n+1}}. $$
And we’re done.
