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. I'm interested in the case when $M$ 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 $M$.

  • $\begingroup$ It seems to me we should be able to actually cover $M$ by a countable collection of closed (compact) balls. $\endgroup$ Apr 4, 2020 at 21:14
  • $\begingroup$ I forgot the disjoint hypothesis in my question.. $\endgroup$
    – Pii_jhi
    Apr 5, 2020 at 8:16
  • 2
    $\begingroup$ 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. $\endgroup$ Apr 5, 2020 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, 2020 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. $\endgroup$ Apr 7, 2020 at 4:45

2 Answers 2


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 $M$.

  • $\begingroup$ 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, 2020 at 8:18

For any smooth Riemannian manifold $(M,g)$ there is a countable disjoint union of balls with complement of measure $0$.

Let $\mu$ be Riemannian measure and for each $p\in M$ let $B_p$ be a small precompact ball centered at $p$ such that $\mu(\partial B_p)=0$ (we can do this because $\{r>0;\mu(\partial B(p,r))>0\}$ is countable).

Claim 1: There is some $k>0$ such that for small $\delta>0$ and for any $q\in B_p$, $\frac{\mu(B(q,\delta))}{\mu(B(q,2\delta))}>k$.

Proof: Let $\varphi:U\to\mathbb{R}^n$ be a chart, where $U$ is a nhood of $\overline{B_p}$. Let $g'$ be the pullback from the metric of $\mathbb{R}^n$ in $U$. Then in some neighborhood $V$ of $\overline{B_p}$, the Riemannian distances $d,d'$ from $g,g'$ respectively are bilipschitz (so the volumes $\mu,\mu'$ are also related by a constant). So if $\delta_0$ is small enough that $d$-balls centered in $B_p$ of radius $\delta_0$ are contained in $V$, then for some big constant $\alpha$ we have that for any $q\in B_p$ and $\delta<\delta_0$, $$B'\left(q,\frac{\delta}{\alpha}\right)\subseteq B(q,\delta)\subseteq B(q,2\delta)\subseteq B'(q,\alpha\delta).$$ So, as $\frac{\mu(B'(q,\alpha\delta))}{\mu(B'(q,\frac{\delta}{\alpha}))}$ is uniformly bounded, we have proved the claim. $\square$

Claim 2: For any open subset $A$ of $B_p$ there is a finite set of disjoint balls $B_1,\dots,B_m$ contained in $A$ such that $\mu(\cup_{i=1}^m B_i)>\frac{1}{2k}\mu(A)$.

Proof: Let $\delta$ be so small that it satisfies the previous claim and such that, if $B:=\{x\in A;d(x,M\setminus A)>\delta\}$, then $\mu(B)>\frac{1}{2}\mu(A)$. Now consider a maximal $2\delta$-separated set $\{x_1,\dots,x_m\}$ in $B$, and let $B_i:=B(x_i,\delta)$. These balls are disjoint, and the balls $B(x_i,2\delta)$ cover $B$, so $\sum_i\mu(B_i)>\frac{1}{k}\sum_i\mu(B(x_i,2\delta))\geq\frac{1}{k}\mu(B)\geq\frac{1}{2k}\mu(A)$. $\square$

We can also ensure that the boundaries of the balls $B_i$ of claim $2$ have measure $0$: if not, note that for each $q\in M$, the set $\{r>0;\mu(\partial B(q,r))>0\}$ is countable, so we can reduce the radii of the balls just a little bit so that the sum of their volumes is still $>\frac{1}{2k}\mu(A)$.

Claim 3: We can cover any open set $X\subseteq B_p$ up to measure $0$ by a disjoint collection of balls contained in $X$.

Proof: Take $A=X$ in the previous claim, and find balls $B_{0,1},\dots,B_{0,n_0}$ with boundary of measure $0$ such that $\mu(\cup_{i=1}^m B_i)>\frac{1}{2k}\mu(X)$. Now let $X_1=X\setminus\cup_i\overline{B_{0,i}}$, so that $\mu(X_1)\leq(1-\frac{1}{2k})\mu(X)$. Applying the same to $X_1$ we can remove from it finitely many balls $B_{1,1},\dots,B_{1,n_1}$ to obtain some open $X_2$ with $\mu(X_2)\leq(1-\frac{1}{2k})\mu(X_1)$. Repeating this step to obtain spaces $X_n$ for each $n$, we get that the balls $\{B_{i,j}\}_{i\in\mathbb{N};j=1,\dots,n_i}$ are pairwise disjoint, and $\mu(X\setminus\bigcup_{i,j}B_{i,j})=\lim_{m\to\infty}\mu(X_m)\leq \lim_{m\to\infty}(1-\frac{1}{2k})^m\mu(X)=0$. $\square$

Claim 4: We can cover $M$ up to measure $0$ using a countable collection of disjoint compact balls.

Proof: Consider the collection of balls $\mathcal{B}:=\{B_p;p\in M\}$. As $M$ is second countable, we can find a countable subcover of $\mathcal{B}$, $(B_n)_{n\in\mathbb{N}}$. Moreover, for each $n$, we can cover $B_n\setminus\bigcup_{i=1}^{n-1}\overline{B_i}$ up to measure $0$ with a countable collection of disjoint compact balls. The union of these countable collections of balls covers $B_n$ up to measure $0$ for all $n$, thus it covers all $X$ up to measure $0$. $\square$


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