Density of Borel set at 0 The Lebesgue density theorem says that if $E$ is a Lebesgue measurable set, then the density of $E$ at almost every element of $E$ is 1 and the density of $E$ at almost every element not in $E$ is 0.
However, is it true that for each $t$ strictly between 0 and 1, there is a Borel set $E$ that has density $t$ at 0?
I have no idea how to construct such a set for a random value of $t$.
Any help would be appreciated.

Consider a sequence of numbers $r_n \searrow 0$ such that $\frac{r_{n-1}}{r_n} \to 1$.  Let $\theta$ be a measure preserving map from $(0,r_1]$ to $\mathbb R^2$ that takes $(\pi r_{n}^2,\pi r_{n-1}^2] \subset \mathbb R$ to $\{x \in \mathbb R^2: r_n < |x| \le r_{n-1}\}$.  Then let $A$ be a 'piece of pie' centered at the origin in $\mathbb R^2$, with angle $\alpha$ at the corner.  Then $\theta^{-1}(A)$ will be a set with density $\alpha/(4\pi)$ at $0$.
This will give densities $0 \le t \le \frac12$.  To get $\frac12 < t \le 1$, simply add $(-\infty,0]$.
