Re: [分析] [分析]測度一題
※ 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)
────────
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
────────
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 →←
--------------------
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
 ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄#
───────────
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)
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 59.112.238.160
推
04/20 10:35, , 1F
04/20 10:35, 1F
討論串 (同標題文章)