Let $n\in \mathbb{N}$, and $H^s$ be the $s$-dimentional Hausdorff measure.
Does there exist $0<t<n$, and $K \subset \mathbb{R}^n$; compact set
such that $H^t(K)=\infty$, and
$H^s(K)=0$ for all $s > t$?
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Yes --- Use a (generalized) Cantor set with nonzero finite measure for a Hausdorff measure function that approaches zero quicker than $x^t$ and slower than any power of $x$ greater than $t,$ such as $-\frac{x^t}{\log x} = \frac{x^t}{\log (1/x)}.$ More precise examples than this were actually given by Hausdorff in his 1918 paper where Hausdorff measure was introduced. The bibliography of this paper might be useful. – Dave L. Renfro Jan 14 '18 at 17:41
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Thank you very much. I will check on it. – user521436 Jan 15 '18 at 00:33