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Let $A$ be the union of closed unit balls around countable many points $\{v_i\}\in\mathbb R^n$. If $\{v_i\}$ are chosen from a fixed unit ball, is it true that the boundary of $A$ has area $< c(n)$? Here $c(n)$ is a uniform constant depending only on $n$ and the "area" means the $(n-1)$-dimensional Hausdorff measure.

Motivation: this is a stronger form of To show that the set point distant by 1 of a compact set has Lebesgue measure $0$. From the answer by Dave L. Renfro, it follow that the area of $\partial A$ is finite (by a theorem of Erdős), since being on the boundary of $A$ implies having distance $1$ from the compact set $K=\overline{\{v_i\}}$.

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Yes, the surface area can be bounded by a constant only depending on $n$. To start with, I'll use the Minkowski content rather than the Hausdorff measure. Then, we can give a simple bound. For a set $S\subseteq\mathbb{R}^n$ and $r > 0$, I will use $S_r$ to denote the set of points $x\in\mathbb{R}^n$ within a distance $r$ of $S$. i.e., such that $\lVert x-y\rVert < r$ for some $y\in S$. The Minkowski content of an (n-1)-dimensional surface $S$ is (using $\mu$ for the Lebesgue measure), $$ M^*(S)=\limsup_{r\to0}\frac1{2r}\mu(S_r). $$ For reasonably regular surfaces, this agrees with other measures of surface area. If $S$ is the set of points $\{v_i\}$ in the question, then $A=S_1$. I will show that the boundary measure is bounded, \begin{align} M^*(\partial S_r)\le\frac nr\mu(S_r).&&{\rm(1)} \end{align} Note, in the case where $S$ is a point, so $S_r$ is a ball of radius $r$, this is an equality and relates the surface area of a ball to its volume. If $S$ is contained in a unit ball, then $A=S_1$ is contained in a ball of radius 2. Hence, $$ M^*(\partial A)\le n2^n\mu(B_1) $$ where $B_1$ is a unit ball. Next, the Hausdorff (n-1)-measure of a set in $\mathbb{R}^n$ is bounded by a multiple ($c_n$, say) of the Minkowski content. See Geometry of Sets and Measures in Euclidean Spaces by Pertti Mattila, page 79 (available on google books). Many thanks to Live Forever for pointing this out in the comment below. So, $$ \mathcal{H}^{n-1}(\partial A)\le c_n M^*(\partial A)\le c_nn2^n\mu(B_1). $$

Now, to prove (1). It is known that if we have a set of unit balls in Euclidean space and continuously move them around such that the distance between the centres of each pair increases, then the volume of their union increases. This is intuitively plausible, as we would expect the volume of their overlap to decrease. The Kneser-Poulsen conjecture states this, and is proven in the case of continuous motions. Note that for $r > 0$, $(rS)_1$ is the union of unit balls about points in $rS$, so $\mu( (rS)_1 )$ is increasing in $r$. Also, $(\partial S_r)_\epsilon\subseteq S_{r+\epsilon}\setminus S_{r-\epsilon}$. So,

\begin{align} \mu((\partial S_r)_\epsilon)&\le\mu(S_{r+\epsilon})-\mu(S_{r-\epsilon})\\ &=(r+\epsilon)^n\mu(((r+\epsilon)^{-1}S)_1)-(r-\epsilon)^n\mu(((r-\epsilon)^{-1}S)_1)\\ &\le(r+\epsilon)^n\mu((r^{-1}S)_1)-(r-\epsilon)^n\mu((r^{-1}S)_1)\\ &=(1+\epsilon/r)^n\mu(S_r)-(1-\epsilon/r)^n\mu(S_r)\\ &=\frac{2n\epsilon}{r}\mu(S_r)+O(\epsilon^3) \end{align} Dividing through by $2\epsilon$ and taking the limit $\epsilon\to0$ gives (1).

George Lowther
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  • Two remarks: (A) You have the complete proof, since Minkowski content dominates Hausdorff measure (it corresponds to covering by balls of equal radius, while Hausdorff measure has no such restriction). A reference is p.79 in Mattila's Geometry of Sets and Measures. (B) You don't need Kneser-Poulsen; the Brunn-Minkowski theorem states that $f(r) = \mu(S_r)^{1/n}$ is a concave function, hence $f'(r)\le f(r)/r$, which gives your estimate (1). –  Feb 05 '16 at 01:06
  • @Live Forever: Thanks! Just checking those points and I'll update the answer – George Lowther Feb 05 '16 at 01:14
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    @Live Forever: The problem I am having trying to apply Brunn-Minkowski, is that we want to show that $\mu(S_{tr+(1-t)s})^{1/n}\ge t\mu(S_r)^{1/n}+(1-t)\mu(S_s)^{1/n}$. To do this with B-M, we want to show that $S_{tr+(1-t)s}\supseteq tS_r+(1-t)S_s$. However, the latter inequality is not true unless $S$ is convex. – George Lowther Feb 05 '16 at 02:20
  • I see your point; I now don't remember why I thought B-M implies that $f$ is concave. But on the other hand, I don't see counterexamples either. I'll ask this as a separate question. –  Feb 05 '16 at 02:31
  • Posted here. Not coincidentally, the concavity claim also appears in this answer, which was me writing anonymously... –  Feb 05 '16 at 02:49
  • I can prove the required inequalities in this question by elementary means, without reference to Kneser-Poulsen, but the concavity you suggested seems rather stronger. – George Lowther Feb 05 '16 at 20:49