→ wuxr :M 大, 我有一個疑問, I1, I2 為什麼可以取到同樣的 05/11 14:55
→ wuxr :length ? 05/11 14:56
推 smartlwj :推ㄧ下convolution 很妙的証法 05/11 15:01
※ 引述《wuxr (wuxr)》之銘言：
: 請教各位先進
: Let A, B are measuralbe set in R
: and mA, mB >0
: Show that the set {a-b |a in A, b in B } contains an open interval
: where m denotes the Lebesgue measure.
(方法 1)
Let E_1 and E_2 be Lebesgue measurable subsets in |R with measure
|E_1|＞0 and |E_2|＞0. Then
E_1 － E_2 ≧ I, I is an interval.
Hint. Vitali lemma and almost every point in a measurable subset is a
point of density.
NOTE. In general, this is true for case in |R^n.
Proof.
Let a and b be the points of density of E_1 and E_2. By definition, there
exist intervals I_1 and I_2 centered at a and b, respectively with length
is 2d so that |E_j∩I_j|≧(3/2)d, j＝1,2. For convenience, E_j∩I_j:＝F_j,
j ＝1,2.
Then |F_1|≧(3/2)d and |F_2|≧(3/2)d. (＊）
F_1 F_2
---------(----a----)-------------------(----b----)---------
Consider the translate F_2 by a － b, denoted by F_3.
If |F_1∩F_3| ＝ 0, then by (＊）|F_1∪F_3|＝|F_1|＋|F_3| ≧ 3d (a).
However, F_1 ∪ F_3 is contained in I_1. It implies that
|F_1∪F_3| ≦ |I_1| ＝ 2d. (b)
From (a) and (b), we get a contradiction. So, |F_1∩F_3：＝ A|＞0.
Hence, we get
(－δ,δ) ≦ A － A
by Vitali's lemma. So, E_1 － E_2 ≧ I, I is an interval. □
(方法 2) 利用 convolution. [有需要再說…]
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※ 編輯: math1209 來自: 114.32.219.116 (05/11 14:08)