Points of intermediate density for a measurable set Let $E\subset [0,1]$ be a Lebesgue measurable set with measure $m(E)=1/2$.
Define the upper and lower densities
\begin{align*}
\theta_*(E,x) &= \liminf_{r\searrow 0} m(E\cap B_r(x))/2r\\
\theta^*(E,x) &= \limsup_{r\searrow 0} m(E\cap B_r(x))/2r,
\end{align*}
where $B_r(x) = (x-r,x+r)$.  Does there necessarily exist a point $x\in I$ such that 
$$
0 < \theta_*(E,x) \leq \theta^*(E,x) < 1?
$$
Better yet, does there exist a constant $c>0$ independent of $E$ such that we can always find a point $x$ with 
$$
c \leq \theta_*(E,x) \leq \theta^*(E,x) \leq 1-c?
$$
So far, I've tried defining $x=\inf\{y; \theta_*(E,y) = \theta^*(E,y) = 1\}$ in hopes of finding a "jump" on the left side, but in general it is possible for $\theta^*(E,x) = 0$.  I've also been able to find a sequence $x_k, r_k$ such that $|E\cap B_{r_k}(x_k)|/2r_k = 1/2$ for all $k$, but this doesn't tell me anything about the density at $x = \lim x_k$.  
 A: This is the answer posted here, pasted on this page for your convenience. I believe $A=E$ in your notation.
Since $\mu(A)>0$ and $\mu(\mathbb{R}\backslash A)>0$, there is a density point $y_1$ in $A$ and $z_1$ in $\mathbb{R}\backslash A$. Choose $r>0$ sufficiently small such that
$$\frac{\mu(A\cap B_{r_1}(y_1))}{2r_1}> \frac{3}{4} \quad \text{and} \quad \frac{\mu(A\cap B_{r_1}(z_1))}{2r_1}< \frac{1}{4}.$$
Define
$$f_r(x):= \frac{\mu(A \cap B_{r_1}(x))}{2r_1}.$$
For each $r>0$, $f_r$ is continuous, and so by Intermediate value theorem there is $x$ between $y_1$ and $z_1$ such that
$$\frac{\mu(A \cap B_{r_1}(x))}{2r_1}=\frac{1}{2}.$$
Next, there are density points $y_2$ and $z_2$ of $A$ and $\mathbb{R}\backslash A$ in $B_{r_1}(x_1)$. Then choose $r_2<r_1/2$ sufficiently small such that $B_{r_2}(y_2), B_{r_2}(z_2) \subset B_{r_1}(x_1)$, and
$$\frac{\mu(A\cap B_{r_2}(y_2))}{2r_1}> \frac{3}{4} \quad \text{and} \quad \frac{\mu(A\cap B_{r_2}(z_2))}{2r_1}< \frac{1}{4}.$$
Again by Intermediate value theorem, there is $x_2$ between $y_2$ and $z_2$ such that
$$\frac{\mu(A \cap B_{r_2}(x_2))}{2r_2}=\frac{1}{2}.$$
Continuing this, we have a sequence of points $\{x_n\}$ and radius $\{r_n\}$ such that
$$\frac{\mu(A \cap B_{r_n}(x_n))}{2r_n}=\frac{1}{2}\quad \text{and} \quad B_{r_n}(x_n)\subset B_{r_{n-1}}(x_{n-1}) \quad \text{and} \quad r_n \downarrow 0,$$
and so $(x_n)$ converges to the point $x\in \bigcap_{n=1}^\infty \overline{B_{r_n}(x_n)}$.
Also, $x$ is not a density point of both $A$ and $\mathbb{R}\backslash A$ since $\mu(A\cap B_{2r_n}(x))\leq 2r_n+ 2r_n/2$ (since half of $B_{r_n}(x_n)$ is in $\mathbb{R}\backslash A$), and so
$$ \frac{\mu(A\cap B_{2r_n}(x))}{4r_n}\leq \frac{3}{4} \text{  for each $n$},$$
which implies $x$ is not a density point of $A$. Similarly, $x$ is not a density point of $\mathbb{R}\backslash A$. This is the required $x$.
