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Let $M$ be a smooth Riemannian manifold with riemannian measure $\mu$. I don't suppose that $M$ is complete. Can we find a finite or countable disjoint collection of open (or closed) and relatively compact geodesic balls$(B_n)_{n\in\mathbb{N}}$ such that :

$$\mu\left(M\setminus\bigsqcup_{n\in\mathbb{N}}B_n\right) = 0$$ ? We might assume that $M$ has a bounded curvature. The $M$ I'm interested in is the intersection of a (complete) submanifold of $\mathbb{R}^d$ with a ball. The compactness restriction ensures that each ball is a "real" ball, and hasn't been cropped by the "edge" of $M$.

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  • $\begingroup$ It seems to me we should be able to actually cover $M$ by a countable collection of closed (compact) balls. $\endgroup$ – Kevin Casto Apr 4 at 21:14
  • $\begingroup$ I forgot the disjoint hypothesis in my question.. $\endgroup$ – Pii_jhi Apr 5 at 8:16
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    $\begingroup$ In the case of $E^n$'s, such packings do exist. For instance, Apollonian packing does the job for the Euclidean plane. Not sure what happens if one requires zero Hausdorff dimension of the residual set. $\endgroup$ – Moishe Kohan Apr 5 at 8:49
  • $\begingroup$ The residual set will contain the boundary of the ball (if we consider open balls), so it will be of Hausdorff dimension $d-1$ at least and I think it has to be strictly greater than $d-1$. In the case of Apollonian packing, which feels close to an"optimal" packing, has Hausdorff dimension $\simeq 1.3$.. $\endgroup$ – Pii_jhi Apr 5 at 10:26
  • $\begingroup$ @Pii_jhi: Of course, I meant closed balls, as in your question. In all packing constructions I know, the residual set has positive H.D. $\endgroup$ – Moishe Kohan Apr 7 at 4:45
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I believe that this is true. If you look at Lemma 1.10. Of Introduction to Smooth Manifolds By John Lee,

Lemma 1.10. Every topological manifold has a countable basis of precompact coordinate balls.

If you follow the proof using an open cover of $M$ by Riemannian normal coordinate charts, you should end up with a countable basis of precompact geodesic balls of $M$ so that the union of all such balls is equal to $M$. Bounded curvature and completeness are not necessary for the proof.

I'm not sure what you mean by the last two sentences, but the fact that $M$ is topologically embedded into $\mathbb{R}^d$ should imply that the intersection of $M$ with a ball in $\mathbb{R}^d$ is an open set in $M$.

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  • $\begingroup$ Thank you for your answer. In fact I forgot the disjointness hypothesis in my question, which makes it less trivial. I believe it is true on R^n and belong to the family of Vitali covering theorem. $\endgroup$ – Pii_jhi Apr 5 at 8:18

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