推 ericakk :謝謝您!! 02/28 10:00
※ 引述《ericakk (代買IPSA單品8折)》之銘言：
: 欲證明:An extended real-valued function f is measurable if and only if
: the sets Α={x屬於χ:f(x)= +∞},Β={x屬於χ:f(x)= －∞} belong to Χ,
: and the real-valued function f1 defined by
: f1(x) = f(x) ,if x not belong to Α∪Β
: = 0 ,if x belong to Α∪Β
: Proof:
: 左至右：If f is in M (χ,Χ), it has already been noted that Α and Β
: belong to Χ, Let α belong to Ｒ and α ≧ 0, then
: {x屬於χ:f1(x)＞α} = {x屬於χ:f(x)＞α}\Α
Suppose x lies in {x屬於χ:f1(x)＞α}.
Since f1(x)>α and α≧0,f1(x)=f(x) by the definition of f1.So f(x)>α.
Again,by definition,x doesn't lie in A∪B.So x doesn't lie in A.
Hence,x lies in {x屬於χ:f(x)＞α}\Α.
Suppose x lies in {x屬於χ:f(x)＞α}\Α.
Since x doesn't lie in A,by the definition of f1,f1(x)=f(x)>α.
Hence,x lies in {x屬於χ:f1(x)＞α}.
: If α < 0, then
: {x屬於χ:f1(x)＞α} = {x屬於χ:f(x)＞α}∪Β
Suppose x lies in {x屬於χ:f1(x)＞α}.
Since α<0,f1(x) is probably zero.
Case 1:x lies in A∪B.
Then x lies in A or B and f1(x)=0.
If x lies in A,then f(x)=+∞>α.
If x lies in B,done.
Case 2:x doesn't lie in A∪B.
Then f(x) is finite and f1(x)=f(x).So f(x)>α.
Hence,x lies in {x屬於χ:f(x)＞α}∪Β.
Suppose x lies in {x屬於χ:f(x)＞α}∪Β.
If x lies in B,then f1(x)=0>α.
If x lies in {x屬於χ:f(x)＞α},
Case 1:x lies in A.
then f1(x)=0>α.
Case 2:x doesn't lie in A.
then f(x) is finite and f1(x)=f(x)>α.
Hence,x lies in {x屬於χ:f1(x)＞α}.
: Hence, f1 is measurable.
: 右至左：if Α and Β belong to Χ, and f1 is measurable, then
: when α ≧ 0, {x屬於χ:f(x)＞α} = {x屬於χ:f1(x)＞α}∪Α
: when α < 0, {x屬於χ:f(x)＞α} = {x屬於χ:f1(x)＞α}\Β
Use the same way as above.
: Hence, f is measurable.
: 我一直看不懂∪跟 \ 怎麼來的？><
: 麻煩請用白話的方式說明一下 謝謝您....
--
律:知道嗎?聽說我們的歌被海外的電視台所錄用耶!看來我們離武道館不遠了
唯:真的嗎?那真的是太好了，我一直夢想能在武道館彈著吉太，好高興
釉:小唯能高興真的是太好了，呵呵~
澪:拜託!那個明明是盜用不是錄用，你們怎麼還這麼高興?
律、唯、釉:啊?什麼? 輕音部
澪:絕望啦!我對盜用錄用分不清楚的輕音部社員們絕望啦! 邁向武道館之路
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.113.178.13
※ 編輯: k6416337 來自: 140.113.178.13 (02/27 13:16)
※ 編輯: k6416337 來自: 140.113.178.13 (02/27 13:17)