On page 24 of Falcolner's The Geometry of Fractal Sets, Falcolner defines the set $F = \{ x \in E : \mathcal{H}^s(E \cap U) < \alpha$ diam$(U)^s$, for all convex sets $U$ containing $x$ such that $0 < $diam$(U) \le p \}$,
where $\mathcal{H}^s$ denotes the $s$-dimensional Hausdorff measure, $E$ is an s-set ($E$ is Borel and $ 0 < \mathcal{H}^s(E) < \infty$), and $p > 0$ is fixed.
He then immediately says that $F$ is a Borel measurable set.I'm unsure how to show that this set is measurable.
Edit:
If it helps any, at this point we also know that
1) For $E$ as above and $r > 0$ fixed, the map $x \to \mathcal{H}^s(E \cap B_r(x))$ is upper semicontinuous, and
2) The maps $x \to \bar{D^s}(E,x)$ and $x \to {D}^s\_(E,x)$ (the upper and lower circular densities) are Borel measurable.