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).

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 *S*_{r} 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 ∪*S*_{r} 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.

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 $\bar B$.

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 *r*_{k} and centres *x*_{k}, then consider the sets
$$
B_t=\bigcup_{k=1}^\infty B(r_k,tx_k)
$$
for real *t*, so that *B*_{1} = *B*. The function $t\mapsto\mu(B_t)$ is increasing in $t\ge0$^{*}. 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: in my initial response, I was thinking that this answer is enough to prove that $\mu(B_t)$ is increasing in $t$. However, as Mizar points out in the comments below, this is not clear. Actually, I don't think we can reduce it to that case. However, the result is still true, by the Kneser-Poulson conjecture. This states that if the centres of set of unit balls in Euclidean space are all moved apart, then the measure of their union increases. Although only a conjecture, it has been proved for continuous motions, which applies in our case. Also, expressing each ball of radius greater than some arbitrarily small $r > 0$ as a union of balls of radius $r$, then it still applies in our case for balls of non equal radii.

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