Existence of a Minimal Cover 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? 

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.
