Let $\mu^*: P(X) \rightarrow [0, +\infty]$ be an outer measure and $E \subseteq X$such that $\mu^*(E)=0$. I want to show that $E$ is $\mu^*$measurable.
For this we have to show that: $$\mu^*(A)=\mu^*(A \cap E) + \mu^*(A\setminus E) \hspace{0.5 cm} (\forall A \subseteq X)$$ But I don't have any idea for doing that.
Do I have to work on different conditions separately? For example; 1. $E = X$, 2. $E = \emptyset$, 3. $E \subset X$ and $E \neq \emptyset$?
By definition $\mu^*$ is $\sigma$-subadditive, and is monotone (see Wikipedia's article for instance). Writing $A = (A\cap E) \cup (A\setminus E) $ and applying the sub-additivity of $\mu^*$ we get $$ \mu^*(A ) \leq \mu^*(A\cap E) + \mu^*(A \setminus E) \leq \mu^*(E) + \mu^*(A) = \mu^*(A), $$ where the second inequality follows by monotonicity of $\mu^*$. Hence $\mu^*(A) = \mu^*(A\cap E) + \mu^*(A \setminus E)$.