# Lower bound for upper $\pi/2$ angular density

This is exercise 2.3 in Falconer's book 'The Geometry of Fractal Sets'.

Let $$E\subset \mathbb{R}^n$$ be an $$\textit{s}$$-set. That is, it is measurable for the s-dimensional Hausdorff measure $$H^s$$ and has $$0 < H^s(E)< \infty$$.

For $$x\in \mathbb{R}^n$$ and a unit vector $$\theta \in \mathbb{R}^n$$, let $$S_r(x,\theta,\pi/2) = x + \lbrace y\in \mathbb{R}^n: y\cdot \theta \geq 0, \|y\| In other words, we intersect the cone of vectors with angle $$\leq \pi/2$$ from $$\theta$$ with the unit ball of radius $$r$$ and translate to $$x$$. This gives a $$\textit{hemiball}$$. $$\overline{D^s}(E,x,\theta,\pi/2):= \limsup\limits_{r\to0} \frac{H^s(E\cap S_r(x,\theta,\pi/2))}{(2r)^s}$$ is the upper angular density with respect to the angle $$\pi/2$$.

Exercise: Show that for $$H^s$$-almost every $$x\in E$$, we have $$2^{-s} \leq \overline{D^s}(E,x,\theta,\pi/2)$$.

The book demonstrates that the similarly defined upper convex density is $$1$$ almost everywhere on $$E$$ so that the upper angular density when we take the full ball (angle = $$2\pi$$) instead of a cone is bounded below by $$2^{-s}$$. I don't see how to imitate the spirit of that proof.

Author showed me this argument. Let $$\alpha < 2^{-s}$$ and $$\rho>0$$. $$F = \lbrace x \in E: H^s(E\cap S_r(x,\theta,\pi/2)) < \alpha(2r)^s \text{ for all } r \text{ with } r \leq 2^{-s}\rho \rbrace.$$ Let $$F' \subset F$$ be any closed set and let $$\varepsilon >0$$. Let $$\lbrace U_i \rbrace$$ be a cover of $$F'$$ by closed convex sets with $$|U_i| < 2^{-s}\rho$$. The absolute value sign means diameter. Moreover, by definition of Hausdorff measure, we can and will require $$\sum |U_i|^s < H^s(F') + \varepsilon.$$
Without loss of generality assume each $$U_i \cap F'$$ is non-empty. By our closed assumption, we choose $$x_i \in U_i \cap F'$$ such that its $$\theta$$ component is minimal. Precisely, extend $$\theta$$ to an orthonormal basis $$\theta_1 = \theta,\dots,\theta_n$$ of $$\mathbb{R}^n$$ and choose $$x_i = \sum c_j \theta_j \in F'\cap U_i$$ with $$c_1$$ minimal. This ensures that $$U_i \cap F' \subset S_{|U_i|}(x_i,\theta,\pi/2) \cap E$$. Now $$H^s(F') \leq \sum H^s(U_i\cap F') \leq \sum H^s(E \cap S_{|U_i|}(x_i,\theta,\pi/2)) \leq \alpha\sum 2^s|U_i|^s \leq \alpha2^s (H^s(F') + \varepsilon).$$ Since $$\varepsilon$$ was arbitrary, $$H^s(F') \leq \alpha2^s H^s(F').$$ Our starting assumption that $$\alpha < 2^{-s}$$ forces $$H^s(F')=0$$ and regularity of Hausdorff measure forces $$H^s(F)=0$$ whence $$\overline{D^s}(E,x,\theta,\pi/2) \geq 2^{-s}$$ almost surely on $$E$$.