A problem regarding regular measure This problem comes from page 99 in Folland's "real analysis: modern techniques and their applications", 2nd edition, as the image below shows.

I tried to prove condition (ii) for n=1 under condition (i), The author says it follows from Theorems 1.16 and 1.18, but I can not see any relation between condition (ii) and the two Theorems mentioned. What $\inf\{...\}$ defines is an outer measure. But, since the collection of all open set, i.e., the topology of $\mathbb R^n$, is not a ring, I can not prove that the outer measure restricted to $\mathbb R^n$ is a measure. In consequence, I can not use subtractivity property of measure to prove condition (ii) using the method similar to that of Theorem 1.14. Could you please help me with this problem? Thanks!
PS: Since this is a specific problem in Folland's book, in addition to the good will to answer, I have to assume the answerer have at hand a copy of Folland's book and, as a must have for owners of this book, its two versions of erratas.
 A: Do the finite case first (by restricting to a compact): $\nu$ is a finite Borel measure on a metric space $X$.
Look at the set $\mathcal{C}$ of $E \in \mathcal{B}$ such that $\nu(E) = \inf_{E \subset U\ \mathrm{open}} \nu(U) = \sup_{K \subset E\ \mathrm{closed}} \nu(K)$.
$\mathcal{C}$ is obviously non-empty and closed under complementation.
Since any open set can be written as the countable union of closed sets (use the metric here), open sets are in $\mathcal{C}$.
Finally, $\mathcal{C}$ is closed under countable union.
Going from the finite case to the "regular" case is achieved by writing $\mathbb{R}^n$ as the union of the closed balls of radii $n$.
EDIT: Let us prove that $\mathcal{C}$ is closed under countable union. Let $E_n$ ($n \geq 1$) be in $\mathcal{C}$, and $E = \bigcup_n E_n$. Let $\epsilon > 0$. For every $n \geq 1$, there is an open set $U_n$ containing $E_n$ s.t. $\nu(U_n \setminus E_n)<\epsilon 2^{-n}$. Let $U = \bigcup_n U_n$. Then $U \setminus E \subset \bigcup_n \left(U_n \setminus E_n \right)$ so $\nu(U \setminus E) < \sum_{n \geq 1} \epsilon 2^{-n} = \epsilon$.
The sup equality is proved in a very similar way, except that you have to choose $m$ such that $\nu\left(E \setminus \bigcup_{n \leq m} \right) < \epsilon$ first (you can only take finite unions of closed sets to get closed sets).
A: oops, I finally know the reason! Condition (i) shows that $\nu$ is finite on bounded Borel sets, so When n=1, Th 1.16 tells us that $\nu$ is a Lebesgue-Stieltjes measure associated with some monotonically increasing and right continuous function F. Thus Th 1.18 can be used directly to get the inf equality of condition (ii). My fault was to take it for granted that the reason lies in the proof and, especially, to overlook the second part of Th 1.16.
