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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 $\epsilon>0$?

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    $\begingroup$ You can have $H^s(E\cap I)=+\infty$ for every open interval $I$. What are you going to do then? $\endgroup$
    – fedja
    Commented May 26, 2018 at 22:56
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    $\begingroup$ @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! $\endgroup$
    – Guo
    Commented May 26, 2018 at 23:15
  • $\begingroup$ 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 $[a,b]$. $\endgroup$
    – fedja
    Commented May 26, 2018 at 23:20

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