I was wondering if somebody could help me out with a solution for the following problem (taken from Royden's Real Analysis, 4e. (ch. 2.4, prob. 18):
Let $E$ have finite outer measure. Show that there is an $F_\sigma$ set $F$ and a $G_\delta$ set $G$ such that $F \subset E \subset G$ and $m^*[F]=m^*[E]=m^*[G]$.
So far I constructed a candidate set $G$ (for each $n \in \mathbb{N}$ there is a countable open interval cover $\{ I_{n,k} \}$ such that $\Sigma_k \ \ell(I_{n,k}) < m^*[E] + \frac{1}{n}$; then set $G:=\cap_n \cup_k I_{n,k}$). Since $G$ is measurable, I can approximate it with an $F_\sigma$ set $F \subset G$ such that $m^*[F]=m^*[G]$, but of course there is no guarantee that $F \subset E$. Is there anyway to salvage this proof by toying with $F$ so that $F \subset E$ but still remains an $F_\sigma$ set? Or should I approach another way?
Thanks in advance for any help!
Best Answer
We need a further condition on $E$, $m^{\ast}[E] < \infty$ alone is not sufficient:
Since $F_\sigma$ and $G_\delta$ sets are Lebesgue-measurable (they are Borel sets), if we have an $F_\sigma$ set $F$ and a $G_\delta$ set $G$ with $F \subset G$ and $m^\ast[F] = m^\ast[G] < \infty$, then $G\setminus F$ is a $G_\delta$ set with $m^\ast[G\setminus F] = 0$. Since the Lebesgue measure is complete, every set between $F$ and $G$ is thus Lebesgue-measurable (a union of an $F_\sigma$ with a null set).
Thus measurability of $E$ is a necessary condition for the conclusion to hold.
By the outer regularity, it is also a sufficient condition: Let $K_n = \{ x : \lvert x\rvert \leqslant n\}$. For every $k$, there is an open set $O_k \supset (K_n\setminus E)$ with $m^\ast[O_k] < m^\ast[K_n\setminus E] + 2^{-k}$. Then
$$F_n = \bigcup_k (K_n\setminus O_k)$$
is an $F_\sigma$ set with $F_n \subset E\cap K_n$ and $m^\ast[F_n] = m^\ast[E\cap K_n]$. Further,
$$F = \bigcup_n F_n$$
is an $F_\sigma$ set contained in $E$ with $m^\ast[F] = m^\ast[E]$.