# 一題外Lebesgue測度證明 (有關開區間和半開半閉區間)

Prove: The outer Lebesgue measure of each of the intervals (a, b), [a, b), (a, b] is equal to b-a.

THE LEBESGUE MEASURE

Denote by R^n the Euclidean space of n dimensions. The points of R^n are written in the form x = (x_1, ..., x_n). By ana open interval we shall mean a set of the form

I_{a, b} = {x = (x_1, ..., x_n); a_i < x_i < b_i for i = 1, ..., n},

where a = (a_1, ..., a_n), b = (b_1, ..., b_n) are points in R^n. The set K of all open intervals forms a sequential covering class of R^n. Let λ be given by λ(Ø)=0 and

λ(I_{a, b}) = Π_{i = 1}^n (b_i - a_i) if a≠b.

The outer measure determined by this pair K, λ (in accordance with Theorem 1 (請參見下方)) is called the Lebesgue outer measure. The (complete) measure determined by this outer measure (in accordance with Theorem 2 (請參見下方)) is called the Lebesgue measure. The measurable sets are called Lebesgue-measurable sets or, simply, Lebesgue sets.

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CONSTRUCTION OF OUTER MEASURES:

Let K be a class of sets (of X). We call K a \"sequential covering class\" (of X) if

(i) Ø∈K, and

(ii) for every set A there is a sequence {E_n} in K such that A⊆∪n=1~∞ E_n.

Let λ be an extended real-valued, nonnegative set function, with domain K, such that

λ(Ø)=0. For each set A (of X), let

μ(A) = inf{Σn=1~∞ λ(E_n); E_n∈K, A⊆∪n=1~∞ E_n}. (1)

Theorem 1: For any sequential covering class K and for any nonnegative, extended real-valued set function λ with domain K and with λ(Ø)=0, the set function μ defined by (1) is an outer measure.

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Theorem 2: Let μ* be an outer measure and denote by A the class of all μ*-measurable sets. Then A is a σ-algebra, and the restriction of μ* to A is a measure.

Update:

Update 2:

Update 3:

m*(I') = l(I') 呢?

### 1 Answer

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• 1 decade ago
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有錯誤煩請指正！這三題一起證,只要證一重要性質就可了,外測度我表成m*Theorem:The outer Lebesgue Measure of any interval is equal To its lengthproof:We first prove that the close finite interval [a,b],m*[a,b]=b-aNow,given any ε>0,(a-ε,b+ε) is containing [a,b],so m*[a,b]≦Σl(a-ε,b+ε)=b-a+2ε,since ε was 任意給的,so m*[a,b]≦b-ait suffices to show that m*[a,b]≧b-aBy definition of out measure,let {In} be collection of open cover of [a,b],we have to show that Σl(In)≧b-asince [a,b] is compact,there exists subcollection {In'} of {In} also covers[a,b],and the sum of length of the subcollection of {In} is not greater than {In},so it suffices to show that Σl(In')≧b-aNow since [a,b] contained in {In'},a is belongs to some interval (a1,b1),and b1≦b,but b1 is not belongs to (a1,b1) so there exists an open  interval (a2,b2) such that a2<b1<b2Continuing in this way,we get a finite sequences(ak,bk),ai<b(i-1)<bi,for i=2,...,kand b<bkso Σl(In')≧(bk-ak)+(bk-1-ak-1)+...+(b1-a1)=bk-(ak-bk-1)-(ak-1-bk-2)-...-a1since ai<b(i-1),so bk-(ak-bk-1)-(ak-1-bk-2)-...-a1>bk-a>b-aHence Σl(In)≧Σl(In')>b-aso m*[a,b]=b-aIf I is finte interval,there exists closed finite nterval J such that J is contained in I and for any ε>0,l(J)>l(I)-εl(I)-ε<l(J)=m*(J)≦m*(I)≦m*(I')=l(I')=l(I)=>l(I)-ε<m*(I)≦l(I)=>m*(I)=l(I) where I' denote the set cluster points of IIf I is infinite interval,for any real number x,there exists closed finte interval J such that J contained in I ,m*(J)=xm*(I)≧m*(J)=l(J)=x,x is any real numberso m*(I)=l(I)=+∞this complete the proof

2006-08-22 15:43:48 補充：

我的英文字拼錯了！應該是closure

A point x is called a closure point of a set B,iff given any ε>0,there exists y屬於B,such that |x-y|<ε

The set of all closure points of B is denoted B'

If B is closed set,then B=B'

2006-08-22 15:44:55 補充：

一般情況下:B包含於B'

2006-08-22 15:50:23 補充：

∵I'is closed set

Source(s): real analysis
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