In Royden's Real Analysis, the problem given on page 43 as problem 18 used to be as follows:
"Let $E$ have finite outer measure. Show that there exists an $F_\sigma$ set $F$ and a $G_\delta$ set $G$ such that $F \subseteq E \subseteq G$ and $m^*(F) = m^*(E) = m^*(G)$."
But according to the errata, the problem has been revised to the following:
"Let $E$ have finite outer measure. Show that there exists a $G_\delta$ set $G$ such that $E \subseteq G$ and $m^*(E) = m^*(G)$. Show that E is measurable if and only if there exists an $F_\sigma$ set $F$ such that $F \subseteq E$ and $m^*(F) = m^*(E)$."
I assume that the problem has been revised because the finite outer measure of a set without the set's measurability doesn't necessarily ensure the existence of such an $F_\sigma$ set.
Likewise, even though it is true that: "If $E$ is a set of real numbers of finite outer measure, then for any $\epsilon > 0$, there exists an open set $O_\epsilon$ covering $E$ such that $m^*(O_\epsilon) < m^*(E) + \epsilon$.",
We cannot assert that: "If $E$ is a set of real numbers of finite outer measure, then for any $\epsilon > 0$, there exists a closed set $F_\epsilon$ contained in $E$ such that $m^*(F_\epsilon) > m^*(E) - \epsilon$."
Am I correct? I also tried to prove the existence of such $F_\sigma$ sets and closed sets if they do exist, but so far I haven't found an approach. Meanwhile, I have observed some people on Math StackExchange assuming the existence of such sets only with the condition that $E$ is of finite outer measure (without the measurability of $E$ guaranteed), so I am confused.
Thank you in advance.
The assertion that $$m^*(E)=\inf\{m^*(U):E\subset U,\quad U\text{ is open} \} $$ is called the outer regularity of $m^*$. On the other hand, its "dual" assertion, that is, $$m^*(E)=\sup\{m^*(K):K\subset E,\quad K\text{ is compact}\} $$ is called the inner regularity of $m^*$. The outer measure $m^*$ has the outer regularity (which follows immediately from its definition), but it does not have the inner regularity in general. In fact, Lebesgue measurable sets are precisely those sets with the inner regularity. (This is still true even if we replace "compact" with "closed".)
So as for your questions, such an $F_\epsilon$ can be found (for all $\epsilon>0$) precisely when $E$ is Lebesgue measurable. In this case, the union of $F_{1/n}$ will be the desired $F_{\sigma}$-set.