Let $(X, \xi, \mu)$ be a measure space and define: $$\xi_\mu := \{A \in \mathcal{P} : \exists B, C \in \xi: \mu(C) = 0 \text{ and } A \,\triangle\, B\subseteq C\}$$ I already showed that $\xi_\mu$ is a $\sigma$-algebra and $\xi \subseteq \xi_\mu$. For $A \in \xi_\mu$ with $B, C \in \xi$ as in the definition we define:
$$\bar\mu(A) := \mu(B)$$
Now I'll need to prove that this is even well defined. Let's consider two $B_1, B_2, C_1, C_2$ with $\mu(C_1), \mu(C_2) = 0$ and:
$$A \,\triangle\, B_1 \subseteq C_1$$ $$A \,\triangle\, B_2 \subseteq C_2$$
but I'm stuck showing $\mu(B_1) = \mu(B_2)$.
Hint: First prove that $$ B_1\triangle B_2\subset (A\triangle B_1)\cup(A\triangle B_2)\subset C_1\cup C_2. $$