Inclusion-Exclusion Principle for Outer Measure

Question

Question:
Let $\mu^*$ be an outer measure on a set $\Omega$ and $E$ be a $\mu^*$-measurable set. Show that $$ \mu^*(A) + \mu^*(E) = \mu^*(A \cap E) + \mu^*(A \cup E) $$ for all $A \subseteq \Omega$.

Attempt:
The fact that $A\subseteq\Omega$ is not necessarily $\mu^*$-measurable means I cannot use countable additivity property.

By Carathéodory criterion and countable sub-additivity, \begin{align} \mu^*(A) + \mu^*(E) &= \mu^*(A \cap E) + \mu^*(A \cap E^c) + \mu^*(E) \\ &\ge \mu^*(A \cap E) + \mu^*(A \cup E) \end{align}

How do I show the reverse inequality?

Answer 1

Combine your first equality with the following: $\mu^{*}(A\cup E)=\mu^{*}((A\cup E) \cap E) +\mu^{*}((A\cup E) \cap E^{c})=\mu^{*}(E) +\mu^{*}(A\cap E^{c})$

Answer 2

By Carathéodory criterion you also know that, because $E \subset \Omega$ is $\mu^*$-measurable then for any $A \subset \Omega$

$$\mu^*(A \cup E) = \mu^*(A \cap E^c) + \mu^*(E)$$

because

\begin{align*} \left( A \cup E \right) \cap E^c &= A \cap E^c \\ \left(A \cup E \right) \cap E &= E \end{align*}

So where you have your inequality could be an equality.