Is there a nonmeasurable subset of $\mathbb{R}^2$ that is $1-$dimensional Hausdorff measurable?

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?

• Isn’t every $H^1$-measurable set a set of $H^2$ measure zero? Jul 24, 2023 at 1:16
• @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) Jul 24, 2023 at 1:35
• Ah, I was assuming finite measure. Jul 24, 2023 at 1:51