[數學普及] Measure Theory (測度論) 簡介 ver. 2

整緊個 blog , 順手改下D小錯, 兩邊會同步更新

4.2個part有D唔明，

"(i) 如果 E_1 ⊆ E_2 ⊆ E_3 ⊆ ... ⊆ E_n ⊆ ...,


依度點解唔直接union哂所有 E_n set先，然後變成Lim n->inf E_n? 而要整個F set出黎，玩自己咁麻煩?

當然我地知道
μ (E_n) = μ (U_{i = 1}^n E_i)
但係 take limit 果陣, 我唔知右邊舊野係咪會變成 μ (U_{i = 1}^∞ E_i)

就好似 y_n = f(x_n) for all n, 如果 lim y_n = y, lim x_n = x
我地只知 lim f(x_n) = y, 但係我地唔知 f (x) = y 架喎

4.2個part有D唔明，

"(i) 如果 E_1 ⊆ E_2 ⊆ E_3 ⊆ ... ⊆ E_n ⊆ ...,


依度點解唔直接union哂所有 E_n set先，然後變成Lim n->inf E_n? 而要整個F set出黎，玩自己咁麻煩?

當然我地知道
μ (E_n) = μ (U_{i = 1}^n E_i)
但係 take limit 果陣, 我唔知右邊舊野係咪會變成 μ (U_{i = 1}^∞ E_i)

就好似 y_n = f(x_n) for all n, 如果 lim y_n = y, lim x_n = x
我地只知 lim f(x_n) = y, 但係我地唔知 f (x) = y 架喎

例如 f(x) = 0 if x =/= 0, f(0) = 1
x_n = 1/n
y_n = 0

Appendix: Proof of Theorem 2.

下文數學高能注意

Remark. 其實measurable functions有一個再 general d 的定義, 不過呢度我地用呢個定義已經足夠啦~~~ (唔服的話, 實質上我地呢度講緊的係 f: (X, A) -> (ℝ, B), B = Borel σ-algebra)

證明含有 "A ⇔ B" 的statement, 我地要證明兩樣野:
(i) A ⇒ B
(ii) B ⇒ A.

(i) 先假設 f measurable, 即係 , s_n simple functions. 利用 limit of sequence 的定義 (ε = (f(x) - a) / 2 > 0),

f(x) > c
⇔ 存在某個 N ∈ ℕ, 令到 s_n (x) > a for all n ≥ N.
⇔ 存在某個 N ∈ ℕ, 令到 for all n ≥ N.
⇔ 存在某個 N ∈ ℕ, 令到
⇔

所以
(*原來我之前漏左講咩叫兩個set 相等, 兩個sets A = B 意思係 A⊆B and A⊇B, 亦即係 x ∈ A ⇔ x ∈ B)

Theorem 1. 話比我地知, 係measurable set, 由於堆measurable sets係一個 σ-algebra, 因此 係 measurable (σ-algebra定義入面的 4, 再用 De Morgan's Law 同第3 就知 任意sequence of measurable sets 的 intersection 都係measurable)

然後 sequence of measurable set 的union係measurable, 所以 is measurable.

------------------------------------------------

(ii) 係開始之前, 我地先學一個係measure theory非常常用的procedure:

先證 non-negative function係岩, 然後因爲任何 function 都可以切成咁: f(x) = f+(x) - f-(x), where f+(x) = max{f(x), 0}, f-(x) = max{-f(x), 0} (簡單黎講 f+ 係 f positive的部份, f- 係 f negative 的部份), 所以所有 function 都岩

Q: 點解唔會出現 infinity - infinity?

好, 而家假設 is measurable for any a ∈ ℝ. 先假設 f 係 non-negative. 我地證明 f 係measurable, 即係要搵一個sequence of simple function, 而呢個sequence 的 pointwise limit 係 f.

我地先解決一個比較簡單的case 先: f is bounded. (Math 哲學: 搞唔掂咪試下 solve 下 d special cases 睇下有乜睇到先)
即係, 搵到一個 M > 0 令到 f(x) ≤ M for all x.

我地可以先將 [0, M] 斬開兩半, 即係 [0, M/2], (M/2, M]
我地定義

即係: , 睇番我地個假設, 所以 s_1 係simple function.
點整 s_2 呢? 將 [0, M] 斬開四等份, 即係 [0, M/4], (M/4, M/2], (M/2, 3M/4], (3M/4, M]
再定義

即係

所以 s_2 都係 simple function. (注意 )
最後定義

所以 s_n 都係 simple.
而 by construction, 0 ≤ f(x) - s_n(x) ≤ 1/2^n for any x ∈ X.
所以 (實際上再強 d, uniformly)

咁 f unbounded 點做呢? Given一個positive integer n, 我地定義

然後 f_n 都係 bounded, 而且都係measurable (點解呢? 諗下呢個set係咩樣? 如果大家都唔明的話我可以再講解下)), 所以對每個 n 我地都可以執到條sequence of simple function , 令到
之後神奇的野就發生啦: 我地定義 就搞掂

最後就要證明 .

Case 1:
如果 f(x) 係 finite 的話, 我地可以搵到一個 integer K 令到 f(x) < K, 即係話 f_n(x) = f(x) for all n ≥ K. 望返個construction, 即係話

for any m, k ≥ K, n ∈ ℕ

即係話

Case 2:
如果 f(x) = ∞ 點算呢, 咁樣的話, 我地知道

Q.E.D.

this reminds me the pain when taking the course

this reminds me the pain when taking the course

(i) 先假設 f measurable, 即係 , s_n simple functions. 利用 limit of sequence 的定義 (ε = (f(x) - a) / 2 > 0),

呢到即係話f(x) > 2a？點解？

f(x) > c
⇔ 存在某個 N ∈ ℕ, 令到 s_n (x) > a for all n ≥ N.

呢兩句都唔明

你計錯數
1. By definition of limit of sequence, there exists N such that
|s_n (x) - f(x)| < (f(x) - a) / 2 for all n ≥ N
拆absolute value 再移項得到
a < (f(x) + a) /2 < s_n (x) < something we dont care for all n ≥ N

2. Typo, 應該係 f(x) > a

但係都係唔明f(x) 點解會大過a

(i) 先假設 f measurable, 即係 , s_n simple functions. 利用 limit of sequence 的定義 (ε = (f(x) - a) / 2 > 0),

呢到即係話f(x) > 2a？點解？

f(x) > c
⇔ 存在某個 N ∈ ℕ, 令到 s_n (x) > a for all n ≥ N.

呢兩句都唔明

你計錯數
1. By definition of limit of sequence, there exists N such that
|s_n (x) - f(x)| < (f(x) - a) / 2 for all n ≥ N
拆absolute value 再移項得到
a < (f(x) + a) /2 < s_n (x) < something we dont care for all n ≥ N

2. Typo, 應該係 f(x) > a

但係都係唔明f(x) 點解會大過a

oh 我想證 f^{-1} (a, infinity] = countable (union and / or interseciton) of some measurable sets
所姒第一句變左 f(x) > a
btw 再提多次個set 我寫錯左, 一陣間出返個post改番

Erratum: 5.2 Appendix

係證明 measurable function 符合 f^{-1} (a, \infty] 係 measurable 果 part 我寫錯左D野 , 係呢度改番 (Exercise: Find out and explain the mistake in my original proof. )

我地證一個set 係 measurable 的話, 通常 technique 都係寫做一堆 COUNTABLY many (即係我地可以用positive integer label哂佢地, 即係可以寫做 E_1, E_2, ...) measurable sets 的 union and/or intersection
而家即係想將 f^{-1} (a, \infty] 寫做以上咁講的樣

而家用一個 measure theory 幾常用的 trick: 塞一個數字落兩個數字中間 (通常同 rational number 有關)

f(x) > a
⇔ 存在 N, K ∈ ℕ, 令到 s_n (x) > a + 1/K (> a) for any n ≥ N
⇔ 存在 N, K, 令到
⇔

插個 1/K 落去的理由:
係我原本的 proof 入面, => 係岩, 但係返轉頭未必岩 (諗番real numbers 個case: x_n > x 只代表 lim x_n ≥ x)

所以我地要格硬攝一個number 令到個 strict inequality 都岩, 即係 x_n > x + y (y > 0) 就會有 lim x_n ≥ x + y > x
但係呢d number 要 countable咁多 (望返 σ-algebra 的定義)
in principle 我地唔用1/K, 用 x_K > 0, {x_K} decreases to 0都可以, as long as 堆數字係countable.

Erratum: 5.2 Appendix

係證明 measurable function 符合 f^{-1} (a, \infty] 係 measurable 果 part 我寫錯左D野 , 係呢度改番 (Exercise: Find out and explain the mistake in my original proof. )

我地證一個set 係 measurable 的話, 通常 technique 都係寫做一堆 COUNTABLY many (即係我地可以用positive integer label哂佢地, 即係可以寫做 E_1, E_2, ...) measurable sets 的 union and/or intersection
而家即係想將 f^{-1} (a, \infty] 寫做以上咁講的樣

而家用一個 measure theory 幾常用的 trick: 塞一個數字落兩個數字中間 (通常同 rational number 有關)

f(x) > a
⇔ 存在 N, K ∈ ℕ, 令到 s_n (x) > a + 1/K (> a) for any n ≥ N
⇔ 存在 N, K, 令到
⇔

插個 1/K 落去的理由:
係我原本的 proof 入面, => 係岩, 但係返轉頭未必岩 (諗番real numbers 個case: x_n > x 只代表 lim x_n ≥ x)

所以我地要格硬攝一個number 令到個 strict inequality 都岩, 即係 x_n > x + y (y > 0) 就會有 lim x_n ≥ x + y > x
但係呢d number 要 countable咁多 (望返 σ-algebra 的定義)
in principle 我地唔用1/K, 用 x_K > 0, {x_K} decreases to 0都可以, as long as 堆數字係countable.

lm

係咪已經陣亡哂

4.2個part有D唔明，

"(i) 如果 E_1 ⊆ E_2 ⊆ E_3 ⊆ ... ⊆ E_n ⊆ ...,


依度點解唔直接union哂所有 E_n set先，然後變成Lim n->inf E_n? 而要整個F set出黎，玩自己咁麻煩?

當然我地知道
μ (E_n) = μ (U_{i = 1}^n E_i)
但係 take limit 果陣, 我唔知右邊舊野係咪會變成 μ (U_{i = 1}^∞ E_i)

就好似 y_n = f(x_n) for all n, 如果 lim y_n = y, lim x_n = x
我地只知 lim f(x_n) = y, 但係我地唔知 f (x) = y 架喎

例如 f(x) = 0 if x =/= 0, f(0) = 1
x_n = 1/n
y_n = 0

係咪有機會 U_{i = 1}^∞ E_i 同 Lim n->inf (U_{i = 1}^∞ E_i) 會有分別?

Inf有D怪,因為一定要用lim黎define

好似你個例子既inf版本會變成
f(x) = 0 if x =/= inf, f(inf) = 1

將個 limit sign 拉入去個 function 呢步有機會出問題
即係問緊 lim f(x_n) 係咪等於 f (lim x_n)
如果 f 唔 continuous 就會有機會出現 lim f(x_n) =/= f (lim x_n)

係measure 呢個case都係 (雖然嚴格上黎講set 無limit, 得liminf 同 limsup, 未學過唔駛理)
U_{i = 1}^∞ E_i intuitively 可以睇做"lim" U_{i = 1}^n E_i
咁樣就會變成問
lim μ (U_{i = 1}^n E_i) =?= μ ( "lim" U_{i = 1}^n E_i)

係咪已經陣亡哂

第三頁就陣亡左
利申:eng仔，grad左十年重識develop 返green theorem 黎用

係fancy 數學黎講, 咩 green's divergence theorem 都係 stokes' theorem 的special case

係咪已經陣亡哂

死亡緊，比少少時間消化下

係咪已經陣亡哂

死亡緊，比少少時間消化下

唔緊要, 可以等大家get到大部份先繼續

有時可能係我表達唔清楚, 仲有咩唔明係度問啦

Erratum: 5.2 Appendix

係證明 measurable function 符合 f^{-1} (a, \infty] 係 measurable 果 part 我寫錯左D野 , 係呢度改番 (Exercise: Find out and explain the mistake in my original proof. )

我地證一個set 係 measurable 的話, 通常 technique 都係寫做一堆 COUNTABLY many (即係我地可以用positive integer label哂佢地, 即係可以寫做 E_1, E_2, ...) measurable sets 的 union and/or intersection
而家即係想將 f^{-1} (a, \infty] 寫做以上咁講的樣

而家用一個 measure theory 幾常用的 trick: 塞一個數字落兩個數字中間 (通常同 rational number 有關)

f(x) > a
⇔ 存在 N, K ∈ ℕ, 令到 s_n (x) > a + 1/K (> a) for any n ≥ N
⇔ 存在 N, K, 令到
⇔

插個 1/K 落去的理由:
係我原本的 proof 入面, => 係岩, 但係返轉頭未必岩 (諗番real numbers 個case: x_n > x 只代表 lim x_n ≥ x)

所以我地要格硬攝一個number 令到個 strict inequality 都岩, 即係 x_n > x + y (y > 0) 就會有 lim x_n ≥ x + y > x
但係呢d number 要 countable咁多 (望返 σ-algebra 的定義)
in principle 我地唔用1/K, 用 x_K > 0, {x_K} decreases to 0都可以, as long as 堆數字係countable.

好似明呢part了
然而 ii 仲未睇

Erratum: 5.2 Appendix

係證明 measurable function 符合 f^{-1} (a, \infty] 係 measurable 果 part 我寫錯左D野 , 係呢度改番 (Exercise: Find out and explain the mistake in my original proof. )

我地證一個set 係 measurable 的話, 通常 technique 都係寫做一堆 COUNTABLY many (即係我地可以用positive integer label哂佢地, 即係可以寫做 E_1, E_2, ...) measurable sets 的 union and/or intersection
而家即係想將 f^{-1} (a, \infty] 寫做以上咁講的樣

而家用一個 measure theory 幾常用的 trick: 塞一個數字落兩個數字中間 (通常同 rational number 有關)

f(x) > a
⇔ 存在 N, K ∈ ℕ, 令到 s_n (x) > a + 1/K (> a) for any n ≥ N
⇔ 存在 N, K, 令到
⇔

插個 1/K 落去的理由:
係我原本的 proof 入面, => 係岩, 但係返轉頭未必岩 (諗番real numbers 個case: x_n > x 只代表 lim x_n ≥ x)

所以我地要格硬攝一個number 令到個 strict inequality 都岩, 即係 x_n > x + y (y > 0) 就會有 lim x_n ≥ x + y > x
但係呢d number 要 countable咁多 (望返 σ-algebra 的定義)
in principle 我地唔用1/K, 用 x_K > 0, {x_K} decreases to 0都可以, as long as 堆數字係countable.

好似明呢part了
然而 ii 仲未睇

ii 會易明d, 比較concrete

有冇可能講到Lp spaces, Riesz representation theorem或者Radon-Nikodym theorem?

學mathematical finance個陣Radon-Nikodym derivatives聽完都係好撚抽象，唔知點解需要有件咁既事
btw 留名再睇

equivalent martingale measure

Q: 點解唔會出現 infinity - infinity?

因爲唔會同時係positive part 同 negative part?

最後定義

所以 s_n 都係 simple.
而 by construction, 0 ≤ f(x) - s_n(x) ≤ 1/2^n for any x ∈ X.

why 1/2^n but not M/2^n ?

咁 f unbounded 點做呢? Given一個positive integer n, 我地定義

然後 f_n 都係 bounded, 而且都係measurable (點解呢? 諗下呢個set係咩樣? 如果大家都唔明的話我可以再講解下))

f_n^-1 (a, inf] = f^-1 (a, inf] ?

有冇可能講到Lp spaces, Riesz representation theorem或者Radon-Nikodym theorem?

你想我講證明? 應該超過九成日陣亡喎
但係應用同堆theorem 大概講緊乜我可以講

Q: 點解唔會出現 infinity - infinity?

因爲唔會同時係positive part 同 negative part?

最後定義

所以 s_n 都係 simple.
而 by construction, 0 ≤ f(x) - s_n(x) ≤ 1/2^n for any x ∈ X.

why 1/2^n but not M/2^n ?

咁 f unbounded 點做呢? Given一個positive integer n, 我地定義

然後 f_n 都係 bounded, 而且都係measurable (點解呢? 諗下呢個set係咩樣? 如果大家都唔明的話我可以再講解下))

f_n^-1 (a, inf] = f^-1 (a, inf] ?

1. Yes
2. Oh Typo 證明你有細心睇 thanks a lot
3. 差少少，記得分case (呢科經常要分case)