c n b n 0. c k 0 x b n < 1 b k b n = 0. } of integers between 0 and b 1 such that x = b k. b k c k c k

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1 1. Exitence Let x (0, 1). Define c k inductively. Suppoe c 1,..., c k 1 are already defined. We let c k be the leat integer uch that x k An eay proof by induction give that and for all k. Therefore c n b n 0. c k {0, 1,..., b 1} 0 x x k c n b n < 1 b k c n b n = 0. Uniquene (in ome ene) Suppoe there are two different equence {c k }, {c k } of integer between 0 and b 1 uch that c k x = b k = c k b k. We may aume that c 1 = c 1,..., c n 1 = c n 1, c n < c n. Then 1/b n c n c n b n = k=n+1 c k c k b k k=n+1 c k c k b k k=n+1 b 1 b k = 1/b n. However, thi mean c n c n = 1 and c k c k c k = b 1 and c k = 0 for all k > n. = b 1 for all k > n. Hence 2. For all k, let C k be the reult after removing ubinterval at the kth tage. Then C k = {x = k c i 3 i + a/3 k : c i {0, 2} for i = 1,, k and a [0, 1]}. If x C k, then clearly x = c i3 i with c i {0, 2}. Next, note that if x = c i3 i, then it i a limit of x k = k c i3 i. But f k (x k ) = k (c i/2)2 i, hence f(x) = lim f k(x k ) = (c i /2)2 i. k 1

2 4. We have D = where D k i the reult after removing ubinterval at the nth tage. Clearly D k+1 = (1 θ) D k. So D = 0. D i cloed becaue all D k are cloed. To how D i perfect, let x D. For every k, ince x D k, let x k be an endpoint (e.g. left endpoint) of an entire cloed ubinterval of D k which x k lie in. Then all x k are in D. Moreover, D k x x k D k = (1 θ) k. Therefore {x k } i a equence in D which converge to x. 5. The reulting et, C, i perfect a in the lat problem. At the kth tage, the total length of interval removed i 2 k 1 δ3 k. Therefore C = 1 2 k 1 δ3 k = 1 δ. Inductively, we can how that after the kth tage, the reulting et cannot contain an interval of length 1/2 k. Thu C cannot contain an interval of length 1/2 k for all k, i.e. contain no interval. 7. We let I = I k. Obviouly I i meaurable and I I k. On the other hand, given ɛ > 0, uppoe that S = {J n } i a cover of I by interval uch that J n < (1 + ɛ) I. For each n, let Jn be an open interval containing I n in it interior uch that Jn < (1 + ɛ) J n. Now the compact et I (becaue I i a finite union of cloed interval!) i covered by thoe Jn. Therefore I can be covered by finitely many of them, ay, {J 1,..., J M }. Clearly N M I k Jn Jn (1 + ɛ) J n (1 + ɛ) 2 I. The above inequality hold for all ɛ > 0. Hence I k = I. 9. Proof: limup E k = + + E k. k=n Since + E k e < +, o ε > 0, N N, uch that + E k e < ε. Hence + k=n E k + k=n < ε. So limupe k < ε. Thu limupe k ha meaure zero. k=n 2

3 10. Proof: The equality i clear when E 1 + E 2 =. Thu we may aume that both E 1 and E 2 are finite. Since E 1 and E 2 are meaurable, So E 1 E 2 + E 1 E 2 = E 1 + E 2. E 1 E 2 = E 1 + E 1 \E 2 E 2 = E 2 E 1 + E 2 \E Proof: Suppoe E i meaurable. Then there i an open et O, uch that O\E e < ε ε > 0 O i open, o O can be written a a countable union of nonoverlapping open interval. Suppoe O = + O i, then E e < + O < +. So Define S = N 1 N N, + i=n O i < ε O i, N 1 = + i=n O i, N 2 = O\E. Then E = (S N 1 ) N 2, and N 1 e, N 2 e < ε. Converely, ε > 0, S, N 1, N 2, uch that E = (S N 1 ) N 2, where S i a finite union of nonoverlapping open interval, and N 1 e, N 2 e < ε. Chooe an open et N uch that N 1 N, N < 2ε. Then S N i open, E S N and S N E e N + N 2 e < 3ε. So E i meaurable. 12. Proof: Firt we how that if E 1 = 0, then E 1 E 2 i meaurable and E 1 E 2 = 0. Define E2 k = [ k, k] E 2. Then E 1 E 2 = E 1 E2 k. E 1 = 0, o ε > 0, there i an open et O, uch that E 1 O and O < ε/k. Suppoe O = + O i. {O i : i N } i a et of nonoverlapping open interval. Suppoe O i = (a i, b i ). Then + (b i a i ) < ε. So O E k 2 = + + O i [ k, k] = 2k ε/k = 2ε. So E 1 E2 k e 2ε. Hence E 1 E2 k e = 0. So E 1 E 2 e = E 1 E2 k e = 0. So E 1 E 2 i meaurable. Now uppoe E 1, E 2 are meaurable. So there are G δ et H 1, H 2, and Z 1, Z 2 with meaure zero, uch that E 1 = H 1 Z 1, and E 2 = H 2 Z 2. So E 1 E 2 = H 1 H 2 (H 1 Z 2 Z 1 H 2 ). H 1, H 2 are G δ et, o i H 1 H 2. 3

4 Z 1 = Z 2 = 0, o H 1 Z 2 Z 1 H 2 = 0. So E 1 E 2 i meaurable, and E 1 E 2 = H 1 H 2. For each k N, we define A k, B k a open et and A k \E 1 < 1/k, B k \E 2 < 1/k, A k+1 A k, B k+1 B k. Note that if I i and J are cloed interval uch that {I i } are nonoverlapping, we have ( I i ) J = I i J = I i J. Denote A = + A k, B + + B k. Then A\E 1 = B\E 2 = 0. Since A k B k = A k B k, o E 1 E 2 = A B = lim A k B k = E 1 E 2. k Proof. Let A E. Firt, there exit F σ et H A uch that A i = H. Then E = H + E \ H H + E \ A e = A i + E \ A e. Next, chooe G δ et G E \ A uch that G = E \ A e. Then A E \ G and E = E \ G + G A i + E \ A e. 16. Proof: Let P be a parallelepiped. Given any ε > 0. We can ue cube of ame ize and ufficiently mall to cover P uch that v(p ) P + ε. Since ε > 0 i arbitrary, we have v(p ) P. Next, note that P = Q i i a nonoverlapping union of cloe cube. However, for each finite N, N Q i P and N Q i v(p ). Which implie P = P = Q i v(p ). Combining the two etimate, we have v(p ) = P. 17. Proof: Suppoe f i the Cantor-Lebegue function, then f i continuou. Suppoe C i the Cantor et and C 0 i the ubet of C which are not endpoint of C k. Then by the definition of f, f(c 0 ) e = 1 > 0. Then we can find a unmeaurable ubet W of f(c 0 ). But f 1 (W ) C 0, o f 1 (W ) = 0 and hence f 1 (W ) i meaurable but it image i not meaurable. 20. Solution: let X 0 [0, 1] be a nonmeaurable et a built in Thm3.38, let X n = {x + 1 n x X 0}. Then {X n } n=0 are dijoint, and all of them are nonmeaurable. Xn e [0, 2] e = 2 But Σ n=0 X n e = + 4

5 21. Proof: let X n be ame a previou quetion. Let E 0 = X n, n=0 Let E i = X n, n=i for all i. Then E 0 E 1 E 2..., and each E i [0, 2] Therefor 1 E i e 2, Therefore lim k E k e 1 > 0 = E k = φ 22. Proof: Let T be the tranformation matrix for the parallelepiped with a fixed orientation, i.e. the matrix of unite bai for the parallelepiped i T I, where I i the unite matrix in M n n. Then T i Lipchitz, det(t ) 0, and every parallelepiped i of the form T I k, where I k i an interval in R n. And non-overlapping parallelepiped correpond to non-overlapping interval. Therefore if we define the outer meaure a E e = inf σ(s) = inf T I k S v(t I k ) where S i any countable union of non-overlapping parallelepiped covering E, then E e = inf v(t I k ) = inf det(t ) v(i k ) = T 1 E e T I k S T I k S So now it uffice to how T 1 E e = E e. Similar a in Thm3.25, we can how that T 1 E e = det(t 1 ) E e = E e 23. Proof: Firt we how that T Z 0 = 0 for any bounded ubet Z 0 of Z. However, T i clearly Lipchitz on Z 0 and hence T Z 0 0 Z 0 = 0. Now if Z i unbounded, then Z = k Z (Z [k, k + 1]). By cae 1, (Z [k, k + 1]) ha meaure 0 for all k. Therefore the countable union till ha meaure 0. I did not read through the next two quetion 25. Proof: Similar a the contruction of quetion 5, we make a little modification. Fix 0 < δ < 1. At tage 1, cut off the middle δ part of [0, 1] a an open interval (1/2 δ/2, 1/2 + δ/2), remaining et i C 1 = [0, (1 δ)/2] [(1 + δ)/2, 1]. Let D 1 = ((1 δ)/2, (1 + δ)/2). For inductive tep, aume at tage n we get a dijoint union of cloed interval C n = K [a i, b i ], and [0, 1] \ C n = D n a a dijoint union of open interval. Then both of the union are finite union. Then, 1) we cut off the middle δ part(a an open interval) of each [a i, b i ] in C n. 2) add in the middle δ part(a a cloed interval) of each (b i, a i+1 ) in D n. 5

6 Then reulting C n+1 i till a finite dijoint union of cloed interval, o i D n+1 a a finite union of dijoint open interval. Let C δ = lim n C n, then C δ i meaurable and 0 < C δ < 1. For any I = (x, y) [0, 1], ince x y = ɛ > 0, we can find N N ufficiently large uch that C N contain an interval [a N i, bn i ] (x, y). Conider E = C δ [a N i, bn i ], then E > 0 and [a N i, bn i ] E > 0. Therefor we finih the proof. 26. Solution: Similar a the contruction of Cantor-Lebegue function. At the firt tage, make the correponding function increae to 2 3 in [0, 1 3 ], and then decreae by 1 3 in [ 1 3, 2 3 ], and then increae by 2 3 in [ 2 3, 1].... At the n-th tage, for each maximal monotone part [a, b] of f, make the correponding function increae by 2 3 (f(b) f(a))(notice thi may be negative) in [a, a+ 1 3 (b a)], and then decreae by 1 3 (f(b) f(a)) in [a+ 1 3 (b a), a+ 2 3 (b a)], and then increae by 2 3 (f(b) f(a)) in [a (b a), b]. At each tage, denote the correponding function a f n. Then f k f k+1 ( 2 3 )k, therefore {f n } uniformly converge to ome continuou f. However, V (f, [a, b]) = up Γ S Γ = lim k + ( 5 3 )k = +. Therefore f i what we want. 6