Prove that $\mu^* (F \setminus E) = 0.$

Question

Let $\mathcal A$ be an algebra of subsets of a set $X$ and $\mathcal S (\mathcal A)$ be the $\sigma$-algebra of subsets of $X$ generated by $\mathcal A.$ Let $\mu : \mathcal A \longrightarrow [0,+ \infty]$ be a measure on $\mathcal A.$ Let $\mu^*$ be the outer measure induced by $\mu.$ Let $E \subseteq X$ and $F \in \mathcal S (\mathcal A)$ be such that $E \subseteq F$ and $\mu^* (E) = \mu^* (F) < + \infty.$ If $\mu^* (G) = 0,$ $\forall$ $G \in \mathcal S (\mathcal A)$ with $G \subseteq F \setminus E$ then show that $\mu^* (F \setminus E) = 0.$

I have proved that the outer measure $\mu^*$ on $\mathcal P(X)$ induced by $\mu$ can be equivalently defined in terms of the restriction $\bar \mu$ of $\mu^*$ to $\mathcal S (\mathcal A)$ as follows $:$ $$\mu^* (A) = \text {inf}\ \left \{\bar {\mu} (B)\ |\ B \in \mathcal S (\mathcal A), A \subseteq B \right \}.$$ Can it help anyway? I hardly believe that this result will hold. But how to find a counter-example?

Thanks in advance.

Answer 1

The claim to prove is wrong, as shows the following example. Let $X=[0,1]$, $\mu$ be the Lebesgue measure on $X$, and $\mathcal A=\mathcal S(\mathcal A)$ be a $\sigma$-algebra of Lebesgue measurable subsets of $X$. Let $E\subset X$ be an arbitrary set such that $E\not\in\mathcal A$. For instance, $E$ can be a Vitali set. Let $F\in\mathcal A$ be a set such that $E\subseteq F$ and $\mu(F)=\mu^*(E)>0$. Clearly, for all $G \in \mathcal S(\mathcal A)$ with $G \subseteq F \setminus E$ we have $\mu^*(G) =\mu(G)=0$. But if $\mu^*(F \setminus E) = 0$ then $F\setminus E\in\mathcal A$, ans so $E\in\mathcal A$, a contradiction.