If μ is a regular Borel measure, show that the class of continuous functions with compact support is densie in L^p(dμ)
(Hint:Use the fact that for 1<=p<∞, the class of simple functions vanishing outside sets of finite measure is dense in L^p(dμ)
It is enought to approximate χ(E),where E is the Borel set with finite measure
Given ε＞０,there exist open set G and closed set F with F is contained in E,E is contained in G, and μ(G\F)<ε,Now use urysonh's Lemma:if F1,F2 are disjoint closed set in R^n,there is continuous f on R^n,with 0<=f<=1,f=1 on F1,f=0 on F2)
- EricLv 61 0 年前最佳解答
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 － f|p dμ ＝ ∫G\F |1E － f|p dμ
≦μ(G\F) ＜ ε.
The result follows. :
2007-06-14 00:52:55 補充：