※ 引述《k8k8 (k8)》之銘言：
: 書本是Zygmund 的measure and integral
: 問題如下
: 1.if E1 and E2 are measurable,E1 E2 為實數R裏的子集
: then E1 X E2 is measurable,and |E1 X E2|=|E1||E2|
: 2.if Z屬於R實數的一個子集 and |Z|=0,then {x^2|x為Z中元素} has zero measure.
2, 如果我沒記錯，大概就跟 MVT 有關系。
1, 是我以前證明的，希望對你有幫助：
Proof. Write E_1 ＝ H_1 ∪ Z_1, and E_2 ＝ H_2 ∪ Z_2, where
H_1 and H_2 are Fσ-sets. In fact,
∞ ∞
H_1 ＝ ∪ A_n and H_2 ＝ ∪ B_m,
n＝1 m＝1
where A_i and B_j are cpt. Let A_0 ＝ Z_1 and B_0 ＝ Z_2.
Then E_1 ╳ E_2 ＝ (H_1 ∪ Z_1) ╳ (H_2 ∪ Z_2)
∞ ∞
＝ (∪ A_n ) ╳ (∪ B_m)
n＝0 m＝0
∞
＝ ∪ (A_n ╳ B_m)
m,n＝0
If we can show that A_n ╳ B_m are measurable sets for all m, and n,
then we complete it. In order to show this, we consider two cases.
(1) A or B has measure zero, and (2) A and B are cpt.
For (2), it is clear that A ╳ B is cpt.
For (1), without loss of generality, say m(A)＝0, and m(B)＜∞.
Since m(A) ＝ 0, given ε＞0, there exists {I_i}, a family of n-dim
intervals such that
∞ ∞
A ≦ ∪ I_i and Σ m(I_i) ＜ ε.
i＝1 i＝1
Since m(B) ＜ ∞, there exists {J_j}, a family of n-dim intervals
such that
∞ ∞
B ≦ ∪ J_j, and Σ m(I_j) ＜ m(B) ＋ 1.
j＝1 j＝1
∞
So, A ╳ B ≦ ∪ I_i ╳ J_j. It implies that
i,j＝1
∞
m(A ╳ B) ≦ Σ m( I_i ╳ J_j )
i,j＝1
∞
＝ Σ m(I_i)．m(J_j)
i,j＝1
∞ ∞
≦ Σ m(I_i) Σ m(J_j)
i＝1 j＝1
＜ ε．(m(B) ＋ 1).
Let ε → 0, we finally get m(A ╳ B) ＝ 0.
By (1) and (2), we know that A ╳ B is measurable.
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