For $n\in\mathbb{N}^*$ and $s\in\mathbb{R}_{\ge 0}$, the $s-$dimensional Hausdorff measure $H^s$ is an outer measure over $\mathbb{R}^n$, and the $\sigma-$algebra of $s-$dimensional measurable subsets of $\mathbb{R}^n$ is given by Carathéodory's criterion. It is well-known that $H^0$ is the counting measure and that $H^n$ is the Lebesgue outer measure over $\mathbb{R}^n$.

Giving a subset of $\mathbb{R}^2$ that is $H^2-$ (i.e. Lebesgue) but not $H^1-$measurable seems natural: take $V$ to be a Vitali set, then $V\times \{0\}$ works. I was wondering if we could have a subset of $\mathbb{R}^2$ that is $H^1-$ but not $H^2-$measurable?

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    $\begingroup$ Isn’t every $H^1$-measurable set a set of $H^2$ measure zero? $\endgroup$ Jul 24, 2023 at 1:16
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    $\begingroup$ @TedShifrin No I don't believe so; for example the whole space $\mathbb{R}^2$ :) (Note also that $H^1$, or generally $H^s$ for all $s\in [0,n)$, is not $\sigma-$finite) $\endgroup$ Jul 24, 2023 at 1:35
  • $\begingroup$ Ah, I was assuming finite measure. $\endgroup$ Jul 24, 2023 at 1:51


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