$\mu^*(C \cap A)=\mu(A) $, when $\mu^*(C)=1$

Question

For the outer measure of a probability measure $\mu$ on $(\Omega, \mathcal A)$ given by \begin{align} \mu^*(A):=\text{inf}\{ \mu (B) \mid B\in \mathcal A, A \subseteq B\}, \end{align} I have to show, that for a set $C \subseteq \Omega$ with $\mu^*(C)=1$, we can obtain $\mu^*(C \cap A)=\mu(A)$ for all $A \in \mathcal A$.

I don't know how to show this. Maybe someone can help me out.

Answer 1

Here's a proof by contrapositive: suppose that $C \subseteq \Omega$ and there is an $A \in \mathcal A$ with $\mu^*(A \cap C) < \mu(A)$. Noting that $C \subset A^c \cup (A \cap C)$, we find that $$ \mu^*(C) \leq \mu^*(A \cap C) + \mu^*(A^c) = \mu^*(A \cap C) + \mu(A^c) < \mu(A) + \mu(A^c) = 1 $$ So, $\mu^*(C)<1$. We have reached the desired conclusion.