Proving $ \lambda^{*}(A \cup B) = \lambda^{*}(A)+\lambda^{*}(B)$

Question

Let $\lambda$ denote Lebesgue measure and $\lambda^{*}$ denote outer measure. Prove that if A and B are subsets of $R^n$ which are separated in the sense that there exists a measurable set C such that $A\subset C$ and $\lambda(B \cap C) = 0$; then $$ \lambda^{*}(A \cup B) = \lambda^{*}(A)+\lambda^{*}(B)$$

I tried the Caratheodory's criterion that $D$ is measurable iff for any $E\subset R^n$, $\lambda^{*}(E)=\lambda^{*}(E\cap D) + \lambda^{*}(E\cap D^c)$. So if we set $D=C,E=B$ we get $\lambda^{*}(B)=\lambda^{*}(B\cap C^c)$. I also tried to established $\lambda^{*}(A\cap B)=0$ but haven't succeeded.

Any help is appreciated!

Answer 1

It is always true that $\lambda^*(A\cup B) \le \lambda^* A + \lambda^* B$, so we only need establish the other inclusion.

Take $D=C$ and $E=A \cup B$ in the formula $\lambda^{*}E=\lambda^{*}(E\cap D) + \lambda^{*}(E\cap D^c)$ to get $\lambda^{*}(A \cup B)=\lambda^{*}((A \cup B) \cap C) + \lambda^{*}((A \cup B) \cap C^c)$.

Now take $D=B, E=C$ in the formula to get $\lambda^* B = \lambda^* ( B \cap C) + \lambda^*(B \cap C^c) = \lambda^*(B \cap C^c)$ (since $\lambda^* ( B \cap C) = 0$, by assumption). Then we have $\lambda^{*}((A \cup B) \cap C^c) \ge \lambda^{*}(B \cap C^c) = \lambda^* B$.

We have $\lambda^{*}((A \cup B) \cap C) \ge \lambda^{*}(A \cap C) = \lambda^* A$, and gathering these together gives $\lambda^{*}(A \cup B) \ge \lambda^*A + \lambda^*B$.