Lebesgue D1D0
BOB3BS
BQB9AFCEC4CECA
http://math.nju.edu.cn/?meijq
BZ(X,ρ)C8ALBJBEB5,A? X,AKCHAAICSBAC8
d(A) = sup{ρ(a1,a2)|a1,a2 ∈ A}.
XCUAICVB2SA8C8CDBLB8AI,BYAYSCUBWB0AJBLBBCMC3BICVBL,BTAUCV
BLB1CBBXCOS CU.BGBY,R
n
CUCMB7A4B2C2CDBLB8AI.
CWCX (LebesgueCJBF),BZS C8 (X,ρ) CUAICDBLB8B2,{Gα}α∈Γ C8S
AICGAWBCATAV,CP?CO λ > 0,C1AHAG S CUCVB2 A AICSBA d(A) CCCN λ
C0,ACGAKA2AZCNBPAW Gα BR.
CZCY,(CKAQCRAP),B4BZB6BMA6A9BH,CPAM?n ≥ 1,?An? S,C1
AH d(A) <
1
n,AO An A6C7BVA2AZCNBPAW Gα BR,CO An CUBUAJ an,CP
{an} C8 S CUAJBL,CLCN S C2CDBLB8CTAI,C6CMC3BICVBL {ank},C1AH
limn→+∞ank = a0 ∈ S,CLCN {Gα}α∈Γ C8 S AICGAWBCATAV,AX?CO α0 ∈ Γ,
C1AHa0 ∈ Gα0,CIAC,?δ > 0,s.t Ba0(δ)? Gα0.
CLCN limn→+∞ank = a0,AXBDBUAAARAFAI nk0 > 2δ,BT
ρ(ank0,a0) < δ2.
ACC0,AM?x ∈ Ank0,CM
A0C5CFASC9A1A5ABA7BK,2006.3
1
ρ(x,a0) ≤ ρ(x,ank0) + ρ(ank0,a0)
≤ d(Ank0) + ρ(ank0,a0)
< 1n
k0
+ δ2
< δ2 + δ2 = δ.
ADAOAnk0? Ba0(δ)? Gα,CQC2CGAWBNAN! CRA3.
2