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?
