rudin 实分析与复分析习题解答.pdf-资料库

juney2009-10642859-4744300845365860402.pdf-第1页.png

第1页 / 共256页

juney2009-10642859-4744300845365860402.pdf-第2页.png

第2页 / 共256页

juney2009-10642859-4744300845365860402.pdf-第3页.png

第3页 / 共256页

juney2009-10642859-4744300845365860402.pdf-第4页.png

第4页 / 共256页

juney2009-10642859-4744300845365860402.pdf-第5页.png

第5页 / 共256页

juney2009-10642859-4744300845365860402.pdf-第6页.png

第6页 / 共256页

juney2009-10642859-4744300845365860402.pdf-第7页.png

第7页 / 共256页

juney2009-10642859-4744300845365860402.pdf-第8页.png

第8页 / 共256页

Solutions to Exercises in Real Analysis Li Gongbao School of Mathematics and Statistics in CCNU

PREFACE The Preface can be written here

CHAPTER ONE 1. Does there exist an inﬁnite σ-algebra which has only countably many members? Solution: No. There exists no such an inﬁnite σ−algebra M which has only countably members. T Aj = ∅(i 6= j), then Proof 1: Claim 1: If X = Sn i=1 Ai, Ai 6= ∅, Ai ∈ M, Ai there exists a A ∈ M, A 6= ∅, s.t. AT Aj 6= ∅ and AT Ac j 6= ∅ for some j ∈ {1, 2,··· , n}.Hence, Aj = (AT Aj)S(AT Ac j ∈ M, then X can be written by Sn+1 i=1 Bj,∅ 6= Bj ∈ M,and Bj are pairwise disjoint. Proof of Claim 1: Put F{A1, A2,··· , An} be the σ-algebra spaned by A1, A2,··· , An.Since Ai are pairwise disjoint,new numbers diﬀerent from Ai can be only producted by union of some Ai’s.However,{S j∈{1,2,··· ,n} Aj} has at most 2n numbers.So F{A1, A2,··· , An} is ﬁnite. AT Aj 6= ∅ and AT Ac implies that AcT Aj = ∅,thus A ⊆ Aj,which contains the relation: By the deﬁnition of M,there exists A ∈ M s.t.A 6∈ F.We claim that j 6= ∅ for some j ∈ {1, 2,··· , n}.If not,then AT Aj 6= ∅ j) and AT Aj ∈ M, AT Ac Aj ⊆ A. [ {j;AT Aj6=∅} In fact,the above inclusion relation is mutual,for if x ∈ A ⊆ X,then x ∈ Aj for some j,and AT Aj 6= ∅,so x ∈ S{j;AT Aj6=∅} Aj,which proves that A ⊆ S{j;AT Aj6=∅} Aj.Hence A = S{j;AT Aj6=∅} Aj ∈ F.A contradiction.This com- def= A1S Ac1.By Claim 1,there exists a C1 ∈ M s.t.C1T Ai 6= X,then X = A1S A2 6= ∅ for some i,say,i = 1.So A1 is the union of C1S A1 and ∅ and C c1 Now we prove exercise 1.Choose A1 be a nonempty proper subset of pletes the proof. T Ai

4 C c1 S A1,both of which are nonempty measurable subsets. Thus by rearranging the subscripts we can write X = A1S A2S A3,where Ai(i = Ai)[(C c A1)[ X = (C1 \ \ ABSTRACT INTEGRATION 1 A2, 1, 2, 3)are nonempty measurable proper subsets of X which are mutually pairwise disjoint.Continuing in this way, we get a countably inﬁnite disjoint collection {Ai}∞ T Aj = ∅ for i 6= j. Put M1 = {Ai}∞ i=1,then M1 ⊆ M.Since M is a σ−algebra and M1 is in- ﬁnitely countable,2M1 is a subcollection of M.Hence the cardinality inequality holds: i=1 with Ai 2M1 ≤ 2M, which is a contradiction since the former cardinality is equal to [0, 1] ,greater than Nwhich is the latter one’s cardinality.(See P25, exercise 3 in the book of Jiang zejian). We have completed the proof 1. Proof 2: We give another way to ﬁnd a countable subset of M by the im- plication of proof 1, the main tool is Zorn’s lemma(See the remark). Since M is countable we denote M by {Ai}∞ i=1.Then M is partially ordered by the set inclusion relation ⊆. If all totally ordered subsets of M are ﬁnite,then X,the whole space,is ob- viously their common upper bound.Put B0 = X and M1 = M−{B0},apparently that all totally ordered subsets of M1 is ﬁnite,hence their exists a maximal element B1 ∈ M1 by Zorn’s lemma. Put M2 = {Ai|Ai ∈ M, Ai B1} ⊆ M1 T Bc1 6= We claim that M2 is inﬁnite.If not,then put C = {Ai|Ai ∈ M1, Ai ∅ and Ai 6= Bc1},is inﬁnite,which will lead to a result that the inﬁnite set {Bi − Ai|Ai ∈ C} ⊆ M2,a contradiction to our assumption.To see this, pick S B1 ∈ M1 an arbitrary Ai ∈ C,then B1 ⊆ Ai S B1 since B1 is the maximal element in M1,which implies that and B1 = Ai S B1 = X,by this,∅ 6= Ai ⊆ B1that contradicts to the deﬁnition of C.So Ai B1 − Ai B1.For if B1 − Ai = ∅,then B1 Ai which implies that Ai = S B1 = X,contradiction to Ai 6= X since Ai ∈ M1.IfB1 − Ai = B1,then Ai S B1 6= X,then Ai S B1.If Ai

REAL AND COMPLEX ANALYSIS A1 = ∅ or Ai Bc1.Therefore the set {Bi − Ai|Ai ∈ C} ⊆ M2 5 We have proved that M2 is inﬁnite in the above.Applying Zorn’s lemma to M2 ,there exists a maximal element B2 ∈ M2 with B2 B1.Put M3 = {Ai|Ai ∈ M, Ai B2} ⊆ M2. Then M3 is inﬁnite if we replace X by B1 and do the same as above.Hence there is a maximal element B3 ∈ M3 with B3 B2.Continuing in this way and we get a countably inﬁnite collection {Bi}∞ i=1 with the property B0 B1 B2 B3 ··· Contradiction to our assumption of M that M has no inﬁnite totally ordered subset. Hence M is bounded to have a totally ordered subset,which we denote (1)If ˜M1 has no upper bound,then there exists a subsequence {Cni} ⊆ by ˜M1 = {Ci}∞ i=1 ⊆ M. {Ci} s.t. ∅ = Ci0 Ci1 Ci3 ··· Cik Ci2 k=1 ⊆ M is pairwise disjoint hence is just what we Cik+1 ··· Then the set {Cik+1 − Cik}∞ want to ﬁnd. k=1 has the property : ··· Dik Dik+1 ··· (2)If ˜M1 has a upper bound D1,then put ˜M2 = ˜M1 − {D1}.Continuing in this way,if for some k, ˜Mk = ˜Mk−1 − {Dk−1} has no upper bound,we have completed the proof; or else,then {Dk}∞ ∅ = Di0 Di3 Di2 k=1 ⊆ M is just what we want,and we have completed Di1 and the set {Dk+1−Dk}∞ the proof 2. Remark:1. We give the proof that 2N = [0, 1] = C here. For any giv- en A ⊂ N, set ϕA(n) = { 1, Let g : A → (0, 1) is given by 0, g(A) = 0. ϕA(1)ϕA(2) · ··, it is obvious that g is one-to-one. So, 2N ≤ [0, 1]. if n ∈ A; if n 6∈ A.

6 ABSTRACT INTEGRATION On the other hand, for any x ∈ (0, 1), set Ax = {r; r ≤ x, r ∈ R}, where R is the set of all the rational in (0, 1). If x, y ∈ (0, 1) and x 6= y, by Berstein’s theorem, the density of rational in R, Ax 6= Ay. So, (0, 1) ≤ R = 2N. So, 2N = [0, 1] = C. 2.Zorn’s lemma:Suppose (X,≺) is a partially ordered set.If every totally or- dered subsets of X has one upper bound,then there exists a maximal element a ∈ X s.t.for anyx ∈ X,if x ≺ a,then x = a. 2 2. Prove an analogue of Theorem 1.8 for n functions. Solution: We have the following result. Theorem 1.80: Let u1, u2, ..., un be real measurable functions on a measurable space X, let Φ be a continuous mapping of Rn onto a topological space Y , and deﬁne h(x) = Φ(u1(x), ..., un(x)) for x ∈ X. Then h : X → Y is measurable. Proof: Since h = Φ ◦ f, where f(x) = (u1(x), ..., un(x)) and f maps X into Rn. Theorem 1.7 shows that it is enough to prove the measurability of f. If R is any open rectangle in Rn with sides parallel to the coordinate super plane, then R = I1×I2×···×In, where Ii = (ai, bi) ⊂ R1 (i = 1, 2, ..., n) are open segments and f−1(R) = u−1 1 (I1)\ 2 (I2)\· · ·\ u−1 u−1 n (In). In fact, for any given x ∈ f−1(R), we have f(x) = (u1(x), ..., un(x)) ∈ i (Ii) for i = 1, 2, ..., n, i (Ii), we have R = I1 × I2 × · · · × In and ui(x) ∈ Ii, then x ∈ u−1 i (Ii). On the other hand, for any x ∈ Tn that is x ∈ Tn xi ∈ u−1 So f−1(R) is measurable for each open rectangle interval R ⊂ Rn as ui are measurable on X. i (Ii), ui(x) ∈ Ii and f(x) = (u1(x), ..., un(x)) ∈ I1 × I2 × · · · × In. i=1 u−1 i=1 u−1 By Lindel˝of’s Theorem, every open set V is a countable union of open interval with sides parallel to the coordinate super-plane, V = S Ri, So f−1(V ) = Sn 2 3. Prove that if f is a real function on a measurable space X such that {x; f(x) ≥ r} is measurable for every rational r, then f if measurable. i=1 f−1(Ri) is measurable as well.

In fact, for any x ∈ {x; f(x) > α}, there is a n0 such that α < rn0 ≤ f(x), therefore x ∈ {x; f(x) > α}. On the other hand, rn > α implies that +∞[ n=1 {x; f(x) ≥ rn} ⊂ {x; f(x) > α}. REAL AND COMPLEX ANALYSIS 7 Proof: For any α ∈ R1, there is a sequence of rational {rn; n = 1, 2, ...} such that r1 > r2 > r3 > ... > rn > ... > α with lim {x; f(x) > α} = n→∞ rn = α. So {x; f(x) ≥ rn}. +∞[ n=1 By the condition, for any given n, {x; f(x) ≥ rn} is measurable, we see that {x; f(x) > α} is measurable for any α ∈ R1. By Theorem 1.12, f is measurable. 2 4. Let {an} and {bn} be sequence in [−∞, +∞], prove the following asser- tions: n→∞ (−an) = − lim inf (a) lim sup n→∞ an; n→∞ (an + bn) ≤ lim sup (b) lim sup n→∞ an + lim sup the form ∞ − ∞; (c) If an ≤ bn for all n, then n→∞ bn provided none of the sums is of n→∞ an ≤ lim inf lim inf n→∞ bn show by an example that strict inequality can hold in (b). Proof: (a) lim sup n→∞ (−an) = inf k≥1 sup{−ak,−ak+1, ...} = inf k≥1 inf{ak, ak+1, ...} = − lim inf {− inf{ak, ak+1, ...}} n→∞ an. = − sup k≥1 (b) lim sup [sup{ak + bk, ak+1 + bk+1, ...}] n→∞ (an + bn) = inf k≥1 k→∞[sup{ak + bk, ak+1 + bk+1, ...}] = lim {an} + sup ≤ lim {bn}] k→∞[sup n≥k n≥k n→∞ [sup{bn}] n→∞[sup{an}] + lim inf = lim n→∞ an + lim sup = lim sup n→∞ bn. {ak} = lim inf k→∞ inf n→∞ an. n≥k {an} ≤ inf (c) lim inf n→∞ an = lim In (b), the strict inequality can hold. For example, let an = (−1)n, bn = k→∞ inf n≥k

n→∞ bn. n→∞ an + lim sup n→∞ an = 1, lim sup n→∞ bn = 1, but lim sup 8 ABSTRACT INTEGRATION −(−1)n, then an + bn = 0, lim sup n→∞ (an + bn) = 0 < 2 = 1 + 1 = lim sup 2 5. (a) Suppose f : X → [−∞,∞] and g : X → [−∞,∞] are measurable. Prove that the sets {x; f(x) < g(x)}, {x; f(x) = g(x)} are measurable. (b) Prove that the set of points at which a sequence of measurable real-valued functions converges(to a ﬁnite limit) is measurable. Proof: (a1) Suppose f, g are measurable on X, then {x; f(x) < g(x)} is measurable. Suppose {rn} is the set of rational in R1, we claim that [{x; f(x) < rn}\{x; rn < g(x)}] {x; f(x) < g(x)} = ∞[ n=1 In fact, x0 ∈ {x; f(x) < g(x)} ⇐⇒ f(x0) < g(x0) ⇐⇒ ∃rn0, such that f(x0) < rn0 < g(x0) ⇐⇒ x0 ∈ S∞ n=1[{x; f(x) < rn}T{x; rn < g(x)}]. For any r ∈ R, {x; f(x) < r} and {x; r < g(x)} are measurable s- ince f, g are measurable on X. So, by the claim we have proved above, {x; f(x) < g(x)} is measurable. (a2) By (a1), {x; f(x) = g(x)} = X − [{x; f(x) < g(x)}S{x; f(x) > (b) Suppose {fn} is a sequence of real-valued measurable functions on g(x)}] is measurable. (X, M)(a measurable space), then by Theorem 1.9(c), for any n, m ∈ N, |fn(x)− fm(x)| is a measurable function as | · | is a continuous function on R1 × R1. So, {x; |fn(x) − fm(x)| < a} is a measurable set for any a ∈ R1. Clearly, the set where {fn(x)} has ﬁnite limit is given by } {x; |fn(x) − fm(x)| < +∞\ +∞[ A = \ k=1 N=1 n,m≥N 1 k by Cauchy’s criterion for convergence. A is a measurable set because for any n, m ∈ N and k ∈ N, {x; |fn(x)− k} ∈ M and M is a σ−algebra. fm(x)| < 1 2 6. Let X be an uncountable set, let M be the collection of all sets E ⊂ X such that either E or Ec is at most countable, and deﬁne µ(E) = 0 in the ﬁrst case, µ(E) = 1 in the second. Prove that M is a σ−algebra in X and

资料库

rudin 实分析与复分析习题解答.pdf

相关推荐

课程资源

热门标签

最新资料