Re: [分析] 實變

[ppia (papayaPaul)]

※ 引述《smartlwj (還有3間)》之銘言：
: 1. Let μ be a measure (on some measure space X) and let f_n, n=1,2,...
:    be a sequence  of real-valued measurable functions on X. Suppose
:    that ofr every ε>0, the sum
:                  ∞
:                  Σ μ{x| |f_n(x)|>ε} is finete.
:                  n=1
:    Prove that f_n converges to zero almost everywhere.
: 2. Let I=(0,1). Suppose {f_n} is a norm-bounded sequence of function in L^2(I)
:    that converges in measure to a function f.
:    (a) Show that f in L^2(I) and ║f║≦ liminf║f_n║
:                                      2               2
:    (b) Show that ║f_n║ converges to ║f║ if and only if ║f_n - f║ → 0.
:                        2                  2                          2
: 請問這兩題該怎麼做呢??
: 第一題我由題目可以知道 f_n 是依測度收斂到0
: 所以存在子列{f_nj}會幾乎處處收斂到0. 然後我想要証明出f_n →0 a.e.
: 由|f_n-0|≦|f_n-f_nj| + |f_nj|  然後我就不知道該怎麼做了
: 第二題的(a)沒想法
:         (b)的(<=)已証出
:              (=>)我想用LDCT 但是 我缺少了 f_n → f a.e.的條件
:                  而題目只有 f_n → f in measure.
:                  我該怎麼做呢??
: 不知道我目前的想法是否有問題？請指教 謝謝<(_ _)>

(a) Let f_nk be a subsequence of f_n such that

                     lim ║f_nk║ = lim inf║f_n║.
                     k→∞            n→∞

For any ε>0, put J_n = {x│|f(x)|>ε}. Since lim μ(J_n) = 0, there exists a
subesequence J_nki of J_nk such that

                          ∞
                          Σ μ(J_nki)≦1.
                          i=1

Put
                                    ∞
                          E_i = I - ∪  J_nks.
                                   s=i

{E_i} is then an increasing sequence of measurable sets with lim μ(E_i) = 1
and E_i ㄈ I-J_nki. Hence

          2             2                2              2      2           2
   ∫  |f|   ≦  ∫  |f|   ≦ ∫|f_nki-f|    + ∫|f_nki|   ≦ ε + ║f_nki║
    E_i         I-J_nki      I-J_nki          I-J_nki
                                                                   2
By the monotone convergence theorem, the first term approachs ║f║  as i goes
to infinity.


(b) (=>) For any ε>0, let J_ns be a subsequence of J_n such that
                          ∞
                          Σ μ(J_ns)≦1.
                          s=1

Put                     ∞
                 C_s =  ∪   J_nt and D_s = I- C_s.
                      t=s+1

           2           2            2           2              +  2
   ∫|f_ns|   = ║f_ns║ - ∫ |f_ns|  ≦ ║f_ns║ - ∫ [(|f|-ε) ]
   C_s                     D_s                      D_s

Hence
                            2          2            + 2
            lim sup ∫|f_ns|    ≦ ║f║ - ║(|f|-ε)║ .
              s→∞  C_s

         2             2             2        2     2           2       2
║f_ns-f║ ≦ ∫|f_ns-f|   +  ∫|f_ns|  + ∫|f| ≦ ε + ∫|f_ns|  + ∫|f|
              D_s             C_s         C_s           C_s         C_s


Therefore               2       2         2            + 2
       lim sup ║f_ns-f║   ≦ ε  +  ║f║ - ║(|f|-ε)║
        s→∞

Since ε was arbitrary, lim ║f_ns-f║= 0. Now that lim ║f_n║=║f║,
lim ║f_n-f║= 0.       s→∞                      n→∞
n→∞

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◆ From: 125.231.213.40
※ 編輯: ppia            來自: 125.231.221.157      (05/16 12:46)
推 math1209    :推…事實上,只要 conv in measure 則可取代逐點收斂
→ math1209    :這樣一來, 本題就顯得簡單許多了.
推 smartlwj    :推一下~感謝p大
→ ppia        :不懂m大的推文 可以解釋一下嗎?
推 math1209    :我的意思是說 舉凡 LDCT, MCT, Fatou's lemma 或者
→ math1209    :更廣義的 LDCT. 這些條件中的 幾乎處處逐點收斂
→ math1209    :被換成 依測度收斂, 其結論也是成立的.
→ ppia        :不知道是不是我會錯意 但是
→ ppia        :fn→f in measure =\=> fn→f a.e.
→ ppia        :抱歉 我沒看清楚推文...沒事了...