Every outer measure on a measure space can be represented as an infimum of a certain set

Question

I know that if we are given a collection $\mathcal{E}$ of subsets of the power set of a set $X$ which includes $\emptyset$ and $X$ and a function $\rho: \mathcal{E}\to [0,\infty]$ such that $\rho(\emptyset)=0$ that $$ \mu^*(A) = \inf \{\sum_{j=1}^\infty\rho(E_j) : E_j \in \mathcal{E}\text{ }and\text{ }A\subset \bigcup_{j=1}^\infty E_j\} $$ is an outer measure.

Is the converse true? Say, given an outer measure $\mu^*$, there exists $\mathcal{E}\subset \mathcal{P}(X)$ and a function $\rho$ where we can write $\mu^*$ as above? In this direction, I know that the collection of $\mu^*$ measurable sets form a $\sigma$-algebra via Carathéodory's Theorem, so I was thinking of letting those sets be my $\mathcal{E}$, but I'm not sure where to go from there.

Answer 1

Let $\rho=\mu^*$. (Which, just to be explicit, entails that $\mathcal E =\mathcal P(X)$.)