I would like to prove this:
$sup_{R\in R(x)} \frac{1}{|R|}\frac{|R \cap \epsilon B|}{|\epsilon B|}$ $\to \frac{1}{|x_1||x_2|}$ , as $\epsilon \to 0^{+}$, $\forall x_1,x_2 \neq 0$
where $R(x)$ is the family of all rectangles in $\mathbb{R}^2$ which contain $x$ and whose sides are parallel to the axes, $\epsilon B$ is the Euclidean ball of radius $\epsilon$ centered in the origin, | | is the Lebesgue measure on $\mathbb{R^2}$ and $x=(x_1,x_2)$.
It would be obvious if I could interchange the supremum with the limit, because in that case I could consider all the rectangles for which the origin is a Lebesgue density point and then obtain
$\frac{|R \cap \epsilon B|}{|\epsilon B|}$ $\to \chi_{R}(0)=1$ as $\epsilon \to 0^{+}$ for all such rectangles $R$
The supremum now would be the inverse of the measure of the rectangle which has $0$ and $(x_1,x_2)$ as opposite vertices, i.e. $\frac{1}{|x_1||x_2|}$.
Can I interchange limit and supremum? Why?