MA 2210 Fall 2015 - Problem 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 µ denote the Lebesgue outer on R, namely for A ⊂ R ( ∞ ∞ ) [ µ(A) = inf ∑ `(Ij) : A ⊂ Ij, Ij ∈ D ∀ j . j=1 j=1 −n #1 Let n ∈ Z. Prove that µ does not change if we restrict to intervals with `(Ij) ≤ 2 , namely ( ∞ ∞ ) (n) (n) [ −n µ(A) = µ (A), µ (A) := inf ∑ `(Ij) : A ⊂ Ij, Ij ∈ D ∀ j,`(Ij) ≤ 2 . j=1 j=1 N −n #2 Let t ∈ R be of the form t = ∑n=−N kn2 for suitable integers N,k−N,...,kN. Prove that µ is invariant under translations by t, namely µ(A) = µ(A +t) ∀A ⊂ R. −n −m Hints. For #2, reduce to the case t = 2 for some n. Then use #1 and that Dm = {I ∈ D : `(I) = 2 } 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 ( ∞ ∞ ) [ ν(A) = inf ∑ |Rn| : A ⊂ R j, R j ∈ O ∀ j n=0 n=0 where |R| is the Euclidean volume of the cube R. #1 Show that ν coincides with the Lebesgue outer measure µ generated by dyadic cubes. You can use the fact, proved in , that µ(R) = |R| 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 ∈ R ,s > 0,A ⊂ R µ(A +t) = µ(A), µ(sA) = sd µ(A). Hint. For #1 there are two inequalities. One is just the ε/2 j enlargement trick. For the other one, assume ν(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 , µ) be a measure space and throughout E j be measurable sets. There holds 1 2

(P1) µ(/0) = 0; (P2) (monotonicity) if E1 ⊂ E2 then µ(E1) ≤ µ(E2); (P3) µ is countably subadditive; [  (P4) If E j ⊂ E j+1 for all j, then µ E j = lim µ(E j) \  (P5) If µ(E1) < ∞ and E j ⊃ E j+1 for all j, then µ E j = lim µ(E j), and the assumption µ(E1) < ∞ is necessary;  (P6) (lower semicontinuity) µ liminfE j ≤ liminf µ(E j);  S  (P7) (upper semicontinuity) limsup µ(E j) ≤ µ limsupE j provided µ E j < ∞. #1 Prove (P4), (P5) #2 Prove (P6).

4 Dynkin systems. A collection F of of X is called a Dynkin system if (1) (nontrivial) /0 ∈ F , (2) (closed under complement) F ∈ F =⇒ Fc ∈ F , (3) (closed under countable disjoint union) if {Fn : n ∈ N} ⊂ F are pairwise disjoint sets, namely n 6= k =⇒ Fn ∩ Fk = /0,then [ F := Fn ∈ F . n∈N Given G ⊂ P(X), we denote by δ(G ) and by Σ(G ) respectively the smallest Dynkin system and the smallest Σ-algebra containing G . #1 Prove that δ(G ) is well defined and that G ⊂ δ(G ) ⊂ Σ(G ). #2 Let F be a Dynkin system with the further property that (π) F,G ∈ F =⇒ F ∩ G ∈ F (stable under finite intersection). Prove that F is a σ-algebra. #3 Prove that if G has property (π) so does δ(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 µ such that µ(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 ∈ D in R we indicate by D(Q) = {Q ∈ D : Q0 ⊂ Q} and by B(Q) the σ-algebra of Borel subsets of Q. Recall that B(Q) = Σ(D(Q)) for instance.

#1 Let µ,ν be two measures on (Q0,B(Q0)) with the property that µ(Q) = ν(Q) for all Q ∈ D(Q0), and µ(Q0) = 1. Prove that µ = ν. #2 Let µ,ν be two measures on (R,B) with the property that µ(Q) = ν(Q) < ∞ for all Q ∈ D. Prove that µ = ν. #3 Let ν be a measure on (R,B) such that ν(Q0) = 1 and n A ∈ B,n ∈ Z =⇒ ν(A + 2 ) = ν(A) Prove that the completion of ν is the Lebesgue measure. 3

Hints. For 1. show that {A ∈ B(Q0) : µ(A) = ν(A)} is a Dynkin system and then a σ-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 µ denote Lebesgue outer measure on R . Show that for all A ⊂ R there exists a Lebesgue measurable set L with A ⊂ L and µ(A) = µ(L). 7. Consider the

PROPOSITION. Denote by µ the Lebesgue (outer) measure. Then (1) if A ∈ L then µ(A) = inf{µ(O) : A ⊂ O, Oopen} (2) if A ∈ L then µ(A) = sup{µ(K) : K ⊂ A, K compact} (3) A ∈ L if and only if for all ε > 0 there exists an open set O such that A ⊂ O, µ(O\A) < ε; (4) A ∈ L if and only if for all ε > 0 there exists an open set O and a closed set C such that C ⊂ A ⊂ O, µ(O\C) < ε, and C can be chosen to be compact if µ(A) < ∞; d (5) if A ∈ L then there exist a set G ∈ Gδ (R ) (countable intersection of open sets) and a d set F ∈ Fσ (R ) (countable union of closed sets) such that F ⊂ A ⊂ G, µ(G\A) = µ(A\F) = 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)∈{0,2} 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 ×{0} ⊂ R2. Explain why E˜ ∈ L (R2)\B(R2). You can use without proof that 2 A ∈ B(R ) ⇐⇒ {x ∈ R : (x,y) ∈ A} ∈ B(R) ∀y ∈ 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 µ(A) ≤ µ(n)(A) since we have less coverings of A (n) at disposal when computing µ . Furthermore, given any cover {Q j} of A by dyadic intervals in −n k −n R, and Q j with `(Q j) > 2 , denote by {Q j : k = 1,...,m( j)d}, 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 {Q j : k = 1,...,m( j)d} 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 µ(A ± +2−n) = µ(n)(A ± 2−n) ≤ µ(n)(A) = µ(A) −n (n) n since for each cover {Q j} of A ± 2 allowed in µ the collection {Q j ∓ 2 } is an allowed cover of A. By symmetry, the reverse inequality also follows. d 2. Let A ⊂ R . We first show that ν(A) ≤ µ(A). To do so we can assume µ(A) < ∞, otherwise there is nothing to prove. Let ε > 0 and {Q j} be a cover of A by dyadic cubes such that

∑|Q j| ≤ µ(A) + ε. 1 Let Q˜ j be an open cube with the same center as Q j and each side of length (1 + ε) d `(Q j). It follows that |Q˜ j| = (1 + ε)|Q j| and {Q˜ j} is a cover of A by open cubes. Thus ˜ ν(A) ≤ ∑|Q j| = (1 + ε)∑|Q j| ≤ (1 + ε)(µ(A) + ε). Letting ε → 0, the right hand side goes to µ(A), and the inequality is proved. To show the opposite inequality ν(A) ≥ µ(A) assume that ν(A) is finite, otherwise there is nothing to prove. Let ε > 0 be given and R j be a cover of A by open cubes with

ν(A) ≥ ∑|R j| − ε.

By using the Whitney covering lemma, for each j we can find a sequence of dyadic cubes Q jk whose disjoint union covers R j. By Lemma 1.3.1 in the notes,

∑|Q jk| = |R j| k

Then {Q jk} is a cover of A by dyadic cubes, thus

µ(A) ≤ ∑|Q jk| = ∑|R j| ≤ ν(A) + ε. j,k j Letting ε → 0 the inequality follows. For the second part, proceed as in 1. but using translation and dilation invariance of the open cubes. 5

3. (P4). Define inductively

B1 = E1, Bk+1 = Ek+1\Ek, k = 2,3,....

It is easy to see that Bk are measurable, pairwise disjoint and B1 ∪ ··· ∪ Bn = En for all n (draw S S a picture); in particular k Bk = E := Ek Therefore, using finite additivity in the first equality and countable additivity in the third equality n ∞ lim µ(En) = lim µ(Bk) = µ(Bk) = µ(E). n→∞ n→∞ ∑ ∑ k=1 k=1 c c For (P5), let us denote by E = ∩Ek and Fk = E1 ∩(Ek) . The sets Fk increase to E1 ∩E . Notice that µ(Fk) + µ(Ek) = µ(E1). Thus c c µ(E1) = µ(E1 ∩ E ) + µ(E) = lim µ(E1 ∩ (Ek) ) + µ(E) = µ(E1) − lim µ(Ek) + µ(E).

Since µ(E1) is finite, we can subtract it to both sides and obtain the clame. T For (P6), let Fn = k≥n Ek. The sequence Fn increases to F = liminfEn. Clearly µ(Fn) ≤ µ(En). So limsup µ(Fn) ≤ liminf µ(En). Thus by (P4)

µ(F) = lim µ(Fn) ≤ liminf µ(En). 4. Part 1. Well-definition: from the , one can verify that the arbitary intersection of Dynkin systems containing G is also a Dynkin system. The inclusion δ(G ) ⊂ Σ(G ) follows because Σ(G ) is a Dynkin system, and contains G . Part 2. We need to show that D is closed under countable union. Since D is closed under complement and intersection it is closed under relative complement. Let En be a sequence of sets of D, we need to show that their union E is also in D. We can define inductively

D1 = E1, Dk+1 = Ek+1\(D1 ∪ ··· ∪ Dk), k = 2,3,....

Then, proceeding inductively, if D1,...,Dk ∈ D, Fk = D1 ∪ ··· ∪ Dk is a disjoint union of sets in D, thus belongs to D, and Dk+1 = Ek+1\Fk belongs to D as well. Since E = ∪Dk, and the union is disjoint, E ∈ D. Part 3. The trick is the following. If D is any Dynkin system, and F ∈ D, then

DF = {G ⊂ X : F ∩ G ∈ D} is also a Dynkin system. This is quite easy to verify and I leave it to you. Now, to finish the problem, denote by D = δ(G ). For every F ∈ G , the assumption implies G ⊂ DF . But then D ⊂ DF by minimality. This means that for every G ∈ D and F ∈ F the intersection F ∩G ∈ D. But then G ⊂ DG for every G ∈ D. But then by minimality D ⊂ DG for every G ∈ D. Ok, this means that D is closed under intersection. 5. Part 1. We claim that it suffices to show that the collection

F = {A ∈ B(Q0) : µ(Q) = ν(Q)} is a Dynkin system. Since by assumption it contains D(Q0), one has the chain

B(Q0) ⊃ F ⊃ δ(D(Q0)) = Σ(D(Q0)) = B(Q0) where we have used in the equality the fact that δ(D(Q0)) is a σ-algebra, since D(Q0) is closed under intersection, and part 3 of Problem 4. To show that F is a Dynkin system, /0 ∈ F 6 is trivial, closure under countable disjoint union follows from countable additivity of µ,ν and complement (relative to Q0) follows from c c A ∈ F =⇒ µ(A ) = µ(Q0) − µ(A) = ν(Q0) − ν(A) = ν(A ) since µ(A) < ∞. Part 2. I write d = 1 to save notation. Otherwise you have 2d octants to play with. Let A be a n n . Ln = A ∩ [−2 ,0),Rn = A ∩ [0,2 ), An = Ln ∪ Rn and the union is disjoint. It is easy to see that the previous part and the assumptions yield µ(Ln) = ν(Ln), µ(Rn) = ν(Rn) so that µ(An) = ν(An). The set An increase to A, thus

µ(A) = lim µ(An) = limn→∞ν(An) = ν(A). n→∞ Part 3. I do the case d = 1. Let n ≤ 0 say Assume ν is translation invariant by 2−n and ν([0,1). −n −n n −n Let Ij = [ j2 ,( j + 1)2 ) for j = 0,...,2 − 1. Then Ij = I0 + j2 , which clearly means ν(Ij) = ν(I0). Then Ij is a disjoint cover of [0,1). Hence 2n−1 n 1 = ν([0,1)) = ∑ ν(Ij) = 2 ν(I0). j=0 But then all dyadic intervals I of length 2−n have ν(I) = 2−n This tells us that for all dyadic intervals of length ≤ 1, ν(I) = |I|. This actually implies ν(I) = |I| for all dyadic intervals. Apply part 2. d 6. (Solution by D. Schwein) If µ(A) is infinite then we may take L = R . Otherwise, suppose that µ(A) is finite. We will construct L as a Gδ set obtained from increasingly tight covers of A. Let n be a positive integer and let In be a countable disjoint cover of A by dyadic cubes such that ∑ |I| ≤ µ(A) + n−1. I∈In S T −1 −1 Let Ln = In and let L = n Ln. Then µ(Ln) ≤ µ(A) + n . Hence µ(L) ≤ µ(A) + n for every n, meaning that µ(L) ≤ µ(A). But A ⊆ L, so that µ(A) ≤ µ(L) and µ(A) = µ(L). 7. A detailed solution can be found on https://www.math.ucdavis.edu/~hunter/measure_theory/measure_notes.pdf 8. I omit the part on the construction of the Cantor set, which can be found in plenty of ref- erences. I only use that C has the cardinality of R to prove that most Lebesgue subsets are non Borel. Since (R,L , µ) is a complete measure space and µ(C) = 0, all subsets of C are Lebesgue measurable (and thus have measure zero). Since P(C) has the same cardinality as P(R), and B and R have the same cardinality it must be P(C) 6⊂ B. 9. Let E ⊂ R be a nonmeasurable . Then E × {0} ⊂ R2 is a subset of R × {0}. It can be seen rather easily that the 2-dimensional outer Lebesgue measure of R × {0} is 0: [−n,n] × {0} has outer measure zero (cover it by the boxes [−n,n] × [−ε/n,ε/n]) and R × {0} can be countably exhausted by these. Then R × {0} is also Lebesgue measurable, since it is closed in R2. But then E × {0} has to be measurable as well, by completeness of Lebesgue measure. Since E × {0} has a section which is not a Borel set of R, it cannot be a Borel set of R2. 7

I did not ask, but it should not be too hard to prove that 2 A ∈ B(R ) ⇐⇒ Ay = {x : (x,y) ∈ A} ∈ B(R) ∀y ∈ R. Note that the same statement is false for L .

BROWN UNIVERSITY MATHEMATICS DEPARTMENT,BOX 1917, PROVIDENCE, RI 02912, USA E-mail address: [email protected] (F. Di Plinio)