Let $(X,d)$ be an $\mathcal{H}^n$-rectifiable metric space, i.e. there exits a collection of Lipschitz maps from measurable subsets of $\mathbb{R}^n$ to $X$ such that $ \mathcal{H}^n(X \backslash \cup\_i f\_i(A\_i)) = 0 $. 


Is it true that for any subset $A \subset X$,
$$ \mathcal{H^n}(A) = \mathcal{H}^n\_\infty (A) \ .$$


The claim is true on $X = \mathbb{R}^n$. Note that the same equality fails horribly if we consider $\mathcal{H}^k$ for $k<n$ -- think of an infinitely long curve inside a bounded set.


If this helps: I am interested in small scales, so, you might consider asymptotic behavior as $\text{diam} (A) \to 0$.


Thanks for your consideration.



Is Hausdorff Measure equal to Hausdorff Content on rectifiable (metric) spaces?
In general, no. For example, $X$ may be a countably infinite collection of lines through the origin in $\mathbb{R}^2$. Then $X$ is $1$-rectifiable.


For any ball $B$ centered at the origin, $B\cap X$ has finite Hausdorff $1$-content but infinite Hausdorff $1$-measure.


If you have some kind of Ahlfors regularity of the measure, then you can get comparability of Hausdorff measure and Hausdorff content.


