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.

I can show that $\mu^* \left ( (F \setminus E)^c \right ) = \mu^* (X).$ — math maniac., May 15 '20 at 15:43
Shouldn't $\mu$ be a pre-measure since it's only defined on an algebra, not a $\sigma$-algebra? — md2perpe, May 17 '20 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 $\varnothing.$ — math maniac., May 18 '20 at 01:58
For more information you can go through the second edition of the e-book An Introduction to Measure and Integration by Inder K. Rana published by American Mathematical Society which is available online in pdf format. — math maniac., May 18 '20 at 02:03
I forgot to mention that measure of the empty set $\varnothing$ is considered to be $0.$ — math maniac., May 18 '20 at 02:06

Alex Ravsky · Accepted Answer · 2020-05-20T06:41:28.707

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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.

edited May 20 '20 at 06:41

answered May 20 '20 at 06:34

Alex Ravsky

106,166

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how did the author miss that? He claimed that in his book. – math maniac. May 20 '20 at 06:52
@mathmaniac. Maybe because the claim is formulated in too general and too abstract form. – Alex Ravsky May 20 '20 at 06:56

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

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