Is Hausdorff Measure equal to Hausdorff Content on rectifiable (metric) spaces?

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 -- 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 $$\text{diam} (A) \to 0$$.

• What if X is a circle in the plane with the metric inherited from the plane? To handle the asymptotic version consider a union of countably many circles with shrinking radii. – Yuval Peres Oct 29 '19 at 17:08
• What do you mean when you write $\mathcal{H}^n_{\infty}$? – Amir Sagiv Oct 29 '19 at 17:09
• Amir, for the definition of Hausdorff content see e.g. math.stonybrook.edu/~bishop/all2.pdf – Yuval Peres Oct 29 '19 at 17:14
• Thanks @YuvalPeres I in fact like your answer better because the problem is not just at a single point in your example, unlike the accepted answer. – Behnam Esmayli Oct 31 '19 at 3:34

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.