※ 引述《math1209 (人到無求品自高)》之銘言：
: ※ 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…]
請教各位先進, 我這樣做可以嗎?
since |A|, |B|>0, there exist open set, G1, G2 containing A, B resp.
s.t |G1|< 3/2 |A|, |G2|< 3/2 |B|
write G1 as a disjoint union of open intervals I_n
there exist N s.t |I_N|<3/2 |I_N∩A|
Similarly, we have open interval J_K s.t. |J_K|< 3/2 |J_K∩B|
Moreover we can ask |I_N|=|J_K|=L
For convenice, I_N=I, J_K=J
I_N∩A=A , J_K∩B=B
hence |A|>2/3 |I|, |B|>2/3 |J|
and say I, J are centered at x. y, x<y
let I', A' be the translated set I ,A by y-x,
claim A'+d intersects B as nonempty for any |d|<L/6
otherwise, A'+d ∪ B is containded in a interval with length L+|d|
i.e |A'+d ∪ B|<L+|d| (1)
on the other hand |A'+d ∪ B|= |A'+d |+|B|=|A|+|B|> 4/3 L (2)
Hence we have a contracdition (1) and (2)
since A'+d=A+(y-x)+d intersects B as nonempty, we are done!
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