? asked in 科學數學 · 1 decade ago

一題度量外測度證明 (和可測開集有關)

Prove: If μ* is an outer measure and if every open set is measurable, then μ* is a metric outer measure.

註:

outer measure的定義:

An extended real-valued set function ν is said to be "countably subadditive" if

ν(∪j=1~∞ E_j) <= Σj=1~∞ ν(E_j)

whenever the E_j and their union belong to the domain of ν.

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An extended real-valued set funciton ν is said to be monotone if ν(E)≦ν(F) whenever E, F are in the domain of ν and E⊂F.

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An "outer measure" is an extended real-valued set function μ having the following properties:

(i) The domain of μ consists of all the subsets of X (universal set).

(ii) μ is nonnegative.

(iii) μ is countably subadditive.

(iv) μ is monotone.

(v) μ(Ø)=0.

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measurable的定義:

Given an outer measure μ, we say that a set E is μ-measurable (or, briefly, measurable) if

μ(A)=μ(A∩E)+μ(A-E)

for any subset A of X.

metric function 和 metric space 的定義:

Suppose that there exists a real-valued function ρ defined for every ordered pair (x, y) of points of the universal set X, having the following properties:

(i) ρ(x, y) ≧ 0, and ρ(x, y) = 0 if and only if x = y.

(ii) ρ(x, y) = ρ(y, x) (symmetry).

(iii) ρ(x, z) ≦ ρ(x, y) + ρ(y, z) (the triangle inequality).

The function ρ is then called a metric function (or a distance function) on X, and the pair (X, ρ) is called a metric space.

open ball 的定義:

For any x∈X, ε > 0, the set

B(y, ε) = {y; ρ(x, y) < ε}

is called the open ball with center y and radius ε, or an ε-neighborhood of y.

開集合的定義:

A set E is called an open set if for any y∈E there is a ball B(y, ε) that is contained in E.

metric outer measure的定義:

Let X be a metric space with metric ρ. An outer measure μ* on X is called a metric outer measure if it satisfies the following property:

(vi) If ρ(A, B) > 0, then μ*(A∪B)=μ*(A)+μ*(B).

Update:

metric outer measure的相關定理:

Lemma: Let μ* be a metric outer measure and let A, B be any sets such that A⊂B, B open, μ*(A)

Update 2:

metric outer measure的相關定理:

Theorem: If μ* is a metric outer measure, then every closed set is measurable.

Corollary: If μ* is a metric outer measure, then every Borel set is measurable.

Update 3:

Lemma: Let μ* be a metric outer measure and let A, B be any sets such that A⊂B, B open, μ*(A)

Update 4:

嗯…那個Lemma不知為何後半都打不進來,現補充如下:

μ*(A)

Update 5:

因為只能輸入純文字,再補充一次:

μ*(A)小於無限大. For any positive integer n, let

A_n={x; x屬於A, ρ(x, B^c)大於等於1/n}. (B^c是B的補集)

Then

lim n趨近無限大 μ*(A_n)=μ*(A) .

1 Answer

Rating
  • Eric
    Lv 6
    1 decade ago
    Favorite Answer

    Lemma.  If A, B are subsets of X with ρ(A,B) > 0, then A ∩ closure(B) = ∅.Proof.  Suppose otherwise; then there would exist a sequence {bk} in B converging to b ∈ A ∩ closure(B). Then ρ(bk,b) → 0, contradicting the fact that ρ(A,B). ∎Proof of claim. Say A, B are subsets of X with ρ(A,B) > 0. Set E = X∖closure(B), which is open. Then E ∩ B = ∅, and in light of the lemma, we know that E contains A. Therefore,(A∪B)∩E = A,(A∪B)∖E = B.Since open sets are measurable,μ*(A∪B) = μ*((A∪B)∩E)+μ*((A∪B)∖E)= μ*(A) + μ*(B). ∎

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