※ 引述《ppia ((= =))》之銘言：
: ※ 引述《keroro321 (日夕)》之銘言：
:
感謝你的回應!!我看了你舉的例子!!,但是有個地方的估算應該是錯了.
: 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
: 1/2^n
: (b) ∫ Fn dm = 1/2^n;
: 0
: (c) Fn(x + 1/2^n) = F(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 with
: k k-1 k
: |g(x)-g(──)|<ε, for any x in [──, ──] , for k=1,2,...,2^n, for all n>N.
: 2^n 2^n 2^n
: Hence for each n > N,
: 1
: ∫ │Fn(x)g(x)-g(x)│dx ≦
: 0
: 2^n k/2^n
: Σ ∫ Fn(x)│g(x)-g(k/2^n)│+│g(x)-g(k/2^n)│dx < 2ε.
: k=1 (k-1)/2^n
######
: 在這兒你的估算不對 Fn(x)│g(x)-g(k/2^n)│+ Fn(x)g(k/2^n) +........
少了 | Fn(x)g(k/2^n)-g(k/2^n)│
這項經過積分加總後影響可是非常大,上述估計沒估到這項 @@
######
Therefore,
: 1 1 1
: lim ∫ Fn g dm = ∫ g dm = ∫ g dμ, which means m =μ.
: n→∞ 0 0 0
還是感謝你的回應!!!!
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