測度收斂但極限不存在

Question

Define functions f_m^n(x) on [0, 1) by f_m^n(x)=1 if x in [(m-1)/n, m/n) and f_m^n(x)=0 otherwise. Enumerate the f_m^n in any way, and call the sequence thus obtained {phi_j}. For instance, phi_1=f_1^1, phi_2=f_2^1, phi_3=f_1^2, phi_4=f_3^1, .... Prove that {phi_j} converges in measure to zero, but lim phi_j(x) does not exist for any x in [0, 1).

註：f_m^n代表函數f後面接著下標為ｍ，上標為ｎ。

Answer 1

你的題目根本就有問題:
因為 {Φ4= 1}= [ 2/1, 3/1)=[2,3) 不在 [0,1)裡面 !

m( {Ψi≠0} )= m( [ (m-1)/n, m/n) ) = 1/n -> 0 as n->oo
而且 [ (m-1)/n, m/n ) 包含於 [ 0, 1)
所以 1 ≦ m ≦ n.
故為 (1,1) -> (1,2)->(2,2)->(1,3)->(2,3)->(3,3)->(1,4)->.........

1. For each 1 > ε> 0 be given, there is a postive integer M such that 1/N<ε.
Let M> N(N+1), then
m( { |Ψi - 0 | ≧ ε) = m ( {Ψi≠0} ) < 1/N <ε if i ≧ M.
By def, Ψi -> 0 in measure.

2. For each x in [0, 1) and postive integer n, there is an postive integer Mn such that 0≦Mn≦n and x in [(Mn-1)/n, Mn/n).
(Because [(1-1)/n, 1/n)∪......∪[(n-1)/n, n/n)=[0,1) )
Let f(n)= Mn + n(n-1)/2 for all potive integer i.
Then {Ψf(n)} is a subsequence of Ψi and Ψf(n) (x) = 1 for all n.
It implies Ψf(n) -> 1 as n->oo
Similar, there is an postive integer Ln such that 0≦Ln≦n and x is not in [(Ln-1)/n, Ln/n).
Let g(n)= Ln + n(n-1)/2 for all potive integer i.
Then {Ψg(n)} is a subsequence of Ψi and Ψg(n) (x) = 0 for all n.
It implies Ψg(n) -> 0 as n->oo
So we have lim Ψi(x) does not exist for any x in [0, 1).

2006-12-01 20:39:29 補充：
Similar, there is an postive integer Ln such that 0≦Ln≦n+1 and x is not in [(Ln-1)/(n+1), Ln/(n+1)).Let g(n)= Ln + n(n+1)/2 for all potive integer i.Then {Ψg(n)} is a subsequence of Ψi and Ψg(n) (x) = 0 for all n.It implies Ψg(n) -> 0 as n->oo

2006-12-01 20:41:47 補充：
要修改的原因是Ψ1(x)=1 for all x in [0,1).當n=1時會找不到對應的M1.