Recently I was trying to solve this problem.
Suppose that $X$ is a set and $\mu^{\star}$ is an outer measure on $2^X$. Let $A \subseteq X$ be a set such that $\mu^{\star}(A)<+\infty$ and suppose that $C \in \Sigma_{\mu^{\star}}$ satisfies $A \subseteq C$ and $\mu^{\star}(C)=\mu^{\star}(A)$. Show that $$ \mu^{\star}(A \cap E)=\mu^{\star}(C \cap E) \quad \forall E \in \Sigma_{\mu^{\star}} . $$
The $\Sigma_{\mu^{\star}}$ is just a short for the set of $\mu^*$ measurable sets as in Caratheodory.
Now I did something similar for measures, not outer-measures.
I looked at the solutions of the author but they are wrong.
Has anyone seen this result before ?
Since $A \subseteq C$, from the monotonicity of $\mu^{\star}$, we have $$ \mu^{\star}(A \cap E) \leqslant \mu^{\star}(C \cap E) $$ and $$ \mu^{\star}\left(A \cap E^c\right) \leqslant \mu^{\star}\left(C \cap E^c\right) $$ (here $E^c=X \backslash E$ ). Also, since $E \in \Sigma_{\mu^{\star}}$, we have $$ \mu^{\star}(A)=\mu^{\star}(A \cap E)+\mu^{\star}\left(A \cap E^c\right) \leqslant \mu^{\star}(C \cap E)+\mu^{\star}\left(C \cap E^c\right)=\mu^{\star}(C) $$ Because by hypothesis $\mu^{\star}(A)=\mu^{\star}(C)$, we conclude that $$ \mu^{\star}(A \cap E)=\mu^{\star}(C \cap E) . $$
I would like to remark that going from the inequalities to the equality, in the end, of what we want to proof took me a while to get convinced but it is just algebra in the end.