Please help me prove the following:
1) Let $A \subseteq \{(x,y)|0< x^2+y^2 < 1 \}$ be a lebesgue measurable set such that every point in $\{(x,y)|0< x^2+y^2 < 1 \}$ is a density point of $A$, then (0,0) is a density point of A as well.
2) Find a set $A$ such that for every n: $\frac{m \bigl(\bigl[0,\frac{1}{2^n}\bigl] \cap A \bigl)}{m\bigl(\bigl[0,\frac{1}{2^n}\bigl]\bigl)}=\frac{1}{2}$.
Couldn't prove the 1st question. Any help? And anyone have an example for the second question? Thanks!
For the first problem, define $B= \{0<x^2+y^2<1\}\setminus A.$ Suppose $m(B)>0.$ Then by Lebesgue's differentiation theorem, a.e. $(x,y) \in B$ is a point of density of $B.$ Let $z_0=(x_0,y_0)$ be one of these. Then it is easy to see
$$ \lim_{r\to 0^+} \frac{m(A\cap D(z_0,r))}{m(D(z_0,r))} = 0.$$
Thus $z_0$ is not a point of density of $A,$ contradiction. This shows a.e. point of $\{0<x^2+y^2<1\}$ is in $A.$ The desired conclusion is immediate from this.