可測度集合的證明

Question

最近被實分析搞的暈頭轉向,下面這一題:If {Ik}k=1N is a finite collection of disjoint intervals,then ∪Ik is measurable and m(∪Ik)=Σm(Ik)

m是Lebesgue measure,這題好像要用數學歸納法

一集合A為可測,Geven any ε>0,there exists an open set O,such that A contained in O,m*(O\A)<ε,m*為外測度

Answer 1

請問 m 是否 Lebesgue measure，
還有你所可以用的結果有哪些？

2006-09-27 01:59:28 補充：
Proof. Let ε > 0, and choose any δ ∈ (0,ε). Set Ok = (ak-δ/N,bk+δ/N) ⊇ Ik.Clearly we haveOk \ Ik ⊆ (ak-δ/N,ak]∪[bk,bk+δ/N)som*(Ok \ Ik) ≤ m*((ak-δ/N,ak]∪[bk,bk+δ/N))≤ m*((ak-δ/N,ak]) + m*([bk,bk+δ/N))= ak-(ak-δ/N) + bk-(bk-δ/N)= δ/N.The set ∪k=1N Ok is open, and∪k=1N Ok \ ∪k=1N Ik ⊆ ∪k=1N (Ok \ Ik)som*(∪k=1N Ok \ ∪k=1N Ik) ≤ m*(∪k=1N (Ok \ Ik))≤ ∑k=1N m*(Ok \ Ik) ≤ δ< ε.Hence, ∪k=1N Ik is measurable. Since the Ik are disjoint and m is a measure, we havem(∪k=1N Ik) = ∑k=1N m(Ik).  ∎