Strong one-sided density without uniform density
SStrong one-sided density without uniform density
Zolt´an Buczolich ∗ , Department of Analysis, ELTE E¨otv¨os Lor´andUniversity, P´azm´any P´eter S´et´any 1/c, 1117 Budapest, Hungaryemail: [email protected] http://buczo.web.elte.hu ORCID Id: 0000-0001-5481-8797Bruce Hanson, Department of Mathematics,Statistics and Computer Science,St. Olaf College, Northfield, Minnesota 55057, USAemail: [email protected]
Bal´azs Maga † , Department of Analysis, ELTE E¨otv¨os Lor´andUniversity, P´azm´any P´eter S´et´any 1/c, 1117 Budapest, Hungaryemail: [email protected] http://magab.web.elte.hu/ andG´asp´ar V´ertesy ‡ , Department of Analysis, ELTE E¨otv¨os Lor´andUniversity, P´azm´any P´eter S´et´any 1/c, 1117 Budapest, Hungaryemail: [email protected] 26 th January, 2021 ∗ This author was supported by the Hungarian National Research, Development and Innovation Office–NKFIH,Grant 124003. † This author was supported by the ´UNKP-20-3 New National Excellence Program of the Ministry for Inno-vation and Technology from the source of the National Research, Development and Innovation Fund, and by theHungarian National Research, Development and Innovation Office–NKFIH, Grant 124749. ‡ This author was supported by the ´UNKP-20-3 New National Excellence Program of the Ministry for Inno-vation and Technology from the source of the National Research, Development and Innovation Fund, and by theHungarian National Research, Development and Innovation Office–NKFIH, Grant 124749.
Mathematics Subject Classification:
Primary : 28A05, Secondary : 28A75.
Keywords: strong one-sided density, uniform density type. a r X i v : . [ m a t h . C A ] J a n bstract In this paper we give an example of a closed, strongly one-sided denseset which is not of uniform density type. We also show that there is a setof uniform density type which is not of strong uniform density type.
Strongly one-sided dense sets and sets of uniform density type (UDT) were in-troduced in [2]. (In Section 2 the reader can find the definition of UDT sets aswell as the definitions of several other less well-known concepts appearing in thisintroduction.) In [2] we showed that UDT sets are strongly one-sided dense. Thereferee of that paper asked whether the reverse implication is also true. In this pa-per, answering this question, we provide an example of a closed, strongly one-sideddense set which is not UDT.The concepts of strongly one-sided dense sets and UDT sets have played animportant role in our research project related to characterizing Lip 1 and lip 1 sets.(These are sets E in R for which there is a continuous function defined on R suchthat Lip f = E , or lip f = E , respectively). In [3] the lip 1 sets were characterizedas countable unions of closed sets which are strongly one-sided dense. Such a nicecharacterization of the Lip 1 sets has not yet been discovered. The main result of[2] states that if E is G δ and E has UDT then there exists a continuous function f satisfying Lip f = E , that is, the set E is Lip 1. On the other hand, in [4] weshowed that the converse of this statement does not hold. There exist Lip 1 setswhich are not UDT. Moreover, in [2] we showed that if E ⊂ R is Lip 1 then E isa weakly dense G δ set. We also showed that there exists a weakly dense G δ set E ⊂ R which is not Lip 1, thus this condition on E is necessary, but not sufficient.Replacing a lim sup set by a lim inf set in the definition of UDT sets oneobtains sets of strong uniform density type, SUDT for short. In this paper, weshow that the SUDT property is strictly stronger than the UDT property by givingan example of a UDT set which is not SUDT. On the other hand, we prove thatany measurable set equals an SUDT set almost everywhere.In [4] we proved that there exists a measurable SUDT set E such that for any G δ set (cid:101) E satisfying | E ∆ (cid:101) E | = 0 the set (cid:101) E does not have UDT. In the same paperwe also showed that modulo sets of zero Lebesgue measure any measurable setcoincides with a Lip 1 set. In fact, these two results imply that there exist Lip 1sets not having UDT, a result we mentioned earlier. We denote the closure of the set A by A , its complement by A c .2 efinition 2.1. The set E is strongly one-sided dense at x if for any sequence r n → + we have max (cid:110) | E ∩ [ x − r n , x ] | r n , | E ∩ [ x, x + r n ] | r n (cid:111) → . The set E is strongly one-sided dense if E is strongly one-sided dense at everypoint x ∈ E . Definition 2.2.
Suppose that E ⊆ R is measurable and γ, δ >
0. Let E γ,δ = (cid:26) x ∈ R : ∀ r ∈ (0 , δ ] , max (cid:26) | ( x − r, x ) ∩ E | r , | ( x, x + r ) ∩ E | r (cid:27) ≥ γ (cid:27) , where | E | denotes the Lebesgue measure of the set E .We say that E has uniform density type (UDT) if there exist sequences γ n (cid:37) δ n (cid:38) E ⊆ (cid:84) ∞ k =1 (cid:83) ∞ n = k E γ n ,δ n .We say that E has strong uniform density type (SUDT) if there exist sequences γ n (cid:37) δ n (cid:38) E ⊆ (cid:83) ∞ k =1 (cid:84) ∞ n = k E γ n ,δ n .Clearly SUDT is a stronger property than UDT.The “big Lip” and “little lip” functions are defined as follows:Lip f ( x ) = lim sup r → + M f ( x, r ) , lip f ( x ) = lim inf r → + M f ( x, r ) , (2.1)where M f ( x, r ) = sup {| f ( x ) − f ( y ) | : | x − y | ≤ r } r . While the definition of the Lip function has a long history, the definition of lipis more recent see [6] and [9]. For some more recent appearances of lip f we referto [1], [7], [8], [5], [10] and [12]. Definition 2.3.
A set E ⊂ R is a Lip 1 (lip 1) set if there exists a continuousfunction f defined on R such that Lip f = E (lip f = E .) Definition 2.4.
Given a sequence of non-degenerate closed intervals { I n } , wewrite I n → x if x ∈ I n for all n ∈ N and | I n | → E is weakly dense at x if there exists I n → x such that | E ∩ I n || I n | →
1. The set E is weakly dense if E is weakly dense at x for each x ∈ E .3 Main results
Theorem 3.1.
There exists a closed, strongly one-sided dense set which does nothave UDT.Proof.
We will define a closed strongly one-sided dense set E with the followingproperty: for any sequences γ n (cid:37) δ n (cid:38)
0, we can find x ∗ = x ( γ n ,δ n ) n ∈ N ∈ E such that x ∗ / ∈ (cid:84) ∞ k =1 (cid:83) ∞ n = k E γ n ,δ n . Hence E cannot be UDT: the points x ( γ n ,δ n ) n ∈ N demonstrate that no sequence γ n , δ n can be a valid choice in the definition.In order to define this set E , we will recursively define sequences. First, wedefine a n = 2 − n . Proceeding recursively, if a n ,...,n k ∈ R is already defined for all n , ..., n k ∈ N , then we define a n ,n ,...,n k ,n = a n ,n ,...,n k − ,n k +1 + 12 n +3 r n ,...,n k , (3.1)where r n ,...,n k = a n ,...,n k − a n ,...,n k − ,n k +1 . To get familiar with this definition, and for later use we calculate a = 2 − , a = 2 − , a = a + 12 ( a − a ) = 2 − + 2 − · − , (3.2) a = a + 12 ( a − a ) = 2 − + 2 − · − , r := a − a = 2 − · − . Figure 1:
The set A and its part with some objects used in the proofNote that a n ,...,n k ,n (cid:38) a n ,...,n k +1 (3.3)and the following conditions are satisfied:4i) a n ,...,n k ,n ∈ ( a n ,...,n k − ,n k +1 , a n ,...,n k ) for any k , any indices n , ..., n k , and n ∈ N ,(ii) a n ,...,n k − ,n k +1 − a n ,...,n k − ,n k +2 = a n ,...,n k − ,n k − a n ,...,n k − ,n k +1 k and any indices n , ..., n k .For each k ∈ N define A k = { a n ,n ,...,n k : n , n , . . . , n k ∈ N } and let A = (cid:83) ∞ k =1 A k . Observe that A ⊂ (0 , ]. The conditions above yield that the set A isdisjoint from all the sets (cid:20) a n ,...,n k − r n ,...,n k , a n ,...,n k − r n ,...,n k (cid:21) ∪ (cid:20) a n ,...,n k + 14 r n ,...,n k , a n ,...,n k + 12 r n ,...,n k (cid:21) . Set α k = 10 − k , γ k = 1 − α k and for n , n , . . . , n k ∈ N define I n ,n ,...,n k ,L = (cid:18) a n ,...,n k − r n ,...,n k , a n ,...,n k − (cid:16) − α k (cid:17) r n ,...n k (cid:19) ,I n ,n ,...,n k ,R = (cid:18) a n ,...,n k + (cid:16) − α k (cid:17) r n ,...,n k , a n ,...,n k + 12 r n ,...n k (cid:19) , and I n ,n ,...,n k = I n ,n ,...,n k ,L ∪ I n ,n ,...,n k ,R . Observe that { I n ,n ,...,n k ,L : k, n , n , . . . , n k ∈ N } ∪ { I n ,n ,...,n k ,R : k, n , n , . . . , n k ∈ N } consists of pairwise disjoint closed intervals. (3.4)We also define J n ,n ,...,n k ,L = (cid:20) a n ,...,n k − r n ,...,n k , a n ,...,n k (cid:21) and J n ,n ,...,n k ,R = (cid:20) a n ,...,n k , a n ,...,n k + 12 r n ,...n k (cid:21) . Then set E = [ − , \ (cid:91) n ,...,n k ∈ N I n ,...,n k . We first demonstrate that E is not UDT. Suppose otherwise, so there exist (cid:101) γ n (cid:37) (cid:101) δ n (cid:38) E ⊂ ∞ (cid:92) k =1 ∞ (cid:91) n = k E (cid:101) γ n , (cid:101) δ n . (3.5)5or each n ∈ N let δ n = min { (cid:101) δ n (cid:48) : (cid:101) γ n (cid:48) < γ n +1 } . Then for all n (cid:48) satisfying γ n ≤ (cid:101) γ n (cid:48) < γ n +1 we have E (cid:101) γ n (cid:48) , (cid:101) δ n (cid:48) ⊂ E γ n ,δ n and thereforefrom (3.5) we get E ⊂ ∞ (cid:92) k =1 ∞ (cid:91) n = k E γ n ,δ n . (3.6)Given an interval J we define E J = | E ∩ J || J | . Then from the definitions it followsthat if J = J n ,...,n k ,L or J = J n ,...,n k ,R , where n k ≥
1, we have E J ≤ − α k < γ k .It follows from this that for each x = a n ,...,n k ∈ A we havemax (cid:26) | ( x − r, x ) ∩ E | r , | ( x, x + r ) ∩ E | r (cid:27) < γ k , for r = | J n ,...,n k ,L | = | J n ,...,n k ,R | = 12 r n ,...,n k . (3.7)Then, by continuity inequality (3.7) holds in a neighborhood of x as well.We now describe how to choose a point x ∗ ∈ E such that x ∗ / ∈ (cid:84) ∞ k =1 (cid:83) ∞ n = k E γ n ,δ n , which will give the desired contradiction to (3.6). To streamline the notation weuse the following convention: given x = a n ,...,n k ∈ A , we define r x = r n ,...,n k = | J n ,...,n k ,L | = | J n ,...,n k ,R | . If n k is sufficiently large, then r x < δ k and hence from(3.7) we know that there is a neighborhood U x of x such that U x ∩ E γ k ,δ k = ∅ . (3.8)Thus, we can choose n ∈ N and a closed interval I containing a n in itsinterior such that n > I ∩ E γ ,δ = ∅ . Proceeding inductively and using(3.8), we choose a sequence of integers { n k } and closed intervals { I k } such thatfor all k ∈ N we have • n k > , • a n − ,n − ,...,n k − − ,n k ∈ int ( I k ) , (keeping in mind (3.3)) • I k ⊃ I k +1 , • I k ∩ E γ k ,δ k = ∅ .Since E is closed, it follows that E ∩ ( (cid:84) ∞ k =1 I k ) (cid:54) = ∅ and for any x ∗ ∈ E ∩ ( (cid:84) ∞ k =1 I k )we have that x ∗ / ∈ (cid:84) ∞ k =1 (cid:83) ∞ n = k E γ n ,δ n . This concludes the argument that E is notUDT.Next we show that all points x ∈ E are strong one-sided density points of E .6irst suppose that x / ∈ A . In this case we claim thatthere is a closed interval I such that x ∈ I ⊂ E (3.9)and therefore E is strongly one-sided dense at x . To see the truth of our claim let J = ∞ (cid:91) k =1 J k , where J k = (cid:110) [ a n ,...,n k − r n ,...,n k , a n ,...,n k + 12 r n ,...,n k ] : n , ..., n k ∈ N (cid:111) . (3.10)For each J = [ a n ,...,n k − r n ,...,n k , a n ,...,n k + r n ,...,n k ] ∈ J we define J (cid:48) = I n ,...,n k .For (cid:15) >
0, define J (cid:15) = { J ∈ J : | J | ≥ (cid:15) } and let E (cid:15) = [ − , \ (cid:91) J ∈J (cid:15) J (cid:48) . Then each E (cid:15) is a finite union of closed intervals and E = (cid:84) (cid:15)> E (cid:15) . Choose (cid:15) = dist ( { x } , A ) . Thus, we have x ∈ I (cid:48) ⊂ E (cid:15) for some closed interval I (cid:48) . It mayhappen that x is an endpoint of I (cid:48) , but I (cid:48) is non-degenerate by (3.4). Observe that E is constructed from E (cid:15) by removing unions of open intervals J (cid:48) ⊂ J , where J iscentered at a point of A and | J | < (cid:15) . Since (cid:15) ≤ dist ( { x } , A ), it follows that thedistance from x to any of the intervals being removed is at least dist ( { x } , A )and therefore there must be a non-degenerate closed subinterval I of I (cid:48) such that x ∈ I ⊂ E .Now suppose that x ∈ A .To finish the proof we need the following technical lemma: Lemma 3.2.
For any n , . . . , n k ∈ N , we let K n ,...,n k = [ a n ,...,n k − ,n k +1 , a n ,...,n k − ,n k ] .Then | E ∩ K n ,...,n k | ≥ (1 − α k ) | K n ,...,n k | for every n , ..., n k ∈ N . (3.11) Proof.
For n ∈ N we set E n = [ − , \ (cid:91) n ,...,n k ∈ N k ≤ n I n ,...,n k . Note that E = (cid:84) ∞ n =1 E n .Let K = K n ,...,n k = [ a n ,...,n k − ,n k +1 , a n ,...,n k − ,n k ]. Consider the set E k ∩ K . Thisset consists of the interval K with two open intervals I n ,...,n k ,L and I n ,...,n k +1 ,R oftotal length α k | K | removed. At the next stage of the construction, to create theset E k +1 ∩ K , for each J ∈ J k +1 such that J ⊂ K we remove two open intervals7f total length equal to 2 α k +1 | J | . Since the intervals in J k +1 are non-overlapping,it follows that the total length of the intervals removed from K at the ( k + 1)ststage of the construction is less than 2 α k +1 | K | . An entirely analogous argumentshows that at the j th stage of the construction where j > k , the total length ofthe intervals removed from K is no more than 2 α j | K | . It follows that | E ∩ K | ≥ (cid:16) − α k − ∞ (cid:88) n = k +1 α n (cid:17) | K | > (1 − α k ) | K | , as desired.Note, first of all, that if x ∈ A , then from the construction of E we see thatthere exists a closed interval I = [ x − δ, x ] such that I ⊂ E and therefore E isstrongly one-sided dense at x .So we may assume that x / ∈ A . Also, note that 0 ∈ [ − , ⊂ E so E is stronglyone-sided dense at 0 and therefore we may assume that x > x ∈ A \ A , we can choose a sequence of indices n , n , . . . and anested sequence of closed intervals K (cid:48) ⊃ K (cid:48) ⊃ K (cid:48) ⊃ . . . such that for each k ∈ N we have x ∈ K (cid:48) k = K n ,...,n k . For each k ∈ N we let r (cid:48) k = | K (cid:48) k | so that r (cid:48) k (cid:38) { x } = ∞ (cid:92) k =1 K (cid:48) k . (3.12)Assume that r (cid:48) k +1 ≤ r ≤ r (cid:48) k . We consider two cases: Case 1:
Assume that r (cid:48) k ≤ r ≤ r (cid:48) k .In this case we first assume that n k >
1. Let K = K n ,...,n k = [ a n ,...,n k − ,n k +1 , a n ,...,n k ]and K ∗ = K n ,...,n k − ,n k − = [ a n ,...,n k , a n ,...,n k − ,n k − ]. Then [ x, x + r ] ⊂ K ∪ K ∗ andit follows from Lemma 3.2 that | E c ∩ [ x, x + r ] | ≤ α k ( | K | + | K ∗ | ) = 6 α k r (cid:48) k ≤ α k r, so we get | E ∩ [ x, x + r ] | ≥ (1 − α k ) r. (3.13)A similar argument gives the same estimate if n k = 1. We leave the details upto the reader. Case 2:
Assume that r (cid:48) k +1 ≤ r ≤ r (cid:48) k .Note that by (3.1), we have | K n ,...,n k , | = r (cid:48) k > r (cid:48) k +1 and therefore n k +1 ≥ x ∈ K (cid:48) k +1 = K n ,...,n k ,n k +1 ⊂ [ a n ,...,n k − ,n k +1 , a n ,...,n k , ], and weobtain[ x, x + r ] ⊂ (cid:104) a n ,...,n k − ,n k +1 , a n ,...,n k , + 164 r (cid:48) k (cid:105) ⊂ [ a n ,...,n k − ,n k +1 , a n ,...,n k , ] .
8o simplify the notation at this point we define a l = a n ,...,n k ,l and L l = [ a l +1 , a l ]so, for example, x ∈ K (cid:48) k +1 = L n k +1 . Let n be the largest value of n such that[ x, x + r ] ⊂ n k +1 (cid:91) l = n L l that is x + r ∈ L n . As r (cid:48) k +1 ≤ r , we have n < n k +1 . Thus x ≤ a n ,...,n k ,n +1 , andhence [ a n ,...,n k ,n +1 , , a n ,...,n k ,n +1 ] ⊂ [ x, x + r ]. Therefore | [ x, x + r ] | ≥ | [ a n ,...,n k ,n +1 , , a n ,...,n k ,n +1 ] | = 1516 r n ,...,n k ,n +1 = 1532 r n ,...,n k ,n > n k +1 (cid:88) l = n n − l r n ,...,n k ,n = 1564 n k +1 (cid:88) l = n r n ,...,n k ,l = 1564 n k +1 (cid:88) l = n | L l | . (3.14)Consequently | E c ∩ [ x, x + r ] | ≤ n k +1 (cid:88) l = n | E c ∩ L l | ≤ by Lemma 3.2 α k +1 n k +1 (cid:88) l = n | L l | ≤ by (3.14) α k +1 · r. and hence | E ∩ [ x, x + r ] | ≥ (cid:16) − α k +1 (cid:17) r. (3.15)Combining the estimates (3.13) and (3.15) and making use of the fact that α k (cid:38) r (cid:48) k → r →
0, we conclude that E is strongly one-sided dense at x , as desired.A question which is similar in nature to the one answered in Theorem 3.1 iswhether or not there are UDT sets which are not SUDT. This question is answeredaffirmatively by the next theorem. Theorem 3.3.
There is a UDT set which is not SUDT.Proof.
First we observe that if F ⊂ R is measurable, and x ∈ I ⊂ F where I is anon-degenerate interval, then x ∈ (cid:84) ∞ k =1 (cid:83) ∞ n = k F γ n ,δ n for any sequences γ n (cid:37) δ n (cid:38) E defined in the proof of Theorem 3.1. This set will bemodified in a set of measure zero to obtain the set E (cid:48) which will be a UDT set ,but not a SUDT set.First we claim that E := (cid:8) x ∈ E : { x } is a component of E (cid:9) is a set of measure 0. (3.16)9rom (3.9) it follows that E ⊂ A .According to Lebesgue’s density theorem, it is enough to prove that A hasno density points. Take an arbitrary x ∈ A . By the argument following theproof of Lemma 3.2 and ending at (3.12), we can suppose that x / ∈ A and wecan choose a sequence of indices n , n , . . . such that { x } = (cid:84) ∞ k =1 K n ,...,n k . As( a n ,...,n k , , a n ,...,n k ) ∩ A = ∅ for every k ∈ N , we have | A ∩ K n ,...,n k | ≤ | [ a n ,...,n k − ,n k +1 , a n ,...,n k , ] | r n ,...,n k = 116 . Hence x cannot be a density point of A . This implies (3.16).Let γ (cid:48) n := 1 − − n and δ (cid:48) n := 2 − n for every n ∈ N , and set E (cid:48) := ∞ (cid:92) k =1 ∞ (cid:91) n = k ( E γ (cid:48) n ,δ (cid:48) n ∩ E ) . By the observation at the beginning of this proof E \ E ⊂ E (cid:48) . Hence (3.16) impliesthat E (cid:48) ⊂ (cid:84) ∞ k =1 (cid:83) ∞ n = k E (cid:48) γ (cid:48) n ,δ (cid:48) n , which means that E (cid:48) is UDT.We claim that for every ( γ n ) ∞ n =1 and ( δ n ) ∞ n =1 with γ n (cid:37) δ n (cid:38)
0, there isan x (cid:48) ∈ E (cid:48) \ (cid:83) ∞ k =1 (cid:84) ∞ n = k E (cid:48) γ n ,δ n , that is, E (cid:48) is not SUDT.Fix ( γ n ) ∞ n =1 and ( δ n ) ∞ n =1 .Let k, n , . . . , n k ∈ N and x ∈ K n ,...,n k ∩ E (cid:48) . As I n ,...,n k − ,n k ,R ⊂ [ x, x +2 r n ,...,n k ]and I n ,...,n k − ,n k +1 ,L ⊂ [ x − r n ,...,n k , x ] , we havemax (cid:26) | [ x − r n ,...,n k , x ] ∩ E | r n ,...,n k , | [ x, x + 2 r n ,...,n k ] ∩ E | r n ,...,n k (cid:27) ≤ r n ,...,n k − | I n ,...,n k − ,n k +1 ,L | r n ,...,n k = 1 − α k r n ,...,n k − ,n k +1 r n ,...,n k = 1 − α k . (3.17)On the other hand, by (3.13) and (3.15),if x ∈ A ∩ K n ,...,n k and r ∈ (0 , r n ,...,n k ), thenmax (cid:26) | [ x − r, x ] ∩ E | r , | [ x, x + r ] ∩ E | r (cid:27) ≥ − α k . (3.18)We will define sequences ( n (cid:48) k ) ∞ k =1 , ( k (cid:48) j ) ∞ j =1 , ( m j ) ∞ j =1 and ( m (cid:48) j ) ∞ j =1 by inductionsuch that the latter three of them are strictly increasing and for every j ∈ N (a) K n (cid:48) ,...,n (cid:48) k (cid:48) j ∩ E ⊂ E (cid:48) γ (cid:48) m (cid:48) j ,δ (cid:48) m (cid:48) j ,(b) K n (cid:48) ,...,n (cid:48) k (cid:48) j ,n (cid:48) k (cid:48) j +1 ∩ E (cid:48) γ mj ,δ mj = ∅ . 10he existence of such sequences would conclude the proof. Notably, as K n (cid:48) ,...,n (cid:48) k ∩ E is a non-empty closed set for every k , and lim k →∞ | K n (cid:48) ,...,n (cid:48) k | = 0, the set ∞ (cid:92) k =1 K n (cid:48) ,...,n (cid:48) k ∩ E consists of a single point. If this point is denoted by x (cid:48) , then by (a), x (cid:48) ∈ (cid:84) ∞ k =1 (cid:83) ∞ n = k ( E (cid:48) γ (cid:48) n ,δ (cid:48) n ∩ E ) = E (cid:48) . However, x (cid:48) / ∈ (cid:83) ∞ k =1 (cid:84) ∞ n = k E (cid:48) γ n ,δ n by (b). Conse-quently, what remains in order to complete the proof is the construction of thesesequences.Set m (cid:48) := 1, k (cid:48) := 10, and n (cid:48) k := 1 for every k ∈ N ∩ [1 , k (cid:48) ]. Recall (3.2). Since δ (cid:48) m (cid:48) = 2 − < − − · = r n (cid:48) ,...,n (cid:48) k (cid:48) and γ (cid:48) m (cid:48) = 1 − − < − · − = 1 − α k (cid:48) ,by (3.18) (a) is satisfied for j = 1.Now, suppose that j ∈ N and we have already defined k (cid:48) , . . . , k (cid:48) j , m (cid:48) , . . . , m (cid:48) j , m , . . . , m j − and n (cid:48) , . . . , n (cid:48) k (cid:48) j .Let m j ∈ N be such that γ m j > − α k (cid:48) j , and m j > m j − if j >
1. Take a largeenough n (cid:48) k (cid:48) j +1 to satisfy 2 r n (cid:48) ,...,n (cid:48) k (cid:48) j ,n (cid:48) k (cid:48) j +1 < δ m j . Then (b) is satisfied by (3.17).Observe that lim i →∞ α k (cid:48) j + i − γ (cid:48) m (cid:48) j + i = lim i →∞ · − ( k (cid:48) j + i ) − ( m (cid:48) j + i ) = 0and lim i →∞ r n (cid:48) ,...,n (cid:48) k (cid:48) j , i δ (cid:48) m (cid:48) j + i = lim i →∞ r n (cid:48) ,...,n (cid:48) k (cid:48) j · − i − m (cid:48) j + i ) = ∞ where 1 i = 1 , . . . , (cid:124) (cid:123)(cid:122) (cid:125) i times . Hence we can select large enough i j ∈ N such that γ (cid:48) m (cid:48) j + i j < − α k (cid:48) j + i j and δ (cid:48) m (cid:48) j + i j < r n (cid:48) ,...,n (cid:48) k (cid:48) j , ij .Let m (cid:48) j +1 := m (cid:48) j + i j and k (cid:48) j +1 := k (cid:48) j + i j . Set n (cid:48) k := 1 for every k ∈ [ n (cid:48) k (cid:48) j +2 , n (cid:48) k (cid:48) j +1 ] ∩ N . Then (a) is true by (3.18). This concludes the proof.As one can expect, each measurable set equals an SUDT set modulo a set ofmeasure zero. We prove this in the next simple theorem. Theorem 3.4. If E ⊂ R is measurable, then there exists E ∗ ⊆ E such that E ∗ has SUDT, and | E \ E ∗ | = 0 . That is, every measurable set equals an SUDT setmodulo a set of measure zero. roof. Fix a measurable set E ⊂ R . By the Lebesgue density theorem, almostevery point of E is a density point of the set, hence we can suppose that E is denseat all of its points.Now fix γ n (cid:37)
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