Measurability from Inner Measure Let $E$ have finite outer measure. Show that there is a $G_\delta$ set $G$ which contains $E$ and has the same outer measure. Then show $E$ is measurable if and only if $E$ contains an $F_\sigma$ set $F$ of the same outer measure.
$\textit{Proof:}$ Choose and open set $\mathcal{O}_k$ such that $m^*(\mathcal{O}_k$ \ $E) < \frac{1}{k}$. Define $G = \cap_{k=1}^{\infty} \mathcal{O}_k$. Then $G$ is a $G_\delta$ that contains $E$ and by monotonicity $m^* (G$ \ $E) \leq m^*(\mathcal{O}_k$ \ $E) < \frac{1}{k}$ so $m^* (G$ \ $E) = 0$ and we've shown the first part. 
Now a set $E$ is measurable provided for any set $A$ we have that 
$$ m^*(A) = m^* (A \cap E) + m^*(A \cap E^C)$$
What is the next step?
Edit: The textbook says the fact that (iii) For each $\epsilon > 0$, there is a closed set $F$ contained in $E$ for which $m^* (E \setminus F) < \epsilon$. Can someone explain which parts of Royden's proof need to be altered to show this? He says it follows from DeMorgan's identities. 

If you are trying to show that the complement of a measurable set is measurable, here's what I would do. You know that for all $K \in \Bbb{N}$ there is an open set $O_n$ such that $m_\ast (\mathcal{O}_K - E) \leq \frac{1}{k}$. Now consider the union 
$$\mathcal{O} = \bigcup_{k=1}^\infty \mathcal{O}_k^c.$$
Each $\mathcal{O}_k^c$ is closed and thus measurable, so that $\mathcal{O}$ is a measurable subset contained in $E^c$. Now write
$$E^c = \mathcal{O} \cup( E^c \setminus \mathcal{O}).$$
Since $\mathcal{O}$ is measurable, can you show that $E^c \setminus \mathcal{O}$ is measurable? Hint: show that it is contained in  a set of measure zero and thus measurable.
