維正 發問時間: 科學數學 · 1 0 年前

一題測度論證明 (測度的總和仍是測度)

Let {u_n} be a sequence (finite or infinite) of measures defined on the same sigma-algebra A. Define sum n=1~無限大 u_n by

(sum n=1~無限大 u_n)(E) = sum n=1~無限大 u_n(E)

for every E 屬於 A. Prove that sum n=1~無限大 u_n is a measure.

註:

[Definiton] A measure is an extended real-valued set function u having the following properties:

(i) The domain A of u is a sigma-algebra.

(ii) u is nonnegative on A.

(iii) u is completely additive on A.

(iv) u(空集合)=0.

已更新項目:

第2點的 mu_n(E) = 0 是不是應該改成 >= 0?

1 個解答

評分
  • Eric
    Lv 6
    1 0 年前
    最佳解答

    Proof. I will verify the four conditions.The domain of ∑μn is A, which is a σ-algebra.If E ∈ A, then μn(E) = 0 for all n, so (∑μn)(E) = ∑μn(E) ≥ 0.If {Ek ∈ A} is a countable collection of disjoint sets, then μn(∪k Ek) = ∑k μn(Ek) for all n, so(∑n μn)(∪k Ek) = ∑n μn(∪k Ek) = ∑n (∑k μn(Ek))= ∑k (∑n μn(Ek))    (valid since the terms are nonnegative)= ∑k (∑n μn)(Ek)Therefore, ∑μn is completely additive.μn(∅) = 0 for all n, so (∑μn)(∅) = ∑μn(∅) = 0.Hence, ∑μn is indeed a measure.

    2006-07-25 01:04:42 補充:

    啊,不好意思

    第二點: μn(E) ≥ 0 而不是 = 0

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