Union of spheres is a null set It is proven that the sphere $S(0^{[k]};a)$ is a null set in $R^k$ for every $a \in R$.
Question: Let $C \subseteq R$ be a null set. Prove that $\bigcup_{a \in C} S(0^{[k]};a)$ is a null set in $R^k$.
My Try: I used the fact $C$ is a null set, in order to group the spheres into donut-shaped groups $\bigcup_{a \in D_n} S(0^{[k]};a)$, when $(B_n) \in \Omega$ is a sequence of boxes and $C \subseteq \bigcup_{B_n \in \Omega} B_n$ and $\Sigma_{B_n \in \Omega} vol(B_n) < \varepsilon$ (definition of null set).
I then tried to cover those donuts in boxes using the fact that $S(0^{[k]};a)$ is a null set, but failed to do so, while keeping the sum of their volume at some barrier.
 A: Let's take an $\varepsilon\gt0$. Note that the function $x\mapsto x^k$ is absolutely continuous as it is an integral of $(k-1)x^{k-1}$. Define the $\sigma_k$ to be the well-known constants (dependent on the dimension $k$) for the volume of a $k$-ball. Explicitly $\sigma_k=\frac{\pi^{k/2}}{\Gamma\left(1+\frac{k}{2}\right)}$ but this is unimportant here. There then exists a $\delta\gt0$ such that any finite partition $(x_i)_{i=1}^n$ whose mesh is less than $\delta$ will satisfy $\sum_{i=1}^{n-1}(x_{i+1}^k-x_i)\lt\frac{1}{2\sigma_k}\varepsilon$. Define $\mu^\ast$ to be the Lebesgue outer $k$-measure and $\mu$ to be the Lebesgue $k$-measure.
By the nullity of $C$, let's take a countable cover of $C$ by $1$-rectangles $\{R_m\}_{m\in\Bbb N}$ such that the sum of their volumes (lengths) is less than $\delta$. Let's term the endpoints of each $R_m$, $d_m$ and $c_m$, i.e. $R_m=[c_m,d_m]$.
The worst case on the size of $\bigcup_{a\in C}S_a$, where by "size" I mean terms of subset containment, is when $C$ completely fills these $R_m$. We have the bound: $$\bigcup_{a\in C}S_a\subseteq\bigcup_{m\in\Bbb N}\bigcup_{a\in R_m}S_a$$And $\mu\left(\bigcup_{a\in R_m}S_a\right)=\sigma_k(d_m^k-c_m^k)$. It follows by the aforementioned absolute continuity and $\sum_{m\in\Bbb N}(d_m-c_m)\lt\delta$ that $\sum_{m\in\Bbb N}(d_m^k-c_m^k)\le\frac{1}{2\sigma_k}\varepsilon\lt\frac{1}{\sigma_k}\varepsilon$ (since it holds for all finite partial sums).
Finally: $$0\le\mu^\ast\left(\bigcup_{a\in C}S_a\right)\le\sigma_k\sum_{m\in\Bbb N}(d_m^k-c_m^k)\lt\varepsilon$$
As $\varepsilon\gt0$ was chosen arbitrarily, we conclude $\bigcup_{a\in C}S_a$ is a null set and hence measurable, so may write $\mu$ rather than $\mu^\ast$.
