I'm stuck on a problem from Cohn's book.
Let $(X,\mathscr{A})$ a measurable space, and let $C$ be a subset of $X$. Let $\mathscr{A}_C$ be the trace of $\mathscr{A}$ on $C$, that is all the subsets $B$ of $C$ of the form $B=A\cap C$ for $A\in \mathscr{A}$. Now suppose that $\mu$ is a finite measure on $(X,\mathscr{A})$. Let $C_1$ a set that belongs to $\mathscr{A}$, and satisfies $\mu(C_1)=\mu^*(C)$. Show that if $A_1$ and $A_2$ belong to $\mathscr{A}$ and satisfy $A_1\cap C=A_2\cap C$, then $\mu(A_1\cap C_1)=\mu(A_2\cap C_1)$.
The only idea I have had is showing $\mu^*(C_1\setminus C)=0$. But I think this is not true. There exists a subset $E$ of $[0,1]$ such that $\lambda^*(E)=1=\lambda^*(E^c)$. Letting $C=E$ and $C_1=[0,1]$,
$$\lambda^*(E)=1=\lambda([0,1]) \text{ and } \lambda^*([0,1]-E)=\lambda^*(E^c)=1\not=0$$
Since the Lebesgue measure is finite in the unit interval, it's not always true that $\mu^*(C_1\setminus C)=0$. So the solution it have to be in other direction. I'd appreciate if someone can help me.
Assuming that $C \subset C_1$ - the exercise makes not much sense and the assertion is in general false without that assumption - we have
\begin{align} \mu(C_1) &= \mu\bigl((C_1 \cap A_k) \cup (C_1 \setminus A_k)\bigr)\\ &= \mu(C_1 \cap A_k) + \mu(C_1\setminus A_k)\\ &\geqslant \mu^\ast(C\cap A_k) + \mu^\ast(C\setminus A_k)\\ &\geqslant \mu^\ast(C)\\ &= \mu(C_1), \end{align}
so we have equality everywhere, in particular $\mu^\ast(C \cap A_k) = \mu(C_1 \cap A_k)$ for $k \in \{1,2\}$.