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

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?

• I can show that $\mu^* \left ( (F \setminus E)^c \right ) = \mu^* (X).$ May 15, 2020 at 15:43
• Shouldn't $\mu$ be a pre-measure since it's only defined on an algebra, not a $\sigma$-algebra? May 17, 2020 at 21:17
• @md2perpe in my book, a measure is defined as a countably additive set function over a collection of subsets of a set $X$ containing the empty set $\varnothing.$ May 18, 2020 at 1:58
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.