How can I prove this statement?
Let $X,Y \subseteq \mathbb{R}$ such that $$\overline{X}\cap Y=X \cap \overline{Y}=\emptyset$$ Prove that $$\mu^{*}(X\cup Y)=\mu^{*}(X)+\mu^{*}(Y)$$
Note: Here $\mu^{*}$ is the outer-measure and $\overline{C}$ is the closure of the set $C$.
My attempt: I know by the sub-addtive property in the outer-measure that $$\mu^{*}(X \cup Y)\leq \mu^{*}(X)+\mu^{*}(Y)$$ Now, by the second inequality, I need to prove that $$\mu^{*}(X\cup Y)\geq \mu^{*}(X)+\mu^{*}(Y)$$ But, I don't know how can I prove that inequality. I think the prove is similar for Carathéodory theorem if I can with the hypothesis to prove that $d(X,Y):=\inf\{|x-y|: x \in X, y\in Y\}>0$, so by Carathéory theorem the prove is immediate, but I don't know how can I approach that.
I think the following proof is rather inelegant, perhaps someone else can come up with a nicer way:
Let $U = \{(a_i, b_i) : i \in \mathbb{N} \}$ be any basic open cover for $X \cup Y$. The idea is that we want to find a tighter subcovers $U_x$ and $U_y$ of $X$ and $Y$ individually. To that end, let $\tilde{U_x} = \{(a, b) : (a, b) \in U, (a, b) \cap X \not = \emptyset\}$ and $\tilde{U_y}$ defined analogously. Suppose $(a, b) \in \tilde{U_x} \cap \tilde{U_y}$. Thus $(a, b) \cap X, (a, b) \cap Y \not = \emptyset$. Since $(a, b) \cap X \subseteq \mathbb{R}\setminus \overline{Y}$, and the latter is open, that means we may cover $(a, b) \cap X$ in countably many open intervals $\tilde{I_n}$, for $n \in \mathbb{R}$ such that each $\tilde{I_n}$ is disjoint from $\overline{Y}$ and thus $Y$. Let $I_n = \tilde{I_n} \cap (a, b)$. This is still an interval and the $I_n$ still cover $X \cap (a, b)$. Moreover, $\Sigma_{n \in \mathbb{N}} \mu^* (I_n) \leq \mu (a, b)$ (for a variety of reasons, but if nothing else, each $I_n$ is a lebesgue measurable subset of $(a, b)$). Now replace the entry $(a, b)$ in $\tilde{U_x}$ with $\{I_n\}_{n \in \mathbb{N}}$. Do this for $\tilde{U_Y}$ as well, and do this for every $(a, b) \in \tilde{U_X} \cap \tilde{U_Y}$. You will end up with a resultant $U_X$ and $U_Y$ which are basic open covers of $X$ and $Y$, and moreover, $\Sigma_{(a, b) \in U_X} (b -a) + \Sigma_{(a, b) \in U_Y}(b-a) \leq \Sigma_{(a, b) \in U} (b-a)$ . The only thing to worry about here is the sets $(a, b)$ which were in both $\tilde{U_X}$ and $\tilde{U_Y}$, but we replaced them with intervals $I^{X}_n$ (in the $X$ case) and intervals $I^{Y}_n$ (in the $Y$ case), which are all subsets of $(a, b)$ and thus add up to having less measure than $(a,b)$.
So then $\mu^{*}(X) + \mu^{*}{Y} \leq \Sigma_{(a, b) \in U_X} (b -a) + \Sigma_{(a, b) \in U_Y}(b-a) \leq \Sigma_{(a, b) \in U} (b-a)$.
Which menas that $\mu^{*}(X) + \mu^{*}(Y) \leq \Sigma_{(a, b) \in U}(b - a) $ for an arbitrary basic open cover $U$ of $X \cup Y$, which means tht $\mu^{*}(X) + \mu^{*}(Y) \leq \mu(X \cup Y)$.
I skipped quite a few details in order to keep it from becoming too long. Hopefully the idea remains clear.