Show that there is an $F_\sigma$ set $F$ and $G_\delta$ set $G$ such that $F \subseteq E \subseteq G \text{ and } m^*(F)=m^*(E)=m^*(G).$ [duplicate]

Solution 1:

The way you find sets $F$ and $G$ in your posting under the assumption that $E$ is measurable (in the sense of Caratheodory) is correct. The problem is with the statement of the problem. I check the statement in the book you quote and it seems surviving author might have missed the measurable adjective for $E$, whether that change in subsequent revisions, I don't know.

Here we show that if $E$ satisfies the conclussion of the problem, then $E$ must be measurable in the sense of Caratheodory). By Caratheodory's theorem, the outer measure $m^*$ extends $m$ to a measure on a $\sigma$-algebra $\mathcal{M}$ that contains the Borel $\sigma$-algebra $\mathscr{B}(\mathbb{R})$ as well as the $m$-null sets (sets with outer measure $0$).

If $F$ is an $F_\sigma$ set and $G$ is a $G_\delta$ set, then both are Borel measurable. So, if in addition, $F\subset E\subset G$, $m^*(F)=m^*(G)=m^*(E)<\infty$, then $m^*(G\setminus F)=0$. Indeed, since $F$ and $G$ are Borel sets $$\mu(G)=\mu^*(G)=m^*(F)+m^*(G\setminus F)=m(F)+m(G\setminus F)$$ Since $m^*(E)<\infty$, $m(G\setminus F)=m(G)-m(F)=0$. This would imply the $m^*$-measurability of $E$ for $$E=F\cup(E\setminus F),$$ $F\in\mathscr{B}(\mathbb{R})\subset\mathcal{M}$, and $E\setminus F\in\mathcal{M}$ on account that $m^*(E\setminus F)\leq m(G\setminus F)=0$.

As a counter example to the statement of the problem as it is written in the book, consider a the classical Vitali set $A$ constructed using the Axiom of choice and the equivalence $r\sim s$ if $r-s\in\mathbb{Q}$. $A$ set is not Lebesgue measurable measurable; $E=A\cap[0,1]$ satisfies $m^*(E)=1$ and $$\sup\{m(F):F\,\text{closed}\,,F\subset E\}=0$$

Solution 2:

Here is the problem statement in my text, fourth edition, 2018 reissue as part of Pearson's modern classic series:

Let $E$ have finite outer measure. Show that there is a $G_\delta$ set $G\supseteq E$ with $m(G) = m^*(E)$. Show that $E$ is measurable if and only if there is an $F_\sigma$ set $F\subseteq E$ with $m(F) = m^*(E)$.

For each natural number $n$, there is a countable collection $\{I_{n,k}\}_{k=1}^\infty$ of open, bounded intervals for which $E\subseteq\bigcup_{k=1}^\infty I_{n,k}$ and $\sum_{k=1}^\infty \ell(I_{n,k}) < m^*(E) + 1/n$. Define $G_\delta$ set $G =\bigcap_{n=1}^\infty \bigcup_{k=1}^\infty I_{n,k}$, which is measurable by Theorem 9. Then for each $n$, $m(G)\le m(\bigcup_{k=1}^\infty I_{n,k}) = \sum_{k=1}^\infty \ell(I_{n,k}) < m^*(E) + 1/n$. Hence, $m(G)\le m^*(E)$. We also have $m^*(E)\le m(G)$ because $E\subseteq G$. Thus, $m(G) = m^*(E)$.

Next, by Theorem 11(iv), $E$ is measurable if and only if there is an $F_\sigma$ set $F\subseteq E$ with $m^*(E\sim F) = m^*(E) - m(F) = 0$. We can use the excision property here because $F$ is measurable by Theorem 9 and has finite outer measure since $F\subseteq E$ and $m^*(E) <\infty$.