# 一題完全測度問題 (和零測度集有關)

Let μ be a complete measure. A set for which μ(N)=0 is called a null set. Show that the class of null sets is a σ-ring. Is it also a σ-algebra?

complete measure的定義：

A measure μ with domain A is said to be complete if for any two sets N, E the following holds: If N⊆E, E∈A and μ(E)=0, then N∈A.

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Theorem 1: Let μ be a measure on a σ-algebra A and let \\bar{A} denote the class of all sets of the form E∪N, where E∈A and N is any subset of a set of A of measure zero. Then \\bar{A} is a σ-algebra and the set function \\bar{μ} defined by

\\bar{μ}(E∪N)=μ(E) (1)

is a complete measure on \\bar{A}.

We call \\bar{μ} the completion of μ.

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σ-ring的定義：

A class R of sets is called a σ-ring if it has the following properties:

(a) Ø∈R.

(b) If A∈R and B∈R, then A-B∈R.

(c) Any countable union of sets A_n (n = 1, 2, ...) which belong to R also belongs to R.

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σ-algebra的定義：

A class R of sets is called a σ-algebra if it has the following properties:

(a) Ø∈R.

(b) If A∈R and B∈R, then A-B∈R.

(c) Any countable union of sets A_n (n = 1, 2, ...) which belong to R also belongs to R.

(d) X∈R.

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measure的定義：

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Let μ be a set function defined on a ring R. We say that μ is \"completely additive\" if

μ(∪n=1~∞ E_n) = Σn=1~∞ μ(E_n)

whenever the E_n are mutually disjoint sets of R such that ∪n E_n is also in R.

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

(i) The domain A of μ is a σ-algebra.

(ii) μ is nonnegative on A.

(iii) μ is completely additive on A.

(iv) μ(Ø)=0.

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### 1 個解答

• 最佳解答

Lemma.  Measures on σ-algebras are countably subadditive.Claim.  The class of null sets forms a σ-ring.Proof.  Let's verify the three criteria of a σ-ring:The empty set.μ(Ø) = 0.Set differences.If A and B are null sets, observe that A∖B ⊆ A, so by the completeness of μ, A∖B is also a null set. Countable unions.Let {An}n∈N be a countable collection of null sets. By subadditivity,0 ≤ μ(∪n An) ≤ ∑n μ(An) = 0and henceμ(∪n An) = 0.  ∎Does the class of null sets form a σ-algebra?In general, no—unless μ(X) = 0, which is only the case when μ = 0.