推 sitma :萬分感謝....真神 04/20 10:35
※ 引述《sitma (sitma)》之銘言：
: m(A)>0,m(B)>0 =>
: 存在一個open interval (c,d)包含於R
: 使得（c，d)包含於A+B
Without losing generality , Assume that 0< m(A),m(B)<∞
證明 : A－B 包含某個區間 . (A,B:Lebesgue measurable)
用下面的lemma大概就是最少預備知識. (這就不證了)
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Lemma : Let E be a measurable subset of |R with m(E)>0 , then
(3.37) the set of differences E-E = {d:d=x-y | x,y屬於E } contains
an interval centered at origin .
R. L. Wheedem and A. Zygmund :"measure and integral," p.46
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Choose closed connected intervals I1,I2 so that
0 < (2/3) * m(I1) ＜ m(A1) (記 m(I1)=s , m(I2)=t )
0 < (2/3) * m(I2) ＜ m(B2)
where A1 = A∩I1 , B2 = B∩I2
( 這個選取是可以的 ,在上面Lemma證明中可以看出,當然也可以metric density來看)
Suppose that I1=[a,b] , t≦s and [s/t]= n . ([]:Gauss's symbol)
[a,b] = [a,a+t]∪[a+t,a+2t]∪..[a+nt,b] .....(1)
Claim : There exists some x 屬於 |R such that
m(A1 ∩ (B2+x))>0 ( B2+x ={ b+x | b 屬於B2 } )
Proof :
if not , ∵(1) ,then
s = m(I1) ≧ m(E1) + [s/t]* m(E2) ( n*t ≦ s ﹤(n+1)*t )
＞ (2/3)s + n*(2/3)t
> (2/3)s + (1/3)s ＝ s →←
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So for some x , we have m(A ∩ (B+x)) ≧ m(A1 ∩ (B2+x))>0
Let T = A ∩ (B+x) , then
A－B = ( (A－(B+x)) + x ) 包含 (T－T)+x
￣￣￣￣￣￣￣￣#
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W. Rudin :"Real And Complex Analysis," (3d ed.)
Chapter 7, Exercise 5
有提示2種方法(第一種基本上就是類似上面)
第二種是用convolution去做 (很簡潔)
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