UA MATH523A 實分析3 積分理論15 乘積測度

一個不願透露姓名的孩子發表於2020-11-24

UA MATH523A 實分析3 積分理論15 乘積測度

從這一講開始,我們將逐步建立Lebesgue積分的重積分計算理論。考慮兩個可測空間 ( X , M ) (X,\mathcal{M}) (X,M) ( Y , N ) (Y,\mathcal{N}) (Y,N)的乘積可測空間 ( X × Y , M ⊗ N ) (X \times Y,\mathcal{M} \otimes \mathcal{N}) (X×Y,MN),這一講的目標是在這個乘積可測空間上建立測度。假設 μ \mu μ ν \nu ν分別是 ( X , M ) (X,\mathcal{M}) (X,M) ( Y , N ) (Y,\mathcal{N}) (Y,N)上的測度,一個最直接的想法就是通過 μ \mu μ ν \nu ν建立乘積空間上的測度。

在測度論基礎部分我們討論過建立測度一般性路徑,首先在可測空間中找一個集系,這個集系是一個elementary family,然後用elementary family匯出一個代數,並在代數上建立pre-measure,用pre-measure匯出外測度,限制在代數生成的 σ \sigma σ-代數上就是一個測度了。這一講我們要建立乘積空間上的測度,也可以按照這個路徑進行。


elementary family匯出代數,並用代數生成乘積Sigma代數
首先定義一個乘積空間 X × Y X \times Y X×Y上的集系,
Σ = { A × B : A ∈ M , B ∈ N } \Sigma = \{A \times B:A \in \mathcal{M},B \in \mathcal{N}\} Σ={A×B:AM,BN}

可以驗證, Σ \Sigma Σ是一個elementary family,我們來簡單論述一下。

註釋1 X X X上的集系 Σ \Sigma Σ是elementary family,如果

  1. ϕ ∈ Σ \phi \in \Sigma ϕΣ
  2. E , F ∈ Σ ⇒ E ∩ F ∈ Σ E,F \in \Sigma \Rightarrow E \cap F \in \Sigma E,FΣEFΣ
  3. E ∈ Σ ⇒ ∃ { A i } i = 1 n ⊂ Σ , E C = ⊔ i = 1 n A i E \in \Sigma \Rightarrow \exists \{A_i\}_{i=1}^n \subset \Sigma,E^C = \sqcup_{i=1}^nA_i EΣ{Ai}i=1nΣ,EC=i=1nAi

第一條顯然成立: ϕ = ϕ × ϕ ∈ Σ \phi = \phi \times \phi \in \Sigma ϕ=ϕ×ϕΣ
考慮第二條: A × B ∈ Σ A \times B \in \Sigma A×BΣ, C × D ∈ Σ C \times D \in \Sigma C×DΣ,
( A × B ) ∩ ( C × D ) = ( A ∩ C ) × ( B × D ) (A \times B) \cap (C \times D)=(A \cap C)\times(B \times D) (A×B)(C×D)=(AC)×(B×D)

因此第二條也成立。
考慮第三條: A × B ∈ Σ A \times B \in \Sigma A×BΣ,
( A × B ) C = ( X × B C ) ⊔ ( A C × B ) (A \times B)^C = (X \times B^C) \sqcup (A^C \times B) (A×B)C=(X×BC)(AC×B)

也就是兩個 Σ \Sigma Σ中的元素的無交併,因此第三條也成立。這樣我們就說明了 Σ \Sigma Σ是一個elementary family,根據Proposition 1.7(參考UA MATH523A 實分析2 測度論概念與定理整理), Σ \Sigma Σ中元素的有限無交併構成一個代數,記為 A \mathcal{A} A,則
A = { ⊔ i = 1 n E i : E i ∈ Σ , n ∈ N . , n < ∞ } \mathcal{A}=\{\sqcup_{i=1}^n E_i:E_i \in \Sigma,n \in \mathbb{N}.,n<\infty\} A={i=1nEi:EiΣ,nN.,n<}

下面我們論證一個命題: σ ( A ) = M ⊗ N \sigma(\mathcal{A})=\mathcal{M} \otimes \mathcal{N} σ(A)=MN
Note that A = { ∪ j = 1 n A j × B j : A j × B j ∈ Σ , n < ∞ } \mathcal{A}=\{\cup_{j=1}^nA_j \times B_j:A_j \times B_j \in \Sigma, n<\infty\} A={j=1nAj×Bj:Aj×BjΣ,n<}.

For any set in A \mathcal{A} A,
∪ j = 1 n A j × B j = ( ∪ j = 1 n A j × Y ) ∩ ( ∪ j = 1 n X × B j ) ∈ M ⊗ N \cup_{j=1}^nA_j \times B_j = (\cup_{j=1}^nA_j \times Y) \cap (\cup_{j=1}^nX \times B_j) \in \mathcal{M} \otimes \mathcal{N} j=1nAj×Bj=(j=1nAj×Y)(j=1nX×Bj)MN

Thus, A ⊂ M ⊗ N \mathcal{A} \subset \mathcal{M} \otimes \mathcal{N} AMN, by Lemma 1.1(參考UA MATH523A 實分析2 測度論概念與定理整理), M ( A ) ⊂ M ⊗ N \mathcal{M}(\mathcal{A}) \subset \mathcal{M} \otimes \mathcal{N} M(A)MN.

For any set in M ⊗ N \mathcal{M} \otimes \mathcal{N} MN, it could be A × Y A \times Y A×Y or X × B X \times B X×B where A ∈ M A \in \mathcal{M} AM and B ∈ N B \in \mathcal{N} BN. Obviously, both A × Y A \times Y A×Y and X × B X \times B X×B belongs to Σ \Sigma Σ. Thus M ⊗ N ⊂ M ( A ) \mathcal{M} \otimes \mathcal{N} \subset \mathcal{M}(\mathcal{A}) MNM(A).

建立乘積測度的第一部分就完成了,這一部分我們完成了找到一個elementary family,根據elementary family匯出代數,並論述這個代數生成的 σ \sigma σ-代數就是乘積 σ \sigma σ-代數。論述最後這個命題是為了我們最後建立的基於pre-measure匯出的測度定義域是elementary family匯出的代數生成的 σ \sigma σ-代數而不是乘積 σ \sigma σ-代數,所以這二者相同能夠保證我們基於pre-measure匯出的測度可以直接用在乘積空間上。

另外,之所以要找 Σ \Sigma Σ這個集系作為建立理論的起點,是因為它構造非常直觀。我們回顧一下從一維圖形到二維圖形的測度,比如線段和矩形,線段的測度是長度;矩形的測度是長乘寬等於面積,也就是兩個線段長度的乘積,而矩形本身也可以看成是這兩個線段的直積。基於這個觀察,我們建立乘積測度的思路就比較清晰了,因此對於 Σ \Sigma Σ集系中的元素,可以表示成兩個可測空間中的元素的直積, A × B A \times B A×B,因此它的測度或許也可以是這兩個元素的測度的乘積,即 μ ( A ) ν ( B ) \mu(A)\nu(B) μ(A)ν(B)。下面我們基於這些觀察建立嚴謹地理論。


A \mathcal{A} A上建立pre-measure,再匯出乘積測度

考慮 A × B A \times B A×B,它可以表示成一列rectangle的無交併,
A × B = ⊔ j = 1 ∞ ( A j × B j ) A \times B = \sqcup_{j=1}^{\infty} (A_j \times B_j) A×B=j=1(Aj×Bj)

用特徵函式表示也就是
χ A × B ( x , y ) = χ A ( x ) χ B ( y ) = χ ⊔ j = 1 ∞ ( A j × B j ) ( x , y ) = ∑ j = 1 ∞ χ A j ( x ) χ B j ( y ) \chi_{A \times B}(x,y) =\chi_A(x)\chi_B(y)= \chi_{\sqcup_{j=1}^{\infty} (A_j \times B_j)}(x,y) \\=\sum_{j=1}^{\infty} \chi_{A_j}(x)\chi_{B_j}(y) χA×B(x,y)=χA(x)χB(y)=χj=1(Aj×Bj)(x,y)=j=1χAj(x)χBj(y)

我們基於特徵函式討論測度的乘積,先做一些計算
μ ( A ) χ B ( y ) = χ B ( y ) ∫ χ A ( x ) d μ ( x ) = ∫ χ A ( x ) χ B ( y ) d μ ( x ) = ∫ ∑ j = 1 ∞ χ A j ( x ) χ B j ( y ) d μ ( x ) = ∑ j = 1 ∞ ∫ χ A j ( x ) χ B j ( y ) d μ ( x ) = ∑ j = 1 ∞ μ ( A j ) χ B j ( y ) \mu(A)\chi_B(y) = \chi_B(y) \int \chi_A(x)d\mu(x) = \int \chi_A(x)\chi_B(y)d\mu(x) \\ = \int \sum_{j=1}^{\infty} \chi_{A_j}(x)\chi_{B_j}(y)d\mu(x) =\sum_{j=1}^{\infty} \int \chi_{A_j}(x)\chi_{B_j}(y)d\mu(x) \\ =\sum_{j=1}^{\infty} \mu(A_j)\chi_{B_j}(y) μ(A)χB(y)=χB(y)χA(x)dμ(x)=χA(x)χB(y)dμ(x)=j=1χAj(x)χBj(y)dμ(x)=j=1χAj(x)χBj(y)dμ(x)=j=1μ(Aj)χBj(y)

倒數第二個等號用了Folland 定理2.15,當被積函式列有界時可以交換級數與積分的次序。
μ ( A ) ν ( B ) = ∫ μ ( A ) χ B ( y ) d ν ( y ) = ∫ ∑ j = 1 ∞ μ ( A j ) χ B j ( y ) d ν ( y ) = ∑ j = 1 ∞ ∫ μ ( A j ) χ B j ( y ) d ν ( y ) = ∑ j = 1 ∞ μ ( A j ) ν ( B j ) \mu(A)\nu(B) = \int \mu(A)\chi_B(y)d\nu(y)=\int\sum_{j=1}^{\infty} \mu(A_j)\chi_{B_j}(y)d\nu(y) \\ = \sum_{j=1}^{\infty} \int\mu(A_j)\chi_{B_j}(y)d\nu(y) = \sum_{j=1}^{\infty}\mu(A_j)\nu(B_j) μ(A)ν(B)=μ(A)χB(y)dν(y)=j=1μ(Aj)χBj(y)dν(y)=j=1μ(Aj)χBj(y)dν(y)=j=1μ(Aj)ν(Bj)

這個結果非常有意思,它想說明的就是我們熟悉的計算面積體積的割補法同樣適用於測度的乘積。基於這個結果,我們可以在 A \mathcal{A} A上定義pre-measure, π : A → [ 0 , ∞ ] \pi:\mathcal{A} \to [0,\infty] π:A[0,]
π ( ⊔ j = 1 n ( A j × B j ) ) = ∑ j = 1 n μ ( A j ) ν ( B j ) \pi(\sqcup_{j=1}^n (A_j \times B_j))=\sum_{j=1}^n \mu(A_j)\nu(B_j) π(j=1n(Aj×Bj))=j=1nμ(Aj)ν(Bj)

下面驗證一下這個定義確實是pre-measure(證明細節如下,分兩部分論述良定義與pre-measure,可以參考一下,但不影響理解建立乘積測度的整體思路)
First of all, check π \pi π is well-defines. For any set in A \mathcal{A} A, there’re two possible representations
∪ j = 1 n A j × B j = ∪ k = 1 m C k × D k \cup_{j=1}^n A_j \times B_j=\cup_{k=1}^m C_k \times D_k j=1nAj×Bj=k=1mCk×Dk

Need to check the two representations will be mapped to the same value under π \pi π.
π ( ∪ j = 1 n A j × B j ) = ∑ j = 1 n μ ( A j ) ν ( B j ) = μ ( ⊔ j = 1 n A j ) ν ( ⊔ j = 1 n B j ) \pi(\cup_{j=1}^n A_j \times B_j)=\sum_{j=1}^n \mu(A_j)\nu(B_j) = \mu(\sqcup_{j=1}^nA_j)\nu(\sqcup_{j=1}^n B_j) π(j=1nAj×Bj)=j=1nμ(Aj)ν(Bj)=μ(j=1nAj)ν(j=1nBj) π ( ∪ k = 1 m C k × D k ) = ∑ k = 1 m μ ( C k ) ν ( D k ) = μ ( ⊔ k = 1 m C k ) ν ( ⊔ k = 1 m D k ) \pi(\cup_{k=1}^m C_k \times D_k)=\sum_{k=1}^m \mu(C_k)\nu(D_k) = \mu(\sqcup_{k=1}^mC_k)\nu(\sqcup_{k=1}^m D_k) π(k=1mCk×Dk)=k=1mμ(Ck)ν(Dk)=μ(k=1mCk)ν(k=1mDk)

Note that ⊔ k = 1 m C k = ⊔ j = 1 n A j \sqcup_{k=1}^mC_k=\sqcup_{j=1}^nA_j k=1mCk=j=1nAj, ⊔ k = 1 m D k = ⊔ j = 1 n B j \sqcup_{k=1}^m D_k=\sqcup_{j=1}^n B_j k=1mDk=j=1nBj, so
π ( ∪ j = 1 n A j × B j ) = π ( ∪ k = 1 m C k × D k ) \pi(\cup_{j=1}^n A_j \times B_j)=\pi(\cup_{k=1}^m C_k \times D_k) π(j=1nAj×Bj)=π(k=1mCk×Dk)

Next, let’s show π \pi π is pre-measure.

Note that
π ( ϕ ) = π ( ϕ × ϕ ) = μ ( ϕ ) ν ( ϕ ) = 0 \pi(\phi) = \pi(\phi \times \phi) = \mu(\phi) \nu(\phi)=0 π(ϕ)=π(ϕ×ϕ)=μ(ϕ)ν(ϕ)=0

So we need to check additivity. For a disjoint sequence of set { ∪ j = 1 n i A j i × B j i } i = 1 ∞ \{\cup_{j=1}^{n_i}A_j^{i} \times B_j^i\}_{i=1}^{\infty} {j=1niAji×Bji}i=1, we want to show
π ( ⊔ i = 1 ∞ ( ∪ j = 1 n i A j i × B j i ) ) = ∑ i = 1 ∞ π ( ∪ j = 1 n i A j i × B j i ) \pi(\sqcup_{i=1}^{\infty}(\cup_{j=1}^{n_i}A_j^{i} \times B_j^i))=\sum_{i=1}^{\infty}\pi(\cup_{j=1}^{n_i}A_j^{i} \times B_j^i) π(i=1(j=1niAji×Bji))=i=1π(j=1niAji×Bji)

Given i i i, we can find a disjoint sequence of rectangles { C k i } k = 1 m i \{C_k^i\}_{k=1}^{m_i} {Cki}k=1mi such that ∪ j = 1 n i A j i = ⊔ k = 1 m i C k i \cup_{j=1}^{n_i}A_j^{i}=\sqcup_{k=1}^{m_i}C_k^{i} j=1niAji=k=1miCki and a disjoint sequence of rectangles { D l i } l = 1 r i \{D_l^i\}_{l=1}^{r_i} {Dli}l=1ri such that ∪ j = 1 n i B j i = ⊔ l = 1 r i D l i \cup_{j=1}^{n_i}B_j^{i}=\sqcup_{l=1}^{r_i}D_l^{i} j=1niBji=l=1riDli and also
∪ j = 1 n i A j i × B j i = ⊔ k = 1 m i ⊔ l = 1 r i C k i × D l i \cup_{j=1}^{n_i}A_j^{i} \times B_j^i = \sqcup_{k=1}^{m_i}\sqcup_{l=1}^{r_i}C_k^i \times D_l^i j=1niAji×Bji=k=1mil=1riCki×Dli

Thus,
π ( ⊔ i = 1 ∞ ( ∪ j = 1 n i A j i × B j i ) ) = π ( ⊔ i = 1 ∞ ( ⊔ k = 1 m i ⊔ l = 1 r i C k i × D l i ) ) = π ( ⊔ i = 1 ∞ ( ⊔ k = 1 m i C k i ) × ( ⊔ l = 1 r i D l i ) ) = ∑ i = 1 ∞ μ ( ⊔ k = 1 m i C k i ) ν ( ⊔ l = 1 r i D l i ) = ∑ i = 1 ∞ μ ( ∪ j = 1 n i A j i ) ν ( ∪ j = 1 n i B j i ) = ∑ i = 1 ∞ π ( ∪ j = 1 n i A j i × B j i ) \pi(\sqcup_{i=1}^{\infty}(\cup_{j=1}^{n_i}A_j^{i} \times B_j^i)) = \pi(\sqcup_{i=1}^{\infty}(\sqcup_{k=1}^{m_i}\sqcup_{l=1}^{r_i}C_k^i \times D_l^i)) \\ = \pi(\sqcup_{i=1}^{\infty}(\sqcup_{k=1}^{m_i}C_k^i) \times (\sqcup_{l=1}^{r_i}D_l^i)) = \sum_{i=1}^{\infty}\mu(\sqcup_{k=1}^{m_i}C_k^i)\nu(\sqcup_{l=1}^{r_i}D_l^i) \\ = \sum_{i=1}^{\infty}\mu(\cup_{j=1}^{n_i}A_j^{i})\nu(\cup_{j=1}^{n_i}B_j^{i}) =\sum_{i=1}^{\infty}\pi(\cup_{j=1}^{n_i}A_j^{i} \times B_j^i) π(i=1(j=1niAji×Bji))=π(i=1(k=1mil=1riCki×Dli))=π(i=1(k=1miCki)×(l=1riDli))=i=1μ(k=1miCki)ν(l=1riDli)=i=1μ(j=1niAji)ν(j=1niBji)=i=1π(j=1niAji×Bji)

寫出pre-measure匯出的外測度,
μ ∗ ( A ) = inf ⁡ { ∑ j = 1 ∞ π ( E j ) : E j ∈ A , A ⊂ ∪ j = 1 ∞ E j } \mu^*(A)=\inf\{\sum_{j=1}^{\infty}\pi(E_j):E_j \in \mathcal{A},A \subset \cup_{j=1}^{\infty}E_j\} μ(A)=inf{j=1π(Ej):EjA,Aj=1Ej}

根據Proposition 1.14 (參考UA MATH523A 實分析2 測度論概念與定理整理), μ ∗ ∣ σ ( A ) \mu^*|\sigma(\mathcal{A}) μσ(A)是一個測度,我們記這個測度為 μ × ν \mu \times \nu μ×ν,稱其為乘積測度,這個測度滿足
( μ × ν ) ( A × B ) = μ ( A ) ν ( B ) (\mu \times \nu)(A \times B) = \mu(A)\nu(B) (μ×ν)(A×B)=μ(A)ν(B)

值得引起注意的是這個建立乘積測度的路徑可以推廣到多個可測空間的乘積。


最後我們再說明一個簡單性質:如果 μ , ν \mu,\nu μ,ν σ \sigma σ-有限測度,則 μ × ν \mu \times \nu μ×ν σ \sigma σ-有限測度。

如果 μ , ν \mu,\nu μ,ν σ \sigma σ-有限測度, X = ∪ j = 1 ∞ A j , Y = ∪ k = 1 ∞ B k X=\cup_{j=1}^{\infty} A_j,Y=\cup_{k=1}^{\infty}B_k X=j=1Aj,Y=k=1Bk, μ ( A j ) < ∞ , ∀ j \mu(A_j)<\infty,\forall j μ(Aj)<,j, ν ( B k ) , ∀ k \nu(B_k),\forall k ν(Bk),k,則
X × Y = ∪ j , k ( A j × B k ) X\times Y = \cup_{j,k}(A_j \times B_k) X×Y=j,k(Aj×Bk)

並且 ( μ × ν ) ( A j × B k ) = μ ( A j ) ν ( B k ) < ∞ (\mu \times \nu)(A_j \times B_k)=\mu(A_j)\nu(B_k)<\infty (μ×ν)(Aj×Bk)=μ(Aj)ν(Bk)<,所以 μ × ν \mu \times \nu μ×ν σ \sigma σ-有限測度。

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