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