Lebesgue measurable subset of $\mathbb{R}$ with given metric density at zero Let $0 \leq \alpha < \beta \leq 1$. I'm looking for an example of a Lebesgue measurable subset $E$ of $\mathbb{R}$ such that
$$\liminf_{\delta \rightarrow 0} \frac{m(E \cap (-\delta,\delta))}{2\delta} = \alpha$$
but
$$\limsup_{\delta \rightarrow 0} \frac{m(E \cap (-\delta,\delta))}{2\delta} = \beta$$
where $m$ is the Lebesgue measure on $\mathbb{R}$.
Can someone give an example?
Thank you,
Malik
 A: Try the duplication across zero of $$\bigcup_{n \geq 1} \left[\frac{1}{(2n)!}-\alpha\left(\frac{1}{(2n)!} - \frac{1}{(2n+1)!}\right),\frac{1}{(2n)!}\right] \cup \left[\frac{1}{(2n+1)!}-\beta\left(\frac{1}{(2n+1)!} - \frac{1}{(2n+2)!}\right),\frac{1}{(2n+1)!}\right].$$
A: Divide the interval $(0,1]$ of radii into intervals which decrease "fast enough". In some of the intervals put some set of density $\alpha$, and in other put some set of density $\beta$. You can fill-in the rest of the details yourself.
A: I think the following construction works: Given $\alpha \leq \beta$, define sequences $(a_n)_n$ and $(b_n)_n$ as follows: $$a_0 = 0;$$ $$b_0 = 1;$$ $$a_n = (\beta/\alpha)b_{n-1};$$ $$b_n = {1-\alpha \over 1-\beta}a_n.$$ Now define $E_n = [a_n, b_n)$ and $E = \bigcup_{n=0}^\infty E_n$.
The sequences were chosen so that $\bigcup_{i=0}^n E_i$ contains (roughly) $\beta$ of $[0, b_n)$ but only $\alpha$ of $[0, a_{n+1})$.  It actually always contains slightly more than this, because it contains all of $[0, 1)$ instead of some crazy fractal pattern inside it, but any finite initial segment doesn't matter to the problem. As $R \rightarrow \infty$, the density of $E \cap [0, R)$ in $[0, R)$ oscillates (linearly!) between the two values, giving the behavior you requested.
But don't take this as gospel. It's been a while since I did any measure theory and I wasn't much good at it even at the time.
