# Is arbitrary union of closed balls in R^{N} Lebesgue measurable?

Is an arbitrary union of non-trivial closed balls in the Euclidean space $\mathbb R^{N}$ Lebesgue measurable? If so, is it a Borel set?

@George

I still have two questions concerning your sketch of proof.

First, how can you guanrantee each of the open balls in the countable union has radius greater than or equal to 1?

Second, I don't know how to use convexity to prove $\mu (B') \leq (1+\epsilon)^{N}\mu(B)$

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Is a point a closed ball? –  Noah Stein Oct 26 '10 at 20:48
A point is a closed ball. Is every set measurable? –  Homology Oct 26 '10 at 20:49
We are considering non-trival closed balls. So a single point is not a closed ball. –  CKD Oct 26 '10 at 21:18
non-Borel measurability is easy enough, for N=2. Consider the union of unit balls centered at Sx{0} for a non-measurable subset S of the reals, and look at its intersection with Rx{1}. Lebesgue measurability looks a bit more interesting. –  George Lowther Oct 26 '10 at 21:25
@George - that set isn't a union of closed balls in R^2, though... –  David Roberts Oct 26 '10 at 22:21

No, in dimension N > 1, it does not have to be Borel measurable. E.g., in 2 dimensions, consider, a non Borel measurable subset of the reals S, and let A be the union of closed unit balls centered at points (x,0) for x ∈ S. The intersection of A with ℝ×{1} is the non-Borel set S×{1}, so A is not Borel.

On the other hand, for N = 1, any union of non-trivial closed intervals is Borel-measurable. If A is such a union and B is the union of the open interiors, then it can be seen that A is just the union of B with (at most countably many) endpoints of connected components of B.

Lebesgue measurability does hold, however. Faisal posted a link for this as I was typing my answer, but I think its still worth giving a brief sketch of the proof I was starting to type (Edit: added more detail, as requested).

1. Reduce the problem to that of balls with at least some positive radius r and within some bounded region. To do this, suppose that S is the set of closed balls and Sr denotes the balls of radius at least r and with center no further than r from the origin. Then, $$\cup S=\bigcup_{n=1}^\infty\left(\cup S_{1/n}\right).$$ As the measurable sets are closed under countable unions, it is enough to show that ∪Sr is Lebesgue measurable for each r > 0. So, we can assume that all balls are of radius at least r and are within some bounded distance of the origin.

2. Let $A$ be the union of the closed balls, and $B\subseteq A$ be the union of their interiors. This is open so, by second countability, is a union of countably many open balls of radius at least r. Also, $A$ lies between $B$ and its closure $\bar B$.

3. Show that the boundary $\bar B\setminus B$ of $B$ has zero measure. If we scale up the radius of each of the countable sequence of open balls used to obtain $B$ by a factor $1+\epsilon$ to get the new set $B^\prime$ then $\mu(B^\prime)\le(1+\epsilon)^N\mu(B)$. Showing this is the tricky part, but it does follow from convexity of the balls: If the balls have radius rk and centres xk, then consider the sets $$B_t=\bigcup_{k=1}^\infty B(r_k,tx_k)$$ for real t, so that B1 = B. The function t → μ(Bt) is symmetric and unimodal (see this answer), so has a unique local minimum at t = 0 and is increasing in |t|. Also, $B^\prime= (1+\epsilon)B_{1/(1+\epsilon)}$ giving, $$\mu(B^\prime)=(1+\epsilon)^{N}\mu(B_{1/(1+\epsilon)})\le(1+\epsilon)^{N}\mu(B)$$ as claimed. As $\bar B\subseteq B^\prime$ we get $\mu(\bar B\setminus B)\le((1+\epsilon)^N-1)\mu(B)$ which can be made as small as we like by choosing ε small.

Edit: having seen Faisal's explanation, the proof I outline here is completely different to his. The result Faisal quotes is a bit more general as it applies to convex sets with nonempty interior, rather than just balls. However, the proof given above also works for symmetric convex sets with nonempty interiors. As every convex set with nonempty interior is a union of (translates of) symmetric ones, this implies the same result

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Can your argument handle an arbitrary union of circles? –  Joel David Hamkins Oct 26 '10 at 23:44
Circles? You mean closed 2-balls? It will handle arbitrary unions of convex sets with nonempty interior, which is precisely the result Faisal just quoted. –  George Lowther Oct 26 '10 at 23:51
Actually, were you asking about my argument above or below the line? –  George Lowther Oct 26 '10 at 23:52
I meant the boundary. In two dimensions, circles (without the interior). –  Joel David Hamkins Oct 26 '10 at 23:58
If S is a non-Lebesgue measurable subset of the positive reals, then the union set of circles centered at the origin with radius in S is not Lebesgue measurable. Neither is the union of unit circles centered at (x,0) for x in S (look at its intersection with Rx{y} for |y| < 1). –  George Lowther Oct 27 '10 at 0:15

A google search reveals that an arbitrary union of (nondegenerate) convex sets is Lebesgue measurable: see

Balcerzak and Kharazishvili. On uncountable unions and intersections of measurable sets. Georgian Math. J. 6 (1999), no. 3, 201–212.

Edit: As requested, here's a summary of the proof.

The authors prove that an arbitrary union of (closed, nondegenerate) $n$-simplices $\{ S_t \}_{t \in T}$ in $\mathbb{R}^n$ is Lebesgue measurable. First a preliminary definition:

A bounded set $X \subset \mathbb{R}^n$ is said to be $\alpha$-regular, for $\alpha$ a positive real number, if $\lambda(X) \geq \alpha \lambda(V(X))$, where $V(X)$ is a closed ball with minimal diameter for which the inclusion $X \subset V(X)$ holds.

Observe that an $n$-simplex is $\alpha$-regular for some $\alpha \in (0,1]$. Thus

$$\bigcup_{t \in T} S_t = \bigcup_{m=1}^\infty \ \bigcup \{ S_t \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular} \}.$$

So in order to show that $\cup_t S_t$ is Lebesgue measurable, it suffices to show that $X_m = \cup \{ S_t \colon S_t \text{ is } \frac{1}{m}\text{-regular} \}$ is Lebesgue measurable for all $m \in \mathbb{Z}_{>0}$. Towards this end, given $x \in S_t$ and $c \in (0,1)$, let $S_t(x,c)$ denote the image of $S_t$ under the map $y \mapsto x + c(y-x)$. Then

$$\mathcal{F}_m = \{ S_t(x,c) \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular}, x \in S_t, c \in (0,1) \}$$

is a Vitali covering of $X_m$. The Vitali covering theorem now takes us home: the countable subcollection $\mathcal{F}_m^\ast \subset \mathcal{F}_m$ produced by the theorem has a Lebesgue measurable union $\cup \mathcal{F}_m^\ast$, which also satisfies

$$\bigcup \mathcal{F}_m^\ast \subset X_m \quad\text{and}\quad \lambda(X_m \backslash \bigcup \mathcal{F}_m^\ast) = 0.$$

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It would be kind of you to summarize the argument. –  Joel David Hamkins Oct 26 '10 at 23:04
Actually, convexity is hardly required here. I think it is enough that, for each t, there is a positive c such that $\mu(S_t\cap B_r(x))\ge cr^N$ for all $x\in S_t$ and $r > 0$. The construction given for the Vitali cover shows this is true for convex sets of nonempty interior (and it is clear for balls). –  George Lowther Oct 27 '10 at 0:58

EDIT: While I say separability may be important in the answer below, I think having a countable basis (second-countable?) is even more important for the answer. END EDIT.

Since R^n is separable ( has a countable dense subset ), the arbitrary union may be replaceable by a countable union. If so, then then set is Borel. I assume that each closed ball has nontrivial radius and can be approximated by a union from a countable collection of closed balls ( or their complements by a countable union of open balls ).

Note that separability is key here, as is the nontriviality of every closed ball in the collection.

You can't approximate an arbitrary closed ball with a union from a countable collection of closed balls. For instance, suppose you were tempted to take as your collection $\mathcal{C}$ the closed balls of rational radius centered at a point with rational coordinates. Let $B$ be an arbitrary closed ball. You would need $B = \bigcup _{B' \in \mathcal{C}, B' \subset B} B'$. It's easy to see that for $x$ in the boundary of $B$, $x$ lies in that union iff $x$ has rational coordinates, but of course most of the points on the boundary won't. –  Amit Kumar Gupta Oct 26 '10 at 23:09