Yahoo Answers: Answers and Comments for 實分析3(Abstraction integration) [數學]
Copyright © Yahoo! Inc. All rights reserved.
https://tw.answers.yahoo.com/question/index?qid=20070613000015KK07312
From Scharze space
zhHantTW
Wed, 13 Jun 2007 18:28:17 +0000
3
Yahoo Answers: Answers and Comments for 實分析3(Abstraction integration) [數學]
292
38
https://tw.answers.yahoo.com/question/index?qid=20070613000015KK07312
https://s.yimg.com/zz/combo?images/emaillogotw.png

From Eric: Proof. Since the class of simple functions va...
https://tw.answers.yahoo.com/question/index?qid=20070613000015KK07312
https://tw.answers.yahoo.com/question/index?qid=20070613000015KK07312
Thu, 14 Jun 2007 00:26:09 +0000
Proof. Since the class of simple functions vanishing outside sets of finite measure is dense in Lp(μ), it suffices to show that the indicator functions of Borel sets can be approximated in Lp(μ) by continuous functions with compact support, since
by linearity, simple functions vanishing outside sets of finite measure can be approximated in Lp(μ) by continuous functions with compact support, and
by density, the result extends to all of Lp(μ).
Let E be a Borel set of finite measure. Given ε ＞ 0, there exist (by the regularity of μ) an open set G containing E and a closed set F contained in E such that μ(G\F) ＜ ε. By Urysohn's lemma, there exists a continuous f:Rn→[0,1] with f ＝ 1 on F and f ＝ 0 on Rn\G. Then
1E － f ＝ 0 on Rn\(G\F)
1E － f ≦ 1 on G\Fand so
∫ 1E － fp dμ ＝ ∫G\F 1E － fp dμ
≦μ(G\F) ＜ ε.
The result follows. :
20070614 00:52:55 補充：
≦ 是≤