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

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.2011-03-13
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    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 (http://library.nu for example).2011-03-13
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    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.2011-03-13
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    ________________2011-03-13
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    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?2011-03-13

2 Answers 2

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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?2011-03-13
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    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?2011-03-13
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    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)?2011-03-13
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    If $E$ is a countable union of closed sets $A_n$, $B_n=A_1 \cup \ldots A_n$ form an increasing sequence of closed sets, and $\bigcup_n B_n = E$, so $\nu(E)=\lim \nu(B_n)$.2011-03-13
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    If $E$ is open, $A_n = \left\{ x \in X\ |\ d(x,X\setminus E) \geq 1/n \right\}$ works.2011-03-13
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    Regarding your second comment, I don't see what is bothering you: the sup is necessarily $\leq \nu(E)$, and the limit (two comments up) shows the equality.2011-03-13
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    I have thought of it, but this limit is a supremum of SOME of closed sets contained in E, but the definition of E is a supremum of ALL closed sets contained in E. We need to prove that these two sup's are equal. It is clear that the former sup <= the latter sup, but how to prove the converse?2011-03-13
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    If we can show that every closed set contained in E is a subset of some $B_n$, then the two sup's are equal. But I have no idea how to prove this. This is what blocked me in my second comment. Maybe the answer hides in your construction of $A_n$, but I need some time to find it.2011-03-13
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    former $\geq \nu(E) \geq$ latter, because $K \subset E$ implies $\nu(K) \subset \nu(E)$.2011-03-13
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    Going from the finite case to the whole of $\mathbb{R}^n$ is essentially the same procedure as in the "stable under countable union" proof.2011-03-13
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    ? the former sup is $\sup\{B_n\}$(=$\lim(\nu(B_n)$ by continuity from below). I need to prove that it is equal to $\sup\{\nu(K)|K\subseteq E, K \rm{closed}\}$ in definition of $\mathcal C$.2011-03-13
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    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 $\sup M$.2011-03-13
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    I do not know how to prove $\mathcal C$ is closed under countable union, as my first comment indicated. The problem is also in the sup part of the definition.2011-03-13
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    I edited the answer. I don't see the problem with the sup.2011-03-13
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    I got that the two sup's are equal. This is my understanding: $\nu(E)$ is an upper bound of $\{\nu(K)|K\subseteq E, K\rm{closed}\}$. Since $\nu(E)$ is finite, if we can find an $\nu(K)>\nu(E)-\epsilon$ for any $\epsilon>0$, the sup is proved, but some B_n is such a closed set. Thank you Plop for your patient reply. There are still other problems, such as how to prove countable additivity, but I have to take a sleep before getting them.2011-03-13
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    Countable additivity is part of the definition of a measure.2011-03-13
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    To establish the sup equality in the proof of countable additivity of $\mathcal C$, I can only prove that there is a $m$ such that $\nu(E)-\sum\limits_{n=1}^m \nu(K_n)<\epsilon$ where $K_n$ is a closed subset of $E_n$ such that $\nu(E_n)-\nu(K_n)$ can be arbitrarily small. But $\nu(\bigcup\limits_{n=1}^m K_n)\le\sum\limits_{n=1}^m\nu(K_n)$, how do we establish $\nu(E-\bigcup\limits_{n=1}^m K_n)=\nu(E)-\nu(\bigcup\limits_{n=1}^m K_n)<\epsilon$?2011-03-14
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    I ommitted the fact that $K_n$'s are disjoint. So $\sum\limits_{n=1}^m\nu(K_n)$ is safely equal to $\nu(\bigcup\limits_{n=1}^m K_n)$. Now I finally understand all the proof! Thank you Plop! By the way, for general $\nu$, I find that the sup equality $\nu(E)=\sup\{\nu(K)|K\subseteq E, K\rm{closed}\}$ seems not hold any more, right?2011-03-14
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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.