I want to construct the following example to use as a counterexample with respect to something involving outer measure.
Let $\mathcal{E} \subset \mathcal{P}(X)$. Let $\rho$ be defined as follows:
\begin{align} \rho(E) = \begin{cases} 0 & E = \emptyset \\ 1 & E = X \\ \frac{3}{4} & E \subset X. \end{cases} \end{align}
Therefore, it is the case that, $\mu^{*},$ the outer measure induced by $\rho$, defined as:
\begin{align} \mu^{*}(A) = \inf\left\{\sum_{1}^{\infty} \mu(E_{j}): \; E_{j} \in \mathcal{E}, \; A \subset \bigcup_{1}^{\infty}E_{j} \right\} \end{align}
is given by \begin{align} \mu^{*}(E) = \begin{cases} 0 & E = \emptyset \\ 1 & E = X \\ \frac{3}{4} & E \subset X. \end{cases} \end{align}
So, it seems that all three axioms of the outer measure are satisfied and $\mu^{*}$ is an outer measure.
However, this would also imply by Caratheodory's theorem that for two disjoint subsets of the interval,
\begin{align} A &= [1.5,2] \\ E &= [1.5,1.7] \\ F &= [1, 1.2]. \end{align} Then \begin{align} .75 = \mu^{*}(E \cup F) < 1.5 = .75 + .75 = \mu^{*}(E) + \mu^{*}(F) \end{align}
However, $E \cap F = \emptyset$ and thus, by countability of measures, since $E, \; F \in \mathcal{M}$ where $(\mathcal{M},\mu)$ is the measure space on the sigma algebra generated by the set of outer measurable sets (via Caratheodory), we should have that $\mu = \mu^{*}|_{\mathcal{M}}$ but clearly don't have \begin{align} \mu^{*}(E \cup F) = \mu^{*}(E) + \mu^{*}(F). \end{align}