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
$$
S_r(x,\theta,\pi/2) = x + \lbrace y\in \mathbb{R}^n: y\cdot \theta \geq 0, \|y\|<r\rbrace.
$$
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.
Please help me.
 A: 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 $E$.
