# density of fractal measures

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 $\epsilon>0$?

• You can have $H^s(E\cap I)=+\infty$ for every open interval $I$. What are you going to do then? – fedja May 26 '18 at 22:56
• @fedja I see. I saw a density theorem in a book of Falconer for $0<\mathcal{H}^s(E)<\infty$. However I did not find any density theorem for the case of infinity. Could you give me a reference, or there is a simple construction? Thanks! – Guo May 26 '18 at 23:15
• You can find an $s$-dimensional set $E_0$ of infinite $H^s$ measure on $[0,1]$, cannot you? Now just take $E=\cup_{a,b\in \mathbb Q, a<b}T_{a,b}E_0$ where $T_{a,b}$ is a linear map of $[0,1]$ onto $[a,b]$. – fedja May 26 '18 at 23:20