Define μ(E) as the number of points in E if E is finite and μ(E)=∞ if E is infinite. Show that μ is an outer measure. Determine the measurable sets.
An extended real-valued set function v is said to be \"countably subadditive\" if
v(∪j=1~∞ E_j) <= Σj=1~∞ v(E_j)
whenever the E_j and their union belong to the domain of v.
An extended real-valued set funciton v is said to be \"monotone\" if v(E)<=v(F) whenever E, F are in the domain of v and E包含於F.
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.
Given an outer measure μ, we say that a set E is \"μ-measurable\" (or, briefly, \"measurable\") if
for any subset A of X
- EricLv 61 0 年前最佳解答
In proving the following two claims, I will use the fact that counting measure is finitely additive for disjoint finite sets.Claim. μ is an outer measure on X.Proof. Let's verify the five conditions.(i) Domain.μ is defined on the domain of the collection of subsets of X.(ii) Nonnegativity.μ is clearly nonnegative.(iii) Countable subadditivity.Let En be a countable collection of subsets of X. There are two cases:If ∑n μ(En) = ∞, then μ(∪n En) ≤ ∞ = ∑n μ(En).Otherwise, ∑n μ(En) < ∞, so En must be finite for every n. Furthermore, μ(En) must be a nonnegative integer for every n, so there exists N ∈ N such that En = ∅ and μ(En) = 0 for all n > N.μ(∪n En) = μ(∪n≤N En) = ∑n≤N (μ(En) - μ((∪j≤n Ej)∖En))≤ ∑n≤N μ(En) = ∑n μ(En).(iv) Monotonicity.If E and F are subsets of X and E ⊆ F, there are two cases:If F is infinite, then μ(E) ≤ ∞ = μ(F).Otherwise, F is finite, and so is E. E has less elements than F, so μ(E) ≤ μ(F).(v) The empty set.
μ(∅) = 0 since ∅ is a finite set of 0 elements.We conclude that μ is an outer measure on X. ∎The measurable sets: which subsets E ⊆ X satisfyμ(A) = μ(A∩E) + μ(A∖E) (*)for all A ⊆ X?Claim. All subsets E of X are measurable.Proof. Let E and A be subsets of X.First, note that A∩E and A∖E are disjoint and A = (A∩E)∪(A∖E).By subadditivity, we haveμ(A) ≤ μ(A∩E) + μ(A∖E).Hence, it is sufficient to prove μ(A) ≥ μ(A∩E) + μ(A∖E).If A is infinite, then μ(A) = ∞, and so this inequality is already satisfied.Otherwise, A is finite, and so A∩E and A∖E are both finite. Therefore, μ(A) = μ(A∩E) + μ(A∖E).Since (*) has been showed for arbitrary subsets E and A of X, we conclude that all subsets of X are measurable. ∎