1(3)将AUB拆成互不相交的集合的并:AUB=AU(B-A)=AUB-AB,再由 概率的可加性可得 (4)因为BCA,所以A=B∪A-B,由此可得 13当A∩B≠时,a(4)共有16个元素。当A∩B=时,a(A)有8个元素。不难写出 14题目中A3={(a,b):a,b∈R} (1)因为切n,ACAn所以有Ac∩An 下面证明∩AnCA。反证法:假设∈∩4n,xA,那么x-a>0,取n=「xa1+1 则x>a+,由此可得xAn,矛盾。所以∩AnCA。由此可得A=∩An。类似可 n=1 得B=UBn (2)因为a∈R(-x,=∪(=n,a,所以A1c叫(A2)。因为va,b∈R(a=(-∞ (-∞,a],所以A2(A1)。综上可得σ(A1)=σ(A2) ∩(-∞,a+) (一n,a+ m=ln ∩UL-n,a+ va,b∈R,(a,b]=(-∞,b-(-∞,a] (a,b) (-x,b-1-(- a,b=(-∞,b 由此可得a(A1)=0(41),2≤i≤5 1.6(3)因为A2k+1CA2k,所以 A ∫U=n/2Axk=(l,l+ 为偶数 Uk=(x+1)2A2k=(,+n+ln为奇数1.1 (3) òA S B ¤p؃8Ü¿µA S B = A S (B − A) = A S B − AB§2d VÇŒ\5Œ" (4) ϏB ⊂ A§¤±A = B S A − B§ddŒ" 1.3 A T B 6= φž§σ(A)
k16‡ƒ"A T B = φž§σ(A)k8‡ƒ"ØJÑ 1.4 K8¥A3 = {(a, b) : a, b ∈ R} (1) Ϗ∀n, A ⊂ An¤±kA ⊂ \∞ n=1 An" e¡y² \∞ n=1 An ⊂ A"‡y{µb∃x ∈ \∞ n=1 An, x /∈ A§@ox − a > 0§n = d 1 x−a e + 1§ Kx > a + 1 n§ddŒx /∈ An§gñ"¤± \∞ n=1 An ⊂ A"ddŒA = \∞ n=1 An"aqŒ B = [∞ n=1 Bn" (2) Ϗ∀a ∈ R,(−∞, a] = [∞ n=1 (−n, a]§¤±A1 ⊂ σ(A2)"Ϗ∀a, b ∈ R,(a, b] = (−∞, b] − (−∞, a]§¤±A2 ⊂ σ(A1)"nþŒσ(A1) = σ(A2) (3) ∀a ∈ R,(−∞, a] = [∞ n=1 (−n, a] = \∞ m=1 (−∞, a + 1 m ) = \∞ m=1 [∞ n=1 (−n, a + 1 m ) = \∞ m=1 [∞ n=1 [−n, a + 1 m ) = [∞ n=1 [−n, a] ∀a, b ∈ R,(a, b] = (−∞, b] − (−∞, a] (a, b) = [∞ n=1 (−∞, b − 1 n ] − (−∞, a] [a, b) = [∞ n=1 (−∞, b − 1 n ] − [∞ m=1 (−∞, a − 1 m ] [a, b] = (−∞, b] − [∞ m=1 (−∞, a − 1 m ] ddŒσ(A1) = σ(Ai), 2 ≤ i ≤ 5 1.6 (3) ϏA2k+1 ⊂ A2k§¤± [∞ k=n Ak = ½ S∞ k=n/2 A2k = ( 1 4 , 1 2 + 1 n ] nóê S∞ k=(n+1)/2 A2k = ( 1 4 , 1 2 + 1 n+1 ] nÛê 1