[分析] Limits of Measureable Functions

[ppia (papayaPaul)]

在　Daniell integral ( http://en.wikipedia.org/wiki/Daniell_integral ) 中,
可測函數定義有以下兩種定義﹔

(a) f is measureable iff for any positive integrable g
           +           -
      min(f ,g) - min(f ,g) = mid(g,f,-g)  is  integrable.

(b) f is measureable iff f is a limit a.e.
                    of a sequence of elementary functions.

一般來說(b)=>(a)但反之不亦然, 在下列條件底下, (a)=>(b)

(*)  There exists an increasing sequence of elementary functions φ_n with

                         lim  φ_n > 0 a.e..
                         n→∞

我要問的是:
 "(a.e.) limits of sequences of measureable functions is also measureable."

這件事對(a)成立,那麼一般來說, 在(*)不成立的狀況下, 上述是否也對(b)成立?

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自問自答... 答案是肯定的

     Let s_n be a sequence of measureable(b) functions tending a.e. to a
function f. Then f is a measureable function(a). Besides, since each s_n is
a.e. the limit of a sequence of elementary functions {σ_mn︱m=1,2,...},
if we let σ_n to be an enumeration of the double sequence {|σ_mn|} and let
φ_n = max(σ_1, σ_2, ...,σ_n),

                     lim  φ_n  > 0 a.e. on E:= supp(f).
                     n→∞


     Put f_n = mid( nφ_n, f, -nφ_n ) → f a.e. and g_n = f_[n+1] - f_n. Then
each g_n is integrable with Σg_n = f a.e.. Since each g_n is integrable, there
corresponds a sequence of elementary functions {ψ_kn︱k=1,2,...} that tends to
g_n a.e. with
                                               n+k
                      ∫|ψ_kn-g_n| dμ < 1 / 2   .

Put ψ_n = ψ_n1 + ψ_n2 + ... +ψ_nn. It remains to show that ψ_n→f a.e..


     Let E be a measure zero set off which
                 ∞
                 Σ  g  = f  and    lim  ψ   = g    hold.
                 n=1  n             k→∞  kn    n

Put E_n = {x︱|f(x)| < nφ_n} - E. Fix some N,  for each x in E_N, g_n(x) = 0
for all n > N. Hence

     ψ (x) + ψ (x) + ... + ψ (x)
       n1      n2             nN

                         —→ g(x)  + g(x) + ... + g(x) = f(x) as n →∞.
                               1       2            N


Put ζ_n = ψ_n - (ψ_n1 + ψ_n2 + ... +ψ_nN).

   │ζ_n│ = │ψ_n N+1 + ... +ψ_nn│ ≦ │ψ_n N+1│ + ... +│ψ_nn│

Since g_n = 0 on E_N for n>N,

        ∫ │ζ_n│dμ ≦ ∫  │ψ    - g  │+ ... + │ψ - g│dμ
         E_N               E_N   n N+1   N+1            nn   n
                                                      n
     ≦ ∫│ψ    - g  │+ ... + │ψ - g│dμ  ≦ 1/2
             n N+1   N+1            nn   n


By Beppo Levi Lemma, Σζ_n converges absolutely a.e. on E_N, which implies
ζ_n→0 a.e. on E_N. Hence ψ_n→f a.e. on each E_N.
※ 編輯: ppia            來自: 125.231.229.38       (06/27 22:22)
推 Vulpix      :為什麼g_n可積啊？
→ ppia        :因為f_n可積
推 Vulpix      :推這篇，Daniel int.也是一個切入實分析的好方向
→ ppia        :很少實分析課本用non-measure approach來介紹積分
→ ppia        :Daniell integral放在泛函理論裡面會顯得比較自然
推 Vulpix      :當然一開始就從可積分切入不太自然，可是我覺得
推 Vulpix      :能用其他角度看實分析很重要，說明不是只有從measure
→ Vulpix      :開始這唯一一條路
推 XCHCH2      :那這跟一般測度論裡面的可測函數定義等價嗎?
推 XCHCH2      :f是可測函數iff f^-(a,∞)是可測集,for all a
推 Vulpix      :是等價的，不過看起來一開始選的基本函數積分不是
→ Vulpix      :習慣的積分的話，也可以構造出其他測度耶
→ ppia        :其實滷乙烯大大的問題比較不那麼直接
→ ppia        :這個問題的癥結在於 兩種理論給的可測集是否一樣
→ ppia        :如果是以(a)當作可測函數的定義的話 那不必然
→ ppia        :如果是(b)的話 那我不是很肯定
→ ppia        :給定elementary sets 我們可以定義elementary fx's為
→ ppia        :charateristic fx of elementary set的線性組合
→ ppia        :如此構造的理論'可測'等價  至少是用(b)作為定義時
→ ppia        :然而若先給定elementary fx 若elementary fx非
→ ppia        :simple fx 要如何定義elementary set 這我就沒想過了
推 Vulpix      :我是看到原po給的wiki連結才這樣說的，裡面提到說
→ ppia        :簡單的說 若ele. fx 非 simple fx. 這個"等價"到底
→ Vulpix      :measure in Daniel sense <=> usual measure
→ ppia        :是什麼意思 並不是很清楚
→ ppia        :那應該是在比較特殊狀況 比如說Lebesgue積分時
推 XCHCH2      :thanks 還第一次有人叫我滷乙烯XD