Show that $\mu^*(M\cup N)=\mu^*(M)+\mu^*(N)$ for outer measure $\mu^*$ Let $\mu:\mathscr{H}\to\mathbb{R}$ be a content where $\mathscr{H}\subset X$ a semiring and $\mu^*$ the outer measure defined by $\mu$.
I have already proven that that for any $A,B\subset X:\quad \mu^*(A\cup B)+\mu^*(A\cap B)\leq \mu^*(A)+\mu^*(B)$ and equality holds if $A\in\mathscr{A_{\mu^*}}$ or $B\in\mathscr{A_{\mu^*}}$ where $\mathscr{A_{\mu^*}}=\{A\subset X \;|\;\mu^*(A\cap B)+\mu^*(A^C\cap B)=\mu^*(B) \;\;\forall B\subset X\}$.
$M, N\subset X$ are given and $A,B\in\mathscr{A_{\mu^*}}$ such that $M\subset A,\; N\subset B$ and $\mu^*(A\cap B)=0$. I now want to prove $$\mu^*(M\cup N)=\mu^*(M)+\mu^*(N)$$
My attempt is to deduce that $\mu^*(M)=\mu^*(M\setminus N)$ from $\mu^*(B\cap M)+\mu^*(B^C \cap M)=\mu^*(M)$ and use this to show that $\mu^*(M)=\mu^*(M\setminus N)$. Then, because $M$ and $M\setminus N$ are disjoint $\Rightarrow \mu^*(M\cup N)=\mu^*((M\setminus N)\cup N)=\mu^*(M\setminus N)+\mu^*(N)=\mu^*(M)+\mu^*(N)$, but I know that the measure of a union of disjoint sets is not necessarily the sum of the measure of those sets, so my proof is probably wrong.
 A: Lemma 1: If $M, E \subseteq X$ and $\mu^*(E)=0$ then $\mu^*(M \cup E)= \mu^*(M)$.
Proof: Note  that
$$ \mu^*(M \cup E) \leq  \mu^*(M) + \mu^*(E)= \mu^*(M)  $$
On the other hand, since $M \subseteq M \cup E$, we have $ \mu^*(M) \leq  \mu^*(M \cup E)$. So, we have $\mu^*(M \cup E)= \mu^*(M)$.
$\square$
Lemma 2: If $N, E \subseteq X$ and $\mu^*(E)=0$ then $\mu^*(N) = \mu^*(N \setminus E)$.
Proof: Immediate from Lemma 1. In fact, note that $\mu^*(N \cap E)=0$ and $N = (N \setminus E) \cup (N \cap E)$ then apply Lemma 1.
$\square$
Now, since $M\subset A,\; N\subset B$ and $\mu^*(A\cap B)=0$, we have $\mu^*(A\cap N)=0$ and, since $A\in\mathscr{A_{\mu^*}}$, we have
\begin{align*}
\mu^*(M \cup N) &= \mu^*((M \cup N)\cap A) + \mu^*(M \cup N)\cap A^c)=\\
&= \mu^*(M \cup (N\cap A) ) + \mu^*( N \cap A^c) = \\
& = \mu^*(M \cup (N\cap A) ) + \mu^*( N \setminus (N \cap A))= \\
&=   \mu^*(M) + \mu^*( N )
\end{align*}
In the last equality, we used that $\mu^*(N\cap A)=0$ and Lemmas 1 and 2.
Remark: Note that we don't need assumption that $B\in\mathscr{A_{\mu^*}}$.
