I have showed in a similar problem that if $A,B \subset \mathbb{R}^n$ are nonempty and disjoint then $\mu^{\ast}(A \cup B) = \mu^{\ast}(A) + \mu^{\ast}(B)$, where $\mu^{\ast}(\cdot)$ is the outer measure. I have come across a similar problem in a book:
$A$ is a measurable subset and $B$ any subset in $\mathbb{R}^n$. Still nonempty and disjoint. Prove that $\mu^{\ast}(A \cup B) = \mu(A) + \mu^{\ast}(B)$, where $\mu(\cdot)$ is the measure.
For the left, we can use Countable Subadditivity and the fact that $\mu(A) = \mu^{\ast}(A)$ (since $A$ given to be m'ble) to show $$\mu^{\ast}(A \cup B) \leq \mu^{\ast}(A) + \mu^{\ast}(B) = \mu(A) + \mu^{\ast}(B)$$
For the right, choose an $\epsilon > 0$ there exists countably many boxes $Q_k$ with $A \cup B \subset \cup Q_k$ such that $\sum_k \mu(Q_k) \leq \mu^{\ast}(A \cup B) + \epsilon$. Like the first problem, you can then create subsequences $Q_k = Q_{k_A} \cup Q_{k_B}$, where $Q_{k_A}$ intersects $A$ and likewise for $Q_{k_B}$. Then each covers $A$ and $B$, respectively. Thus, the result follows if we assume that $\mu(A) + \mu^{\ast}(B) = \mu^{\ast}(A) + \mu^{\ast}(B)$.
I don't see how assuming that $A$ is measurable changes anything?
If what you say you proved previously were true then yes, assuming $A$ measurable would add nothing. But in fact what you say you proved previously is false; what assuming $A$ measurable adds is it makes the statement correct.
Counterexample to the supposed previous result:
Some technicalities will be simpler if instead of $\Bbb R^n$ we consider subsets of the circle $$T=\{e^{it}:t\in\Bbb R\}.$$Define a subgroup $$G=\{e^{2\pi iq}:q\in\Bbb Q\}.$$Define an equivalence relation on $T$ by saying $\alpha\sim\beta$ if $\alpha/\beta\in G$. Let $A\subset T$ consist of one element chosen from each equivalence class.
For $r\in G$ let $$rA=\{r\alpha:\alpha\in G\}.$$Note that $$T=\bigcup_{r\in G}rA$$is a disjoint union.
Suppose what you said is true. By induction the same holds for the union of $n$ disjoint sets. Note as well that $\mu^*(rA)=\mu^*(A)$. So if $\mu^*(A)>0$ it follows from our assumed (false) finite additivity, plus monotonicity, that $\mu^*(T)=\infty$, contradiction; otoh if $\mu^*(A)=0$ then countable subadditvity shows that $\mu^*(T)=0$, another contradiction..