Riez表示定理怎么证明的啊,能否给出证明啊,我证了很长时间未果,我这里资料少,书找不到.证不出来的滋味很难受啊,尤其当你想知道证明的时候. 萍踪侠侠能否给个证明啊!!
,我不知道这个定理是不是有很多版本,我先把我看到的定理叙述一下:
对于一个紧致集且是Hossdof的集合M上的连续函数(实的复的都可)构成的线性空间,其上的模就是最常见的那个,F是其上的一个线性泛函,那么一定存在M上的一个(可加)测度或者说存在一个可加函数U(这个可加函数看成那个测度),使得F对M上的任一连续函数A的作用等于A在这个集合M上对于这个测度U的积分!

先谢谢了!

如果这个泛函叫做 I, 那么定义开集 U 的测度为 m(U) = sup { I(f)： 0<=f<=1, supp(f)包含于 U },
然后用标准的外测度方法把 m 扩张到相应的可测集族上面。

刚才说的证明可能有漏洞。应该是这样，先证明所谓 “正泛函” 的表示定理，就是说如果 I 是一个正泛函，即，
只要 f>=0 就有 I(f)>=0， 那么 I 可以表示成对某一正测度的积分 （证明方法如前一贴所述）。然后对一般的线性泛函 I 证明所谓 Jordan 分解， I = (I+) - (I-), I+ 和 I- 是两个正泛函。这样就证明 C(M)* 是所有全变差有限的 “符号测度” 的空间 （因为 M 是拓扑空间，这里的测度实际上满足某些跟拓扑有关的条件，这种测度叫 Radon 测度）

资料应该到处都是，比如 严加安著 《测度论讲义》，各地图书城肯定有得卖。

谢谢季候兄啊，这个想法我好象有过，但是这样的定义，一开始就给不出个测度来，原因是这样的，你刚开始定义的m就不满足下面的性质：如果一个开集序列U（i）都是开集，当然这个序列的并也是开集，但不能保证，这个序列的并的m(你上面定义的)测度等于U（i）的测度之和，而这是测度一开始定义中就应该满足的一个条件，（这是某本教才中看到的，我只看过一本书，我不知道是不是（还没想过）在别的书中把这个条件可以除去，而直接考虑后面的外测度和测度的延拓）。

我这里确实找不到测度专门的书，我有这样的书，放家里了，呵呵，我在的这个城市书店里卖理科专业书集很少，比较郁闷，
不知道我上面那篇回复说的对不，如果对的，那肯定是可以把我看到的测度首先要满足的那个条件先除去，或者用别的什么来代替，然后直接考虑外测度。

显然满足啊

上面那个求和是个无限求和，不是有限的啊！！

呵呵，好好想想什么叫函数项级数

这个我确实想过了,其实我就是在这里出了问题,别的都好说.你也仔细想想啊,要不把你的证明过程,大概的写上来啊,很想看看啊,
不是函数项级数的问题!在这里对有限项满足,和的函数等于函数的和,但是无限的就没法说了,可数的不相交开集的m测度之和与可数的不相交开集并的m测度没有任何联系啊.

接上面,原因是线性泛函,不一定满足这个性质,就是可数可加性啊,是吧?

哦，不好意思，我又信口开河了......

其实是这样， supp(f) 是闭集，因为 M 紧，所以 M 里的闭集也紧。如果 supp(f) 包含于 Ui 的并，那么由紧性，可以找到有限个 Ui 包含 supp(f), 也就是说 f 可以写成有限和， 这样就没有无穷的问题了。

证明细节比较多，写在这里是不可能的。

网上的免费书籍很多，比如这个 http://www.essex.ac.uk/maths/staff/fremlin/mt.htm
不过需要用 Latex 编译一下

季候兄,你这次说的确实对了,哈哈,谢谢了啊．
能不能在顺便给出 集合论中Zorn引里的证明啊，就是那个关于半序集中有最大元的那个定理！

Zorn 引理是选择公理的一种等价形式，怎么 “证明”？

啊,是这样啊,真没想到,不过觉得好象是可证明的,而且名字怎么还叫引理,真是没想到是个公理.
记得引理叙述成:
任何一个半序集,如果对于它的每一个全序子集都存在最大元,那么这个半序子集必有最大元.

任何一个半序集,如果对于它的每一个全序子集都存在最大元,那么这个半序子集必有最大元.
~~~~~~~~~应该是 “都有上界”

>>Zorn 引理是选择公理的一种等价形式，怎么 “证明”？

Coincidentally I read a 1996 interview of the famous statistician Lucien Le Cam several weeks ago, he happened to mention Zorn's lemma ---

http://projecteuclid.org/Dienst/UI/1.0/Journal?authority=euclid.ss&issue=1009212241

A conversation with Lucien Le Cam. By Grace L. Yang; Statistical Science, 14(2): 223-241 (May 1999)

On page 225, Le Cam asserted that one can indeed prove Zorn's lemma by using the Axiom of Choice. But this answer to feiguohai's question has to be found in Bourbaki's book written in French: "General Topology". I'm not sure if 季候风 can find that book for you and reproduce the "proof" in this forum, hehe.

------------------------------------------------------------------------------
...

Yang: That was your first exposure to Bourbaki. Many of us are curious about how you got such an abstract way of thinking and writing statistics.

Le Cam: Yeah. It has intersection signs and union signs and things. I had never seen that before. And it was a book of results without proofs, just the statements. One of them was Zorn's lemma. So you have the axiom of choice and you can prove Zorn's lemma out of the axiom of choice. And somehow I was unable to do that. I sent a postcard to Hermann, Bourbaki's publisher. They sent me the first volume of Bourbaki's Topologie Generale.

...
------------------------------------------------------------------------------

呵呵，等价命题之间的证明并不难。

我知道的等价命题有：

选择公理：可以从一族非空集合的每一个中取出一个元素。

Zorn 引理：如果一个偏序集的每一个全序子集都有上界，那么这个偏序集有一个极大元。

Hausdorff 极大原理：任一偏序集有一个极大全序子集。

良序定理：每一个集合有一个良序。（一个 “良序” 就是一个偏序，使得每个子集都有极小元）

其它等价命题还有无数。

以上四个命题之间的等价性是很有意思的逻辑游戏，有兴趣的同修可以不看书本自己试试。

To feiguohai: If you are really interested to know the details of the proof of the Riesz Representation Theorem, the best book for you to read might be Rudin's "Real and Complex Analysis" ---

http://www.qiji.cn/eprint/abs/780.html

Both Chinese and English versions are available and I highly recommend you to read the English version. In the English edition, it took Rudin 7 pages (pp.41-47) to prove this theorem. So I guess you may have to spend several hours to understand the proof.

I can never afford the time to read through this kind of mathematical proof, that's why I think Rudin's books are only suitable to students majored in mathematics. I occasionally use Rudin's books as "dictionaries" or "manuals", the only reason I read "Real and Complex Analysis" at all is because Gauge mentioned Lebesgue Measure in his post on uniform distributions. The book provides very good coverage on this topic.

I also like Rudin's analogy between the concepts for continuous vs. measurable spaces. His description kind of confirmed my thoughts on the connection when I read Casella & Berger's Chapter 1 last year:

-------------------------------------------------------------------------------
http://www.changhai.org/bbs/load_article.php?fid=5&aid=1153705369

另外，Casella & Berger成功展示了用Kolmogorov的三条公理来定义概率的优越性。当然这个定义依赖于sigma algebra (又称Borel field，但不同于Borel sigma algebra)的定义。我在研究sigma
algebra的定义时自然想到了它和topology的定义之联系，在此用英文表达一下，请数学专家指出该观点是否正确：

The definition of a sigma algebra seems to be more specific than that
of a topology when the two concepts are referring to the same space
set S. The "closed under complementation" requirement doesn't appear
in the definition of a topology on the same space. So a sigma algebra
(or Borel field) is always a topology on the same set S, whereas a
topology is not necessarily a sigma algebra.

Interestingly, the definition of "Borel sigma algebra" seems to form a
bridge between topology and probability theory ---

[Definition] Borel sigma algebra is a sigma algebra generated by the
open sets (or equivalently, by the closed sets) of a space S.
-------------------------------------------------------------------------------

Rudin systematically built up the analogy between the following concepts:

topology <--> sigma algebra

topological space <--> measurable space

open set <--> measurable set

continuous function <--> measurable function

I really like Rudin's clarity in this kind of conceptual development. But I also like to complain that his books tend to be too dense, I had to skip many sections of no interest to me. He attempted to pack too many contents in a limited space ...

The definition of a sigma algebra seems to be more specific than that
of a topology when the two concepts are referring to the same space
set S. The "closed under complementation" requirement doesn't appear
in the definition of a topology on the same space. So a sigma algebra
(or Borel field) is always a topology on the same set S, whereas a
topology is not necessarily a sigma algebra.

Interestingly, the definition of "Borel sigma algebra" seems to form a
bridge between topology and probability theory ---
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


这两种结构的定义模式有些相似，但是它们是相对独立的结构。它们满足的公理有很大的不同。
一般的 Sigma 代数并不能构成拓扑，因为它仅要求满足 “可数并封闭” 性质，而拓扑要求
任意个（可以是不可数个）开集的并还是开集；一般的拓扑也不可构成 Sigma 代数，因为
拓扑只要求 “有限交” 封闭，而不一定具有 Sigma 代数的 “可数交封闭” 性质。在非常简单的
情况下它们可能有这种关系，比如平凡拓扑 {空集，全集} 和离散拓扑 {所有子集} 是 Sigma 代数。

“可数可加” 是最关键的概念，是 “测度” 这个概念赖以生存的基础，如果不是因为这个限制，
平面里的直线就会有非零的测度，我们对于测度的原始概念（长度，面积，体积，概率...）就会被颠覆。

Borel 代数的概念是为了在拓扑空间上讨论跟背景拓扑想关的测度问题而存在的，这里拓扑是已经
被指定的，作为背景，而讨论的所有 “Borel 测度” 都是跟这一固定拓扑相关的。

>>一般的 Sigma 代数并不能构成拓扑，因为它仅要求满足 “可数并封闭” 性质，而拓扑要求
任意个（可以是不可数个）开集的并还是开集；一般的拓扑也不可构成 Sigma 代数，因为
拓扑只要求 “有限交” 封闭，而不一定具有 Sigma 代数的 “可数交封闭” 性质。

Thanks for the insightful comments pointing out what I overlooked. Let me reorganize the key conceptual differences between topology and sigma algebra:

(1) Closure under subset unions

* Sigma Algebra: closed under countable unions of measurable sets ("countable" means "countably infinite" in more precise language)

* Topology: closed under arbitrary unions of open sets ("arbitrary" can be finite, countable, or uncountable, which means "uncountably infinite")

(2) Closure under subset intersections

* Sigma Algebra: closed under countable intersections of measurable sets

* Topology: closed under finite intersections of open sets

I did pay attention to the word "countable" in "countable additivity" as one of Kolmogorov's three axioms of the measure of probability. But I didn't pay enough attention to the word "countable" in the definition of sigma algebra or the words "finite" and "arbitrary" in the definition of topology. It's very easy to miss the importance of these adjectives upon first reading. Your clarifications helped me a lot, I'll dig a little deeper on the concept of "sigma addivity" (same as "countable additivity") given your emphasis on its fundamental significance in the context of measure theory.

BTW, Wikipedia did sketch out a proof of Zorn's Lemma from the Axiom of Choice, so feiguohai doesn't have to find the Bourbaki book to satisfy his curiosity ---

===========================================
http://en.wikipedia.org/wiki/Zorn%27s_lemma

Sketch of the proof of Zorn's lemma (from the axiom of choice)

A sketch of the proof of Zorn's lemma follows. Suppose the lemma is false. Then there exists a partially ordered set, or poset, P such that every totally ordered subset has an upper bound, and every element has a bigger one. For every totally ordered subset T we may then define a bigger element b(T), because T has an upper bound, and that upper bound has a bigger element. To actually define the function b, we need to employ the axiom of choice.

Using the function b, we are going to define elements a0 < a1 < a2 < a3 < ... in P. This sequence is really long: the indices are not just the natural numbers, but all ordinals. In fact, the sequence is too long for the set P; there are too many ordinals, more than there are elements in any set, and the set P will be exhausted before long and then we will run into the desired contradiction.

The a's are defined by transfinite induction: we pick a0 in P arbitrary (this is possible, since P contains an upper bound for the empty set and is thus not empty) and for any other ordinal w we set aw = b({av: v < w}). Because the av are totally ordered, this works just fine.

This proof shows that actually a slightly stronger version of Zorn's lemma is true:

If P is a poset in which every well-ordered subset has an upper bound, and if x is any element of P, then P has a maximal element that is greater than or equal to x. That is, there is a maximal element which is comparable to x.
==============================================

I also like the following joke about the equivalent axioms mentioned by 季候风:

"The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" --— Jerry Bona

(This is a joke: although the axiom of choice, the well-ordering principle, and Zorn's lemma are all mathematically equivalent, most mathematicians find the axiom of choice to be intuitive, the well-ordering principle to be counterintuitive, and Zorn's lemma to be too complex for any intuition.)