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Motivation: Some sets have Hausdorff dimension $\alpha$ but have zero $\alpha$-dimensional Hausdorff measure. These sets may have another dimension function; i.e., a function $h:[0,\infty)\to[0,\infty)$ such that if we change the definition of the Hausdorff measure by replacing $R^{\alpha}$ with $h(R)$ (where $R$ denotes the radius of a ball in a covering), the value of the Hausdorff measure is positive and finite.

Question: Does there exist a set which has no dimension function?

In page 68 of the book of Rogers, it is said that `It is unusual for a set to have such a function', but no hint is given.

Ali Khezeli
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1 Answers1

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Yes.

See Construction of null sets with prescribed Hausdorff dimension and generalizations and the following papers and their references:

Borel sets which are null or non-${\sigma}$-finite for every translation invariant measure by Márton Elekes and Tamás Keleti (2006).

Visible and invisible Cantor sets by Carlos Cabrelli, Udayan B. Darji, and Ursula Molter (2013). arXiv.org version

The second paper shows that, for any Polish metric space $X$ (e.g. $X$ can be ${\mathbb R}^n),$ most compact sets in the sense of Baire category (for the hyperspace of nonempty compact sets) have a positive and finite measure for some translation invariant Borel (outer) measure (the measure can vary with the set, of course). However, it is not known whether, for most compact sets in this sense, the measure can always be chosen from among the Hausdorff $h$-measures for some Hausdorff measure function $h$ (i.e. it is not known whether most compact sets in this sense have a dimension function).

The second paper also shows that each of the collections of compact sets --- (a) those compact sets having positive and finite Hausdorff $h$-measure for some Hausdorff measure function $h,$ and (b) those compact sets having zero or infinite measure for every translation invariant Borel measure (this collection of compact sets is a subset of those compact sets that do not have a dimension function) --- is dense in the same hyperspace. So in this sense, compact sets that have a dimension function and compact sets that do not have a dimension function each exist in great profusion.

Dave L. Renfro
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    The paper says that "It remains open whether a generic compact subset of $[0,1]$ has a dimension function". It only proves that every generic compact subset $E$ is "visible"; i.e., has a translation-invariant measure $\mu$ such that $0<\mu(E)<\infty$. – Ali Khezeli Nov 04 '18 at 10:01
  • Yes, you are correct . . . I overlooked an important distinction made in this paper --- The existence of a translation invariant measure vs. the existence of a specific type of such measure, namely a Hausdorff $h$-measure for some Hausdorff measure function $h.$ I'll fix my answer. – Dave L. Renfro Nov 04 '18 at 10:24