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.
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(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.