維正 發問時間: 科學數學 · 1 0 年前

一題外測度問題 (有關外測度的描述) part 2

Let the universal set X be a noncountable space. Let K consist of X, Ø, and all the one-point sets. Let λ(X)=1, λ(E)=0 if E≠X. Describe the outer measure.

註:

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屬於K, and

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

這個定理或許用得到:

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.

已更新項目:

請問一下,什麼是potential covers呢?

1 個解答

評分
  • Eric
    Lv 6
    1 0 年前
    最佳解答

    Claim. μ(E) = 1 if E is uncountable, μ(E) = 0 if E is countable (finite or countably infinite).Proof.The empty set.As for all outer measures, μ(Ø) = 0.Countable sets.If A is a countable subset of X and (an) is an enumeration of the elements in A, consider the cover of A defined by En = {an} for each n. If A is finite, define En = Ø for n > |A|. Clearly, ∑n λ(En) = 0. Hence, μ(A) = 0.Uncountable sets.If A is an uncountable set, consider the potential covers {En} of A. Clearly, Ek = X for some k, for otherwise a countable collection of finite sets cannot cover A. Also, ∑n λ(En) = 1 + ∑n≠k λ(En) ≥ 1, and the left hand side is minimized when En = Ø for n ≠ k. Hence, μ(A) = 1. ∎

    2006-08-09 00:23:52 補充:

    potential: 潛在的, 可能的

    cover: 覆蓋, a collection of sets {En} such that A⊆∪En

    If A is an uncountable set, consider the potential covers {En} of A.

    若 A 為不可數集合,我們來考慮有可能覆蓋 A 的{En}。

    那句可以換成

    If A is an uncountable set, let {En}⊆K be a cover of A; that is, A⊆∪En.

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