Uniform estimate for the boundary area of the union of closed unit balls 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 $K=\overline{\{v_i\}}$.
 A: 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).
