Consider a topological space $D^n$ homeomorphic to the closed unit ball of $\mathbb{R}^n$. Suppose that $D^n$ is equipped with a topogically compatible metric that is intrinsic.
Is the $(n+1)$-dimensional Hausdorff measure of $D^n$ equal to zero?
(A very close question to this would be: does $D^n$ have Hausdorff dimension $n$?)
The case $n=1$ is straightforward, because it is known that an intrinsic metric on a space homeomorphic to an interval makes the space isometric to an interval, so the $2$-dimensional Hausdorff measure would be 0.
Theorems of dimension theory that I've been looking at don't seem to apply here, and finding a counterexample seems complicated, since I'm assuming it would be fractal in nature.
Any help would be appreciated!
Added: Also, does the situation change if one assumes that $D^n$ is homeomorphic to the open unit ball instead of the closed unit ball?
[Upgrading from comments]
In the case of an open ball, the answer is generally no, at least for $n\geq 3$. In dimension $3$, a counterexample is any open subset homeomorphic to $D^3$ in the Heisenberg group $\mathbb H$, equipped with its Carnot–Carathéodory (CC) metric $d_{CC}$. Since the CC metric is intrinsic, and Ahlfors $4$-regular (balls of radius $r$ have Hausdorff $4$-measure comparable to $r^4$), every open subset of $\mathbb H$ has positive Hausdorff $4$-measure, and since $\mathbb H$ is a manifold, there is an open subset homeomorphic to the open ball.
We must be slightly cautious here, since strictly speaking, once we restrict to an open set $U\subset \mathbb H$, our original metric $d_{CC}$ need not be intrinsic any more (a shortest path may leave the open set), however, we can consider the induced length metric $d_{CC}'$ given by taking shortest paths in $U$. It is not hard to show this is locally isometric to $d_{CC}$, so Hausdorff measure will be the same under either metric.
In dimensions higher than $3$, one could consider other Carnot Groups, or note that once we have a locally $4$-regular metric $d$ on $D^3$, then if $d_e$ is the Euclidean metric on $D^k$, then the metric $d'$ on $D^{3+k}\cong D^3\times D^k$ given by $d'((p_1,q_1),(p_2,q_2))=d(p_1,p_2)+d_e(q_1,q_2)$ is locally $(4+k)$-regular, so we have counterexamples for all dimensions greater than $3$.
The case of closed balls is a little more subtle, since a little work would need to be done to ensure that $d_{CC}$ induces a topologically equivalent length metric on $\overline{U}$. I believe this is almost certainly true for reasonable open subsets $U$ of $\mathbb H$, but possibly not for an arbitrary open subset, if the boundary is sufficiently wild.
Finally, I don't know a counter-example in the case $n=2$, though I suspect there is one.