Existence of a Minimal Cover

Question

I'm well aware that for the sequence $x_n=\frac{1}{n}$, $\text{inf }x_n=0$ but $0 \notin (x_n)$. This made me think about something similar but when we are no longer thinking about existence of a number in a sequence but something a bit different. Consider the definition of exterior measure for a set $E \subset\mathbb{R}^d$,

$$m_*(E)=\text{inf }\sum_{j=1}^\infty |Q_j|$$

where the infimum is taken over all countable coverings $E\subset \bigcup_{j=1}^\infty Q_j$ by closed cubes. Does this necessarily imply there is a cover $\{Q_\alpha^*\}_\alpha$ such that

$$\text{inf }\sum_{j=1}^\infty \left|Q_j\right|=\sum_{j=1}^\infty \left|Q_j^*\right|$$

I suspect this is not always the case. I'm wondering about the following:

$1.$ Can one find examples of a set $E\subset \mathbb{R}^d$ (nonfinite) such there is such a "minimal" cover $\{Q_\alpha^*\}$?

$2$. What conditions - if any - on $E$ force there to always be such a "minimal" cover?

$3$. What about when one removes the restriction of $E \subset \mathbb{R}^d$? Are there cases where one clearly can/can't find such "minimal" covers for general metric spaces?

Answer 1

If $X$ is an uncountable null set then it cannot have a minimal cover. A fat Cantor set cannot have a minimal cover either since every non-trivial closed cube will leak out of it.

If $X \subseteq \mathbb{R}^d$ has finite outer measure and $\{Q^{\star}_n : n \geq 1\}$ is a minimal cover of $X$, then $X$ is a full outer measure subset of $Y$ where $Y = \bigcup \{Q^{\star}_n : n \geq 1\}$. Isn't this a characterization of such sets - namely sets with full outer measure in a countable union of cubes? Or sets which are a union of countably many full outer measure subsets of cubes?

On real line this just amounts to being a full outer measure subset of a union of an open set and a countable set.

Answer 2

1. If $E$ is the countable union of disjoint cubes of course you have that the minimum is reached. This is also true if the cubes have intersection of zero measure (instead of being really disjoint) so that they can touch on the sides. This remains also true if you remove from $E$ a set of measure zero.

2. sufficient conditions are explained in 1. I think that they might also be necessary... but I haven't investigated the details.

3. what are cubes in a general metric space? You should use balls, or refer to the diameter of covering sets, as is done with the definition of Hausdorff measure...

Answer 3

In particular, there is a minimal cover for any open set. This follows since any open set can be partitioned into a countable union of half-open cubes of the form $\bar{x}/2^i + [0,1/2^i)^k$ where $\bar{x} \in \mathbb{Z}^k$. The closures of these cubes would give the minimal covering.