Existence of covers for outer measure $\mu^*$ In the proof of countable subaddivity of $\mu^*$ defined by $$\mu^*(E) = \inf\left\{\sum_{j=1}^\infty\rho(E_j): E_j \in \varepsilon, E \subset \cup_{j=1}^\infty E_j\right\}.$$
Let $$\mu^*(E_j) = \sum_{k=1}^\infty\rho(E_j^k)$$ where 
$$E_j \subset \cup_{k=1}^\infty E_j^k \subset \varepsilon, E_j \notin \varepsilon$$
Since $\mu^*(E_j)$ is the $inf$ over possible coverings of $E_j$,
$$\sum_{k'=1}^\infty\rho(E_j^{k'}) \le  \mu^*(E_j) + \frac{\epsilon}{2^j}$$
I understand why we can write the inequality (call it Ineq1), based on the property of inf in $\mathbb{R}$.
I also get that there always exists at least one cover, the whole set $X \in \Large\varepsilon$. What guarantees the existence of any other cover that is neither $\cup_{k=1}^\infty E_j^k$ nor $X$? My understanding is that the existence of such covers depends on the elementary family $\Large\varepsilon$ on which $\rho$ is defined, irrespective of the property of $inf$ in $\mathbb{R}$. 
What if $\Large\varepsilon$ is not rich enough, such that for some $j$ the possible covers of $E_j$ are only: $$\cup_{k=1}^\infty E_j^k \ or\ X$$ 
Then equality holds for the cover $\cup_{k=1}^\infty E_j^k$ and the only other cover is $X$ for $\epsilon > 0$.
If $$\frac{\epsilon}{2^j} \le \rho(X) - \mu^*(E_j) $$ then Ineq1 is not satisfied. If we just use $\cup_{k=1}^\infty E_j^k$ as cover, we get $\epsilon \ge 0$.
What I am missing? In short, what guarantees the existence of covers that make the inequality hold for every $\epsilon > 0$?
 A: The extension of the measure $\rho:\mathcal{E}\to [0,\infty]$ to the power set of $X$, with $A\subset X$, is defined by
$$ \mu^*(A)=\inf \left\{ \sum_{j=1}^\infty \rho\left(E_j\right):E_j\in\mathcal E \text{ and } A\subset \bigcup_{j=1}^\infty E_j \right\} $$
Fix an $A \subset X$.
First, notice that the construction of the outer measure $\mu^*$ does not guarantee that you can find a collection of sets $\left\{ E_j \right\}_{j=1}^\infty$ (each of which is in $\mathcal E$, and whose union contains $A$) such that sum of the measures of these $E_j$'s attains the value $\mu^* (A)$. (To see this, think about the infimum of the set $(0,1)$.)
Thus, for any collection $\left\{E_j\right\}_{j=1}^\infty\subset\mathcal E$ such that $A\subset \bigcup_{j=1}^\infty E_j$, we can only guarantee that
$$ \mu^* (A) \leq \sum_{j=1}^\infty \rho\left(E_j\right) $$
where equality is not necessarily attained.
However, by the definition of the infimum, we should always be able to find some collection of $E_j$'s whose total measure (or, more precisely, sum of measures) is arbitrarily close to $\mu^* (A)$. If not, then $\mu^* (A)$ cannot be the greatest lower bound of the set defined above.
To be more specific, for any $\epsilon>0$, we can always find some collection $\left\{E_j^\epsilon \right\}_{j=1}^\infty$ such that
$$ \sum_{j=1}^\infty \rho\left(E_j^\epsilon\right) - \mu^*(A) \leq \epsilon $$
which is all you need for Folland 1.10.
