Let $M$ be a smooth (second-countable) manifold and let $g, g'$ be smooth Riemannian metrics on $M$ which induce metrics (as in "metric space") $d$ and $d'$ on $M$. Fix $j ≥ 0$ and let $H$ and $H'$ denote the corresponding $j$-dimensional Hausdorff measures on $M$.
Is it true that a subset $A ⊂ M$ is measurable w.r.t. $H$ if and only if it is measurable w.r.t. $H'$?
I know that the notion of Hausdorff nullsets does not depend on the choice of the Riemannian metric (see lemmata 1 and 1' in Sard's Hausdorff Measure of Critical Images on Banach Manifolds (1965)) but is this true for the notion of measurability, too?
My idea for a proof was to use the fact that the metrics $d$ and $d'$ are locally equivalent and, in fact, are "closer and closer to each other" the smaller the neighborhood in question but I can't seem to get the inequalities (and countable unions) tweaked in the right way.
This is true at least for sets of finite measure. That is, if $A$ is $H$-measurable and $H(A)<\infty$, then $A$ is $H'$-measurable. The proof is simple: there exists a Borel set $B$ containing $A$ such that $H(A)=H(B)$ (take a sequence of covers of $A$ by sets whose diameters witness the Hausdorff measure of $A$, and you can assume those sets are closed since that doesn't change their diameters; see Why is Hausdorff measure Borel regular? for details). Since $H(A)<\infty$, this means $B\setminus A$ is $H$-null. That is, $A$ differs from a Borel set $B$ by an $H$-null set. Since every Borel set and every $H$-null set is also $H'$-measurable, this means $A$ is also $H'$-measurable.
For sets of infinite measure it might be true but I would not be surprised if it is quite difficult to prove. In particular, it is no longer true that every measurable set differs from a Borel set by a null set (at least, it is clearly false for $j=0$).