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.

  • $\begingroup$ I think your definition of the Hausdorff dimension is wrong... it should read $\dim_{\mathcal{H}}E = \inf\{0\leq s < +\infty:\mathcal{H}^s(E)=0\} = \sup\{s:\mathcal{H}^s(E)=+\infty\}$. $\endgroup$ – yousuf soliman Feb 25 '18 at 6:12
  • $\begingroup$ @yousufsoliman it won't exceed n $\endgroup$ – yaoliding Feb 25 '18 at 6:51
  • $\begingroup$ Sure... but you have a sup when it should be an inf $\endgroup$ – yousuf soliman Feb 25 '18 at 6:52
  • $\begingroup$ @yousufsoliman You're right $\endgroup$ – yaoliding Feb 26 '18 at 4:38

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).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.