Let $s\in (0, 1)$ be a real number. Let $E\subset [0, 1]$ be a Borel set whose Hausdorff dimension is given by $s$. Assume that $\mathcal{H}^s(E)=+\infty$, that is, the $s$-dimensional Hausdorff measure of $E$ is $+\infty$.

We define lower density $$ \underline{D}^s(E, x):=\underline{\lim}_{r\to 0} \frac{\mathcal{H}^s(E\cap (x-r, x+r))}{(2r)^s} $$ Similarly we define upper density $$ \overline{D}^s(E, x):=\overline{\lim}_{r\to 0} \frac{\mathcal{H}^s(E\cap (x-r, x+r))}{(2r)^s} $$ Define a set $$ \widetilde{E}:=\{x\in [0, 1]: \overline{D}^s(E, x)>0\}. $$ My question is, do we have $$ \mathcal{H}^{s+\epsilon}(\widetilde{E})=0, $$ for every $\epsilon>0$?