# 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.

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More work should be put in this question before it is suitable for MSE. First, there is no condition (i) and no condition (ii) either in the page you reproduce. Second, to refer to Theorems 1.14, 1.16 and 1.18 like you do is to ask the reader to have Folland's book at hand. – Did Mar 13 '11 at 9:58
There is a typo: the two bullets should be replaced by (i) and (ii). I had planned to upload Folland's book (and its errata), but I'm afraid of violation of copyright; these can be found easily from internet (library.nu for example). – zzzhhh Mar 13 '11 at 10:25
I second Didier's complaint. The problem is that your question is about a specific technical detail and cannot be answered without some effort like downloading the book, which I don't want to do. So please do formulate at least the statements of 1.14,1.16 and 1.18 in order to motivate us to answer. Towards an answer: I cannot see how the case $n = 1$ should be any different from the general case. In fact, it is not hard to show that a finite Borel measure on a metrizable space satisfies (ii) and thus (ii) is easily reduced to (i) by a standard exhaustion argument. – t.b. Mar 13 '11 at 10:37
________________ – zzzhhh Mar 13 '11 at 10:47
As I tried to indicate, I would do this in two steps: 1. prove it for finite measures, 2. reduce to 1. by exhaustion because (i) implies that your measure is $\sigma$-finite. With which of the two points do you have trouble? – t.b. Mar 13 '11 at 11:10

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).

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I can only prove that $\mathcal C$ is closed under complementation. And I can only prove that every open set can be written as a countable union of closed sets by Lindelof peoperty of $\mathbb R^n$ because it is second-countable. But from your reply it seems to hold for any metrizable space, how to prove it? – zzzhhh Mar 13 '11 at 17:47
I don't understand either why if we can write open set $E$ as a countable union of closed sets, then $v(E)=\sup\{\nu(K)|K\subseteq E, K\rm{ closed}\}$ so that open sets are in $\mathcal C$. I can express $v(E)$ as a sup of some closed sets formed by finite unions of the countable closed sets obtained above, but how to prove that these two $\sup$'s are equal? – zzzhhh Mar 13 '11 at 17:48
I also do not know how to extend the finite case to general case. Could you please explain these in more detail? What is more, I think this may not be the intended method of the author (maybe it is what we will see in section 7.2?). Could someone who used this books as a textbook in real analysis course please tell me how the author actually deduces the condition (ii)? – zzzhhh Mar 13 '11 at 17:48
If $E$ is open, $A_n = \left\{ x \in X\ |\ d(x,X\setminus E) \geq 1/n \right\}$ works. – Plop Mar 13 '11 at 18:41
Let $M = \left\{ \nu(K)\ |\ K\ \mathrm{closed}\ \subset E \right\}$. $M$ is not empty, and bounded from above by $\nu(E)$. If $M$ is bounded from above by a real number $x$, $\nu(B_n) \leq x$ for all $n$, so passing to the limit, $x \geq \nu(E)$. So $\nu(E)$ is the smallest element bounding $M$ from above, which by definition is $\sup M$. – Plop Mar 13 '11 at 19:10

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.

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