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Problem Set 1 This Problem Set Is Due on Friday, September 25 MA 2210 Fall 2015 - Problem set 1 This problem set is due on Friday, September 25. All parts (#) count 10 points. Solve the problems in order and please turn in for full marks (140 points) • Problems 1, 2, 6, 8, 9 in full • Problem 3, either #1 or #2 (not both) • Either Problem 4 or Problem 5 (not both) • Problem 7, either #1 or #2 (not both) 1. Let D be the dyadic grid on R and m denote the Lebesgue outer measure on R, namely for A ⊂ R ( ¥ ¥ ) [ m(A) = inf ∑ `(Ij) : A ⊂ Ij; Ij 2 D 8 j : j=1 j=1 −n #1 Let n 2 Z. Prove that m does not change if we restrict to intervals with `(Ij) ≤ 2 , namely ( ¥ ¥ ) (n) (n) [ −n m(A) = m (A); m (A) := inf ∑ `(Ij) : A ⊂ Ij; Ij 2 D 8 j;`(Ij) ≤ 2 : j=1 j=1 N −n #2 Let t 2 R be of the form t = ∑n=−N kn2 for suitable integers N;k−N;:::;kN. Prove that m is invariant under translations by t, namely m(A) = m(A +t) 8A ⊂ R: −n −m Hints. For #2, reduce to the case t = 2 for some n. Then use #1 and that Dm = fI 2 D : `(I) = 2 g is invariant under translation by 2−n whenever m ≥ n. d d 2. Let O be the collection of all open cubes I ⊂ R and define the outer measure on R given by ( ¥ ¥ ) [ n(A) = inf ∑ jRnj : A ⊂ R j; R j 2 O 8 j n=0 n=0 where jRj is the Euclidean volume of the cube R. #1 Show that n coincides with the Lebesgue outer measure m generated by dyadic cubes. You can use the fact, proved in class, that m(R) = jRj for all cubes R. #2 Use part (1) to obtain that the Lebesgue outer measure is translation invariant and re- d d spects dilations, in the sense that, for all t 2 R ;s > 0;A ⊂ R m(A +t) = m(A); m(sA) = sd m(A): Hint. For #1 there are two inequalities. One is just the e=2 j enlargement trick. For the other one, assume n(A) is finite. You can use the Whitney Lemma 1.3.3 from the notes to reduce a cover of A by open cubes to a disjoint dyadic cover with lesser total volume. 3. Consider the following PROPOSITION. Let (X;F ; m) be a measure space and throughout E j be measurable sets. There holds 1 2 (P1) m(/0) = 0; (P2) (monotonicity) if E1 ⊂ E2 then m(E1) ≤ m(E2); (P3) m is countably subadditive; [ (P4) If E j ⊂ E j+1 for all j, then m E j = lim m(E j) \ (P5) If m(E1) < ¥ and E j ⊃ E j+1 for all j, then m E j = lim m(E j), and the assumption m(E1) < ¥ is necessary; (P6) (lower semicontinuity) m liminfE j ≤ liminf m(E j); S (P7) (upper semicontinuity) limsup m(E j) ≤ m limsupE j provided m E j < ¥. #1 Prove (P4), (P5) #2 Prove (P6). 4 Dynkin systems. A collection F of subsets of X is called a Dynkin system if (1) (nontrivial) /0 2 F , (2) (closed under complement) F 2 F =) Fc 2 F , (3) (closed under countable disjoint union) if fFn : n 2 Ng ⊂ F are pairwise disjoint sets, namely n 6= k =) Fn \ Fk = /0,then [ F := Fn 2 F : n2N Given G ⊂ P(X), we denote by d(G ) and by S(G ) respectively the smallest Dynkin system and the smallest S-algebra containing G . #1 Prove that d(G ) is well defined and that G ⊂ d(G ) ⊂ S(G ). #2 Let F be a Dynkin system with the further property that (p) F;G 2 F =) F \ G 2 F (stable under finite intersection). Prove that F is a s-algebra. #3 Prove that if G has property (p) so does d(G ). d 5 Uniqueness of Lebesgue measure. Let Q0 = [0;1) . In this problem, we prove that the d Lebesgue measure on R is the unique complete measure m such that m(Q0) = 1 which is translation invariant. Its solution is divided into three parts. 0 For simplicity we work in d = 1. For a dyadic cube Q 2 D in R we indicate by D(Q) = fQ 2 D : Q0 ⊂ Qg and by B(Q) the s-algebra of Borel subsets of Q. Recall that B(Q) = S(D(Q)) for instance. #1 Let m;n be two measures on (Q0;B(Q0)) with the property that m(Q) = n(Q) for all Q 2 D(Q0), and m(Q0) = 1. Prove that m = n. #2 Let m;n be two measures on (R;B) with the property that m(Q) = n(Q) < ¥ for all Q 2 D. Prove that m = n. #3 Let n be a measure on (R;B) such that n(Q0) = 1 and n A 2 B;n 2 Z =) n(A + 2 ) = n(A) Prove that the completion of n is the Lebesgue measure. 3 Hints. For 1. show that fA 2 B(Q0) : m(A) = n(A)g is a Dynkin system and then a s-algebra via Exercise 4. For 2. use 1. and exhaustion. For part 3. verify the assumptions of part 2. by using dyadic translation invariance. d d 6. #1 Let m denote Lebesgue outer measure on R . Show that for all A ⊂ R there exists a Lebesgue measurable set L with A ⊂ L and m(A) = m(L). 7. Consider the PROPOSITION. Denote by m the Lebesgue (outer) measure. Then (1) if A 2 L then m(A) = inffm(O) : A ⊂ O; Oopeng (2) if A 2 L then m(A) = supfm(K) : K ⊂ A; K compactg (3) A 2 L if and only if for all e > 0 there exists an open set O such that A ⊂ O; m(OnA) < e; (4) A 2 L if and only if for all e > 0 there exists an open set O and a closed set C such that C ⊂ A ⊂ O; m(OnC) < e; and C can be chosen to be compact if m(A) < ¥; d (5) if A 2 L then there exist a set G 2 Gd (R ) (countable intersection of open sets) and a d set F 2 Fs (R ) (countable union of closed sets) such that F ⊂ A ⊂ G; m(GnA) = m(AnF) = 0: #1 Prove (2), (3). #2 Prove (4), (5). d d 8 B(R ) 6= L (R ), I. Define the n-th truncated Cantor set as " n n # [ −n −n −n Cn = ∑ an3 ;3 + ∑ an3 : n (a1;:::;an)2f0;2g j=1 j=1 T Then C = n≥0 Cn is called the Cantor middle thirds set. We have seen in class that C is compact and uncountable1. #1 Show that C contains no intervals: thus C is called nowhere dense or meager. #2 Show that C has zero Lebesgue measure. Sketch the construction a set C1 still satisfying 2 1 #1 but whose Lebesgue measure is 2 . #3 Use point 1. and a cardinality argument to prove that B(R) 6= L (R). You can use without proof that B(R) has the same cardinality as R. See Folland, Prop 1.23, p. 43 for a proof. d d 9 B(R ) 6= L (R ), II (d ≥ 2). #1 Let E ⊂ R be a non Lebesgue measurable set, E˜ = E ×f0g ⊂ R2. Explain why E˜ 2 L (R2)nB(R2). You can use without proof that 2 A 2 B(R ) () fx 2 R : (x;y) 2 Ag 2 B(R) 8y 2 R: 1Moreover, every point of C is an accumulation point for C, but you don’t need to write the proof of this fact here. 4 Solution to selected problems 1. For the first part, it is immediate to see that m(A) ≤ m(n)(A) since we have less coverings of A (n) at disposal when computing m . Furthermore, given any cover fQ jg of A by dyadic intervals in −n k −n R, and Q j with `(Q j) > 2 , denote by fQ j : k = 1;:::;m( j)dg, where m( j) = log2(`(Q j)=2 ), −n the dyadic intervals of sidelength 2 contained in Q j. Since these cover Q j, and since m( j)d k ∑ `(Q j) = `(Q j) k=1 −n k we can replace each Q j with `(Q j) > 2 by fQ j : k = 1;:::;m( j)dg without altering the prop- erties of the cover. For the second part, induction reduces rather easily to the case t = ±2−n for some N. One observes that if Q is a dyadic interval with `(Q) ≤ 2−n then Q ± 2−n is also a dyadic interval. Then, one has, by the first part, say m(A ± +2−n) = m(n)(A ± 2−n) ≤ m(n)(A) = m(A) −n (n) n since for each cover fQ jg of A ± 2 allowed in m the collection fQ j ∓ 2 g is an allowed cover of A. By symmetry, the reverse inequality also follows. d 2. Let A ⊂ R . We first show that n(A) ≤ m(A). To do so we can assume m(A) < ¥, otherwise there is nothing to prove. Let e > 0 and fQ jg be a cover of A by dyadic cubes such that ∑jQ jj ≤ m(A) + e: 1 Let Q˜ j be an open cube with the same center as Q j and each side of length (1 + e) d `(Q j).
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