I have the following exercise I wish to solve:
Let $(X,S,\mu)$ be a measure space s.t $\mu(X)=1$.
Let $\mu^{*}$be defined on $X$ by:
$\forall E\subseteq X:\,\mu^{*}(E):=\inf\{\sum_{i=1}^{\infty}\mu(A_{i})\,\mid\, A_{i}\in S,E\subseteq\cup A_{i}\}$
Let $E\subseteq X$ s.t $\mu^{*}(E)=1$, prove that if $A,B\in S,A\cap E=B\cap E$ then $\mu(A)=\mu(B)$
What I tried is to write $A=(A\cap E)\cup(A\cap E^{c}),B=(B\cap E)\cup(B\cap E^{c})$ and then apply $\mu^{*}$on both of them.
Since $A\cap E=B\cap E$ I would have had $\mu^{*}(A\cap E)=\mu^{*}(B\cap E)$ and I hoped that $\mu^{*}(E)=1\implies\mu^{*}(E^{c})=0$ and from $\mu^{*}$being monotone that means that $\mu^{*}(A\cap E^{c})=\mu^{*}(B\cap E^{c})=0$
hence $\mu^{*}(A)=\mu^{*}(B)$ but $A,B\in S$ hence $\mu^{*}(A)=\mu(A),\mu^{*}(B)=\mu(B)$ and than I would be done.
But from this post is turns out that $\mu^{*}(E^{c})$ doesn't have to be $0$ and so my argument is not valid.
Can someone please suggets how to solve this problem ? what I tried was my only idea.