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^* 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: 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.

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.

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 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.

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 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 B₁ = 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.

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 all $x\in S$. The intersection of $A$ with $\mathbb{R}\times \{1\}$ is the non-Borel set $S \times \{1\}$, 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 $B$.

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 $S_r$ 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 $\cup S_r$ 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.

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 $r_k$ 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.

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 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.

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 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 B₁ = B. The function t → μ(B_t) 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.

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 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.

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 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 B₁ = 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.