Approximating a bounded measurable set by closed cubes. Suppose $\mu$ is a regular borel measure on ${\mathbb{R}}^d$. Let $E$ be a bounded measurable set. Does there exist a finite union of closed cubes ${\cup}C_j$ such that $\mu(({\cup}C_j)\Delta E)\leq\epsilon$? Using the regularity condition, I am trying approximate open sets by closed cubes but I am unable to progress.
 A: In terms of approximating open sets by closed cubes: 
Suppose $\mathcal{U}$ is a bounded open set. I claim that there is a countable disjoint union of cubes (open or closed, since the boundary of a closed cube is a zero set) whose measure differs from $\mathcal{U}$ by a zero set. In other words, $\mu (\mathcal{U} \Delta (\sqcup \mathcal{C}_i)) = 0$. 
The way we proved this in class was extremely pictorial, so it helps to draw a picture on graph paper:  Superimpose a unit grid onto your open set, $\mathcal{U}$. Keep all the interiors of the unit cubes contained in $\mathcal{U}$, and throw away the rest. Of the rejected cubes, refine your grid by halving each cube horizontally and vertically; in other words, you consider a dyadic cube of side length $\frac{1}{2}$. Again, keep the interiors of cubes contained in $\mathcal{U}$ and reject the rest. Then refine your grid again, halving each of the rejected cubes both vertically and horizontally to obtain dyadic cubes of side length $\frac{1}{4}$, and accept the interiors of cubes contained in $\mathcal{U}$. Repeat the process until you cover $\mathcal{U}$ completely. The boundaries of the cubes clearly form a zero set, so the countable union of them is also a zero set. 
What I am not entirely sure about is whether every bounded measurable set can be approximated by closed cubes. You might have to figure out the details of this part yourself since I'm not very familiar with Borel measurability (I learned measure theory using Lebesgue measurability). I have a feeling you can apply the regularity condition here; namely, for every $E$, $\exists$ $\mathcal{B} \subset E$ st $\mu \mathcal{B} = \mu E$, and $\mathcal{B} = G_{\delta\sigma\delta ... \delta\sigma\delta...}$, so $\mathcal{B}$ is generated by open sets. 
