推 jacky7987 :長的蠻像LDCT的XD 11/19 20:51
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※ 編輯: znmkhxrw 來自: 1.171.7.198 (11/27 00:59)
※ 引述《kkk3388 (haha)》之銘言：
: 我想三天了
: 拜託版上高手 解答一下
: (1){g_n} be integrable on E and positive ,
: which converges to an integrable function g
: (2){f_n} be measurable functions on E such that |f_n|<g_n，
: and fn converges to f.
: If lim Sgn = Sg， then Sf=limSfn
: n->infinity ~(->這個符號是積分)
: <hint>use fatou`s lemma
: 這是中山大學2011博班考題
: 萬分感激 謝謝
│f_n│＜ g_n => -g_n ＜ f_n ＜ g_n
=> { 0 ＜ f_n + g_n ＜ 2g_n ---(1)
{ 0 ＞ f_n - g_n ＞ -2g_n ---(2)
by fatou's lemma
for (1), ∫liminf (f_n + g_n) ≦ liminf∫(f_n + g_n) ---(1')
~~~~~~~~~~~~~~~~~~~~
↓(equals)
∫(f + g)
for (2), ∫limsup (f_n - g_n) ≧ limsup∫(f_n - g_n) ---(2')
~~~~~~~~~~~~~~~~~~~~
↓(equals)
∫(f - g)
since ∫f_n is bounded
by <lemma>(see p.s.)
(1') = liminf∫(f_n + g_n) = (liminf∫f_n) +∫g
(2') = limsup∫(f_n - g_n) = (limsup∫f_n) -∫g
Hence ∫(f + g) ≦ (liminf∫f_n) +∫g
=>∫f ≦ liminf∫f_n
and ∫(f - g) ≧ (limsup∫f_n) - ∫g
=>∫f ≧ limsup∫f_n
Hence lim∫f_n = ∫f #
<lemma> if a_n is bounded, b_n is convergent sequence with finite value L
then liminf(a_n + b_n) = (liminf a_n) + L
and limsup(a_n + b_n) = (limsup a_n) + L
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有錯請指證~
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◆ From: 1.171.8.55
※ 編輯: znmkhxrw 來自: 1.171.8.55 (11/19 18:50)