If an outer measure adds over two $\mu^*$-finite sets, does it add over subsets of those sets? This question occurred to me when I was reading the definition of a metric outer measure, but I couldn't think of an answer.
Let $X$ be a set and let $\mu^* : \mathcal{P}(X) \to [0,\infty]$ be an outer measure on $X$. Suppose that $E,F \subset X$ are sets of finite outer measure (Edited, in light of discussion in comments)  such that $\mu^*(E \cup F) = \mu^*(E) + \mu^*(F)$. If $E' \subset E$ and $F' \subset F$, does it follow that $\mu^*(E' \cup F') = \mu^*(E') + \mu^*(F')$? What if I also take $E$ and $F$ to be disjoint?
 A: I don't know the answer in the case of a general outer measure. But
most reasonable outer measures arise as in the Caratheodory construction
(cf. https://en.wikipedia.org/wiki/Pre-measure and https://en.wikipedia.org/wiki/Carath\%C3\%A9odory\%27s_criterion),
i.e. we have a (semi)ring $R$ of subsets of $X$ and a premeasure
$\mu_{0}:R\to\left[0,\infty\right]$ such that the outer measure $\mu^{\ast}$
is given by
$$
\mu^{\ast}\left(E\right)=\inf\left\{ \sum_{n}\mu\left(A_{n}\right)\,\middle|\, A_{n}\in R\text{ with }E\subset\bigcup_{n}A_{n}\right\} .
$$
The Caratheodory theorem then states that each of the sets in $R$
is $\mu^{\ast}$-measurable. Thus, using the definition of $\mu^{\ast}$,
it is not hard to see that for any $E\subset X$ with $\mu^{\ast}\left(E\right)<\infty$,
there is a $\mu^{\ast}$-measurable set $E^{\ast}\subset X$ with
$E\subset E^{\ast}$ and $\mu^{\ast}\left(E\right)=\mu^{\ast}\left(E^{\ast}\right)$.
In fact, we can take $E^{\ast}$ to be a countable intersection of
countable unions of sets in $R$.
Now, we have 
\begin{eqnarray*}
\mu^{\ast}\left(E\right)+\mu^{\ast}\left(F\right) & \overset{\text{assumption}}{=} & \mu^{\ast}\left(E\cup F\right)\\
 & \overset{E\cup F\subset E^{\ast}\cup F^{\ast}}{\leq} & \mu^{\ast}\left(E^{\ast}\cup F^{\ast}\right)\\
 & \leq & \mu^{\ast}\left(E^{\ast}\right)+\mu^{\ast}\left(F^{\ast}\right)\\
 & = & \mu^{\ast}\left(E\right)+\mu^{\ast}\left(F\right).
\end{eqnarray*}
Since the first and last term coincide (and are finite!), we must
have equality in every step. Hence, $\mu^{\ast}\left(E^{\ast}\right)+\mu^{\ast}\left(F^{\ast}\right)=\mu^{\ast}\left(E^{\ast}\cup F^{\ast}\right)$.
Now, $E'\subset E\subset E^{\ast}$ and $F'\subset F\subset F^{\ast}$,
so that we can assume $E,F$ to be measurable to begin with. Because
of
$$
\mu^{\ast}\left(E\right)+\mu^{\ast}\left(F\right)=\mu^{\ast}\left(E\cup F\right)=\mu^{\ast}\left(E\right)+\mu^{\ast}\left(F\right)-\mu^{\ast}\left(E\cap F\right),
$$
we thus see that $\mu^{\ast}\left(E\cap F\right)=0$ (note that the
above only holds since $E,F$ are measurable of finite measure). Thus,
by changing $E,E'$ and $F,F'$ by null-sets, we can assume that $E,F$
are disjoint.
Now, by definition of $\mu^{\ast}$measurable sets, we get
\begin{eqnarray*}
\mu^{\ast}\left(E'\cup F'\right) & = & \mu^{\ast}\left(\left[E'\cup F'\right]\cap E\right)+\mu^{\ast}\left(\left[E'\cup F'\right]\cap E^{c}\right)\\
 & = & \mu^{\ast}\left(E'\right)+\mu^{\ast}\left(F'\right),
\end{eqnarray*}
where the last step used the disjointness of $E,F$ which implies
$F'\subset F\subset E^{c}$ and $E'\subset E\subset F^{c}$.
As noted above, I am (like the OP) still very interested and unsure
about the general case of outer measures which do not come from a
premeasure.
A: I think I have a counterexample for general outer measures. Let $X$ be a set with at least 3 elements $x,y,z$.  One can just use $X = \{x,y,z\}$. Define a finite outer measure $\mu^* : \mathcal{P}(X) \to [0,\infty]$ by
$$ \mu^*(E) = \begin{cases}
0 & \text{ if } E = \varnothing \\
1 & \text{ if } |E| =1 \text{ or } 2 \\
2 & \text{ if } |E| \geq  3 \\
\end{cases}.$$
Obviously $\mu^*$ is monotone, so the main thing is to see that $\mu^*(\bigcup A_n) \leq \sum \mu^* (A_n)$ holds. If the RHS is $\geq 2$, then this is trivial. If the RHS is $0$, then all the $A_n$ are empty, so the LHS is $0$ too. If the RHS is $1$, then exactly one of the $A_n$ has $\mu^*(A_n) =1$, and the rest are empty, so the LHS equals $1$ as well. 
Let $E = \{x,y\}$ and $F= \{z\}$. Observe that $\mu^*(E \cup F) = 2 = 1+1 = \mu^*(E) + \mu^*(F)$. On the other hand, taking $A=\{x\}$ and $B = \{z\}$, one has $\mu^*(A \cup B) = 1$ and $\mu^*(A) + \mu^*(B) = 1+1 =2$.

Added: Relating to PhoemueX's answer, note that this outer measure $\mu^*$ is pathological in the following sense: there exist sets $E \subset X$ such that
$$ \mu^*(E) \neq \inf \{ \mu^*(F) : F \supseteq E, F \text{ is } \mu^*\text{-measurable} \}.$$
