※ 引述《keroro321 (日夕)》之銘言：
: 這問題出自一道習題
: ( Exercise 9 pp.133 W.Rudin:"Real and complex analysis"3d ed.)
: Suppose that { Fn(x) } is a sequence of positive continuous functions
: on I=[0,1] ( Fn≧0 ) , that μ is a positive Borel measure on I, and
: that ( m : Lebesgue measure on I )
: (1) lim Fn(x)→0 a.e.[m]
: (2) ∫Fn(x)dm = 1 for all n
: (3) lim∫g Fn(x)dm =∫g dμ for every "continuous function g(x) on I"
: (All integrals are over I)
: Does it follow that
: μ nad m are mutually singular (μ⊥m )
: 我有一些例子,是符合上述條件而且
: μ nad m are mutually singular (μ⊥m )
: 所以我"猜測"應該是μ⊥m.
: 但是我還沒能證出我要的結果
: 請各位大大給一些提示,困擾了好久.....介.介
這應該是不對
Put I_n,k = [(k-1)/2^n,k/2^n], where k = 1,2,...,2^n.
Let Fn be a sequence of non-negative continuous functions defined on [0,1]
satisfying:
1-1/2^n
(a) Fn = 0 on [ 0, ────] ;
2^n
(b) ∫ Fn dm = 1/2^n;
I_n,k
(c) Fn(x + 1/2^n) = Fn(x), for all x with 0 ≦ x ≦ 1-1/2^n.
∞
m({x│Fn(x)!=0, for all n≧N.}) ≦ Σn=N 1/2^n →0 as N→∞.
Hence Fn→0 a.e. [m]. By (b) and (c), ∫ Fn dm = 1.Thus porterties
(1) and (2) are fullfilled. [0,1]
Next, for any continuous g, for any ε>0, there corresponds some N such that
whenever n>N, |g(x)-g(k/2^n)|<ε, for any x in I_n,k , for k=1,2,...,2^n.
Hence for each n > N,
1
│∫ Fn(x)g(x)-g(x)dx│
0
2^n
≦ Σ │∫Fn(x)g(x)-Fn(x)g(k/2^n) dx + ∫-g(x) + Fn(x)g(k/2^n) dx│
k=1 I_n,k I_n,k ───(A)────
(A) = g(k/2^n)/2^n
=∫g(k/2^n) dx
I_n,k
2^n
≦Σ ∫ Fn(x)│g(x)-g(k/2^n)│+│g(x)-g(k/2^n)│dx < 2ε.
k=1 I_n,k
Therefore,
1 1 1
lim ∫ Fn g dm = ∫ g dm = ∫ g dμ, which means m =μ.
n→∞ 0 0 0
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◆ From: 125.231.221.192
※ 編輯: ppia 來自: 125.231.211.133 (04/03 20:25)
※ 編輯: ppia 來自: 125.231.211.133 (04/03 21:49)