Show that the union over a collection of compact cubes in $\mathbb{R}^n$ is Lebesgue measurable

Let $\mathcal{K}$ be a (not necessarily countable) collection of compact cubes in $\mathbb{R}^n$. Show that $\cup\{K:K\in \mathcal{K}\}$ is a Lebesgue set (Measurable with respect to the Lebesgue measure).

If the union is countable the argument is quite simple: each $K\in \mathcal{K}$ is closed and thus a Borel set (i.e. belongs to the Borel $\sigma$-algebra), and since $\sigma$-algebras are closed under countable unions, $S=\cup\{K:K\in \mathcal{K}\}$ is a Borel set. Since every Borel set is Lebesgue measurable, we are done.

Thus, WLOG we can assume that $\mathcal{K}$ is an uncountable family. The hint given is to use the Vitali Covering theorem. I am not quite sure how to work this out yet but have a few ideas. My main idea is to use the fact that every Lebesgue measurable set can be written as the union of a Borel set and a set of Lebesgue measure zero, and every set that can be written in such a form is Lebesgue measurable.

We can take $V=\{C\subset \mathbb{R}^n: C$ is a compact cube s.t. $C\subset K$ some $K\in \mathcal{K}\}$. This gives a Vitali covering of $S$, and so by the Vitali covering theorem there exists a sequence of pairwise disjoint cubes $C_1,C_2,...\in V$ such that $\cup_j C_j$ contains $\lambda_n$-almost every point of $\mathcal{K}$.

Then $\mathcal{K}=\mathcal{K}\cap \left( \cup_j C_j\right)\cup \left(\mathcal{K}-\cup_j C_j \right)=\cup_j C_j \cup \left(\mathcal{K}-\cup_j C_j\right)$ (The last equality holding by how $V$ is defined). Since $\cup_j C_j$ is a Borel set (being a countable union of closed sets), and $\mathcal{K}-\cup_j C_j$ is a Lebesgue null set, $\mathcal{K}$ is a Lebesgue set.

• What do you mean by a cube? Is a single point set a cube ($\{x\} = [x_1,x_1]\times \cdots \times [x_n,x_n]$)? – copper.hat Nov 19 '13 at 6:24
• You can assume the cubes are non-degenerate. So a cube in $\mathbb{R}^n$ is a set of the form $C=I_1 \times \cdots \times I_n$ where each interval $I_i$ has the same length $>0$ – CWsl2 Nov 19 '13 at 6:27

Any union $U$ of non-degenerate cubes can be written as a union of countably many cubes, by taking all cubes in $\mathbb{R}^n$ that are a subset of $U$, whose central points have rational coordinates, and whose edges have rational length.