# Inclusion-Exclusion Principle for Outer Measure

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?

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

• That was helpful! Thank you!
– Tham
Sep 22, 2021 at 11:44

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.

• Oh that's smart! Thank you!
– Tham
Sep 22, 2021 at 11:44