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For a set $E\subseteq\mathbb R^n$, the $s$-dimensional Hausdorff outer measure is defined as $\mathcal H^s(E)=\lim\limits_{\delta\to0}\mathcal H^s_\delta(E)$, where $\mathcal H^s_\delta(E)=\inf\left\{\sum_{k=1}^\infty (diam F_k)^s:E\subseteq\bigcup_{k=1}^\infty F_k,diam F_k<\delta\right\}$.

The Hausdorff dimension of $E$ is defined as $\dim E=\inf\{s<n:\mathcal H^s(E)=0\}$

Suppose $E$ is a Borel set (hence is truly Hausdorff measurable) and has dimension $s$. Define a Borel measure $\mu_E$ by $\mu_E(F)=\mathcal H^s(E\cap F)$.

Does $\mu_E$ a sigma-finite measure?

It's obvious that $\mu_E$ is finite if $\mathcal H^s(E)<\infty$, but in general, I don't know how "big" for $E$ can be.

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  • $\begingroup$ I think your definition of the Hausdorff dimension is wrong... it should read $\dim_{\mathcal{H}}E = \inf\{0\leq s < +\infty:\mathcal{H}^s(E)=0\} = \sup\{s:\mathcal{H}^s(E)=+\infty\}$. $\endgroup$ Feb 25, 2018 at 6:12
  • $\begingroup$ @yousufsoliman it won't exceed n $\endgroup$
    – Liding Yao
    Feb 25, 2018 at 6:51
  • $\begingroup$ Sure... but you have a sup when it should be an inf $\endgroup$ Feb 25, 2018 at 6:52
  • $\begingroup$ @yousufsoliman You're right $\endgroup$
    – Liding Yao
    Feb 26, 2018 at 4:38

1 Answer 1

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The Liouville numbers are uncountable and are of Hausdorff dimension zero. Since the $\mathcal{H}^0$ is the counting measure, this shows that $\mu_E$ is not $\sigma$-finite where $E$ denotes the set of Liouville numbers in $\mathbb{R}$. You can find a proof of this in Oxtoby's Measure and Category (Theorem 2.4).

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