Genericity and UD-random reals

Avigad introduced the notion of UD-randomness based in Weyl’s 1916 definition of uniform distribution modulo one. We prove that there exists a weakly 1-random real that is neither UD-random nor weakly 1-generic. We also show that no 2-generic real can Turing compute a UD-random real. 2000 Mathematics Subject Classification 03D32 (primary); 03D80 (secondary)


Introduction
In [1] Avigad introduced a randomness notion based on uniform distribution modulo 1, as described by Weyl in [10].A sequence x i of real numbers is said to be uniformly distributed modulo one in the case that for every interval I ⊆ [0, 1], we have where µ is Lebesgue measure and [x i ] denotes the fractional part of x i .Weyl proved that if a i is a sequence of distinct integers, then for almost every real x, the sequence a i x is uniformly distributed modulo one [10].
The definition below considers recursive sequences a i of distinct integers, observing that there are countably many such sequences and that therefore almost every x ∈ [0, 1] has the property that a i is uniformly distributed modulo one: Definition 1.1 [1] A real x ∈ [0, 1] is UD-random if and only if for every recursive sequence a i of distinct integers, a i x is uniformly distributed modulo one.
We know several facts about the relationship between UD-randomness and other randomness notions, all due to Avigad's original paper.For instance, every Schnorr random real is UD-random; however, there are UD-random reals that are not weakly 1-random [1].This is surprising, since weak 1-randomness is such a weak randomness notion: not only is every Schnorr random weakly 1-random, but so (by a result of Kurtz) is every weakly 1-generic [6].Avigad observed that, in consequence, there are weakly 1-random reals that are not UD-random, since a UD-random must satisfy the law of large numbers, and no real that is even weakly 1-generic does.
In this paper we investigate the relationship of the UD-random reals to the weakly 1-random reals and the weakly 1-generic reals further, and determine when a generic real can Turing compute a UD-random.There are very few conclusions that can be drawn about Turing degrees of UD-randoms from existing results.We can conclude, for instance, by a result of Nies, Stephan, and Terwijn, that every high Turing degree contains a UD-random because every high Turing degree contains a Schnorr random real [8] and that there is therefore a low UD-random and a hyperimmune-free UD-random using various basis theorems.However, we cannot use any of our standard results involving the coincidence of randomness notions on given classes of Turing degrees.For instance, in the hyperimmune-free degrees, weak 1-randomness is equivalent to Martin-Löf-randomness [8], but we cannot conclude that this also holds of UDrandomness, since there are UD-randoms that are not weakly 1-random.We also know that Schnorr and Martin-Löf randomness coincide outside the high degrees [8], but we cannot conclude that UD-randomness coincides with these notions there as well.
We describe the weakly 1-random degrees further using UD-randomness in Section 2, and in Section 3 we discuss the relationship between the Turing degrees of UD-randoms and generic reals.Section 4 is dedicated to open questions.

Basic observations and notation
Our notation is standard and follows Soare [9] and Downey and Hirschfeldt [3]; we add that we denote the concatenation of two finite binary strings σ and τ by σ τ .
Since we often discuss finite binary strings in the context of uniform distribution modulo one, we make the following convention: We identify a binary string σ with the binary representation of a real on the unit interval.Of course, a finite string also denotes a real number -a dyadic rational number, in fact (the finite string σ represents the same real as σ 0).From time to time we will apply certain functions (especially multiplication) which will take σ outside the unit interval, but since these functions are always postcomposed with the fractional part function, no ambiguity will result.
Fact 1.2 A real X is UD-random if and only if it is UD-random with respect to dyadic intervals.
Proof One implication is obvious.For the other, suppose X is not UD-random.Suppose, in particular, that for an interval I and for arbitrarily large n we have Now we can find I such that I has dyadic endpoints and µ(I I ) < .Then for arbitrarily large n we have 2 Categorizing weakly 1-random reals Since the UD-random reals and the weakly 1-generic reals are disjoint, one can reasonably ask whether the UD-random weakly 1-random reals and the weakly 1generic reals partition the weakly 1-random reals or whether there is a weakly 1-random real that is neither UD-random nor weakly 1-generic.We answer this question with the following theorem.
Theorem 2.1 There is a weakly 1-random real that is neither UD-random nor weakly 1-generic.
Proof To construct such an X ∈ 2 ω , we must build not only X itself but also a dense Σ 0 1 set of strings S that X avoids (witnessing that X is not weakly 1-generic) and an infinite recursive sequence a i witnessing that X is not UD-random with respect to the interval ( 12 , 1).We must also satisfy the following requirement for each Σ 0 1 set W e to ensure that X is weakly 1-random: Without loss of generality, we take W e to be a subset of 2 <ω rather than ω .This finite injury construction requires us to keep track of two sequences of finite binary strings for each e: σ e,s s and τ e,s s .Each τ e,s is a string that is our guess for X under the assumption that no requirement that has higher priority than R e will ever act after stage s, and the σ e,s are extensions of the τ e,s that we construct in tandem with the a i -sequence to ensure that X will not be UD-random.We also build an auxiliary sequence n e,s s that will keep track of the measure we need to activate requirement R e at stage s, and we will also record the length of the a i -sequence at the end of each stage and denote this by s .
In general, our strategy is this.At stage s, we have a list of the requirements with indices below s that have already acted.For every other index e < s, we will keep track of those e < e such that R e has already acted, and we will have an approximation σ e,s−1 to X based on this list.If our list already includes all e < e such that R e will act at any point, then our guess will be correct.We will work in the cylinder above σ e,s−1 and wait for a lower-priority W e to cover a certain fraction of this cylinder.If this happens, we will act to satisfy R e and extend our σ e,s−1 to a string τ e,s to meet W e .Then we will extend τ e,s to σ e,s and extend our sequence of a i s accordingly to make sure that 1  2 < [a i σ e,s ] < 1 for no more than one-fourth of all a i in our sequence.To build a dense set of strings S that X avoids, we will add strings to S every time a new requirement acts to ensure that each string at a new, higher level is extended by an element of S. To this end, we will also keep track of the number of times we have acted to satisfy a requirement: we define t s to be this number at the end of each stage s.
Construction Set S 0 = ∅ and a i = , and let σ e,0 = for all e, and 0 = t 0 = 0.At stage s, suppose we have a list 0 ≤ e 0 < e 1 < e 2 < . . .< e k < s such that R ei has acted already for each i.First, determine whether there is an e less than s such that and R e is not currently satisfied.If there is not, let S s = S s−1 and set σ e,s = σ e,s−1 for all e and go on to the next stage.If there is such an e, choose the least one and act to satisfy R e .
We begin by choosing a τ e,s in W e that extends σ e,s−1 and avoids S s−1 ; we choose the leftmost shortest such string.We check to see whether If this is the case, set σ e,s = τ e,s 0 and let a s−1 +1 = 2 m , where m is the position of this last 0. Then the fractional part of a s−1 +1 σ e,s is less than 1  2 , and, since On the other hand, if is not less than 1  4 , find the least N such that Take σ e,s = τ e,s 0 N and set a s−1 +i = 2 s+i for all 1 ≤ i ≤ N .For each i such that s−1 < i ≤ s−1 + N , the fractional part of a i σ e,s is less than 1  4 , so 4 once more.At this point we say that R e has been met.However, we are not done with stage s: we must extend our set S to ensure its density.To do this, consider each string of length |σ e,s |.If such a string has an extension in S s−1 , we need not extend it again; otherwise, extend it to a string τ long enough that the measure of [τ ] is no more than ).This guarantees that the total measure that S consumes above each string of length |σ e,s | is no more than half the available measure.
At this point, we are ready to end this stage.We have already determined σ e,s .For e > e, reinitialize σ e ,s by setting it equal to σ e,s .If we acted to satisfy a requirement at this stage, define t s = t s−1 + 1.
Let X = lim e,s σ e,s .Verification Lemma 2.2 The limit X = lim e lim s σ e,s exists.
Proof We proceed by induction.Consider e, and let s 0 be the first stage such that all e < e that will ever act already have.If we never act to satisfy R e , then σ e,s 0 = σ e,t for all t > s 0 .If we act to satisfy R e at a later stage s 1 , then we never reinitialize σ e,s 1 and we have σ e,s 1 = σ e,t for all t > s 1 .Thus, σ e = lim s σ e,s exists.Since in our construction, whenever e 1 > e, we have σ e ≺ σ e 1 , the limit X = lim e σ e exists as well.
Lemma 2.3 Each requirement R e is satisfied and thus X is weakly 1-random.
Proof If µ([W e ]) = 1, then R e is satisfied trivially.Suppose µ([W e ]) = 1, and suppose that s 0 is the first stage such that all e < e that will ever act already have.After s 0 , the string σ e,s will never be initialized again, so σ e,s 0 = σ e,t for all t ≥ s 0 .Since µ([W e ]) = 1, there will be some stage ).At this stage, we act to satisfy R e and will define σ e,s 1 so [σ e,s 1 ] is contained in [W e ].After this stage, σ e,s will never be initialized again, so we will have σ e = σ e,s 1 , and since Lemma 2.4 X is not weakly 1-generic.
Proof We first show that the set of strings S we constructed is a Σ 0 1 dense set as required.It is clear from the construction that it is Σ 0 1 .Furthermore, at each stage s at which we add strings to S, we extend all the strings of length |σ e,s | for the appropriate e.Since the lengths of the σ e,s that we work with are unbounded, we will extend every element of 2 <ω by an element in S at some point.
Wesley Calvert and Johanna N.Y.Franklin Now we show that X does not meet the Σ 0 1 set of strings S we constructed.We note that the measure of the strings above any σ e,s−1 is no more than (in fact, is strictly less than) Since at each stage at which we set new values of a σ e,s , we know that Inequality 1 holds, we can see that there is an string τ extending an element of W e,s such that [τ ] ∩ [S s−1 ] = ∅, and we choose such a string as our τ e,s .Therefore we have [σ e,s ] ∩ [S s ] = ∅ as well, and since every σ e avoids S, it follows that X avoids S as well.
Lemma 2.5 X is not UD-random.
Proof We show that the sequence a i we built witnesses that X is not UD-random.First, we note that a i is recursive and infinite by construction: at every point at which we act to satisfy a requirement, we add at least one more element to this sequence, and since there are infinitely many W e with measure 1, we act to satisfy a requirement infinitely often.It is clear that a i is recursive.Furthermore, it is clear that all the elements of a i are distinct: each time we add a new element, it is larger than all the previous terms.
Lemma 2.2 shows us that X = lim e σ e .Each time we act to meet a requirement R e at a stage s, we also define an initial part of a i such that [a i σ e,s ] is between 1 2 and 1 for fewer than 1  4 of the i < s .When this σ e,s is the actual value of σ e , we have a value s such that [a i σ e , s] = [a i σ e ] = [a i X] for all i < s .Since the values of s associated with true values of σ e form an unbounded sequence, we know that there are infinitely many s such that 4 and thus X is not UD-random with respect to the sequence a i and the interval ( 12 , 1).
This completes the proof of the Theorem.
Since our choice of the interval ( 1 2 , 1) was arbitrary, we have the following result: Porism 2.6 If a real X is not UD-random, a recursive sequence can be found witnessing this for any interval I ⊆ [0, 1].

3 Genericity and UD-random reals
There has been a great deal of work relating, on the one hand, Turing degrees of Martin-Löf random, Schnorr random, and recursively random reals, and, on the other, the Turing degrees of generic reals by, for instance, Demuth and Kučera [2], Kautz [5], and Nies, Stephan, and Terwijn [8].
Consider the following question: Given n, is it possible for a (weakly) n-generic real to Turing compute a UD-random even though no (weakly) n-generic real can be UD-random itself?The answer is certainly yes for weak 1-generics, since Kurtz has shown that every hyperimmune degree contains a weakly 1-generic real [6,7], and there are Schnorr random reals that are hyperimmune.In fact, even a 1-generic real can be Turing equivalent to a UD-random real, since there is a high 1-generic.
Here, we provide a proof that this bound is tight: No 2-generic can Turing compute a UD-random real.Franklin proved in [4] that non-high 1-generics cannot Turing compute Schnorr random reals, which implies that no 2-generic can Turing compute a Schnorr random real.She also proved that no 1-generic can tt-compute a Schnorr random real.However, we cannot adapt the first argument to the UD-random case because it relies on the fact that outside the high degrees, Martin-Löf randomness and Schnorr randomness are equivalent.No such result is known for UD-randomness.Proof Let G be a 2-generic real, and let X be a non-recursive real that is Turing reducible to G via a Turing functional Ψ (if X is recursive, then there is nothing to be done).We wish to show that X is not UD-random.
We begin by considering the following statements: • Ψ Z is total.
• There are infinitely many n for which Each of these is a Π 0,Z 2 statement, so G must force each of these statements to either be true or false.Since X = Ψ G , it follows that G must force the first to be true.Also, X is not recursive and thus not cofinite, so that G must force the second statement to be true as well.Therefore, we may suppose that p is an initial segment of G forcing the truth of both statements.Our forcing conditions will be the set P = {p i ∈ 2 <ω | p i ⊇ p}.We consider the set T = {r i | r i = Ψ pi }.Since p is enough to force totality, every element r i in T will have a proper extension r j in T , so we can interpret T as an infinite recursively enumerable binary tree with X as one of its paths.If a real Ψ Z is not UD-random, the following statement P α,β must be true for some recursive sequence of distinct integers a i and the reals α and β (where 0 ≤ α < β ≤ 1): To show that X is not UD-random, it will be enough to set α = 1 2 , β = 1, and = 1 4 and build a recursive sequence a i of distinct integers such that for every r i in T , there is an extension r j in T such that for some n > |r i |, we have: This guarantees that we can never force the fraction above to converge to the "correct" limit 1  2 , and since G is 2-generic, G must force instead that Ψ G is not UD-random.To construct such a sequence a i , we proceed in stages.At each stage, we have a finite sequence of a i s and add at least one element to it.At stage 0, our sequence is the empty sequence.At each subsequent stage s we deal with p s .Suppose we have already defined the finite initial segment a i i<M of our a-sequence.Look for an r j such that p j ⊇ p s and such that there is an n such that |{i<M | 1 2 <[air j ]<1}| M+n < 1 4 and r j contains a 0 at n positions k 0 , k 2 . . .k n−1 such that a i < 2 k for all i < M , ≤ n.
Such an r j must exist, since we forced every extension of p to compute an r i ∈ T with infinitely many zeroes.Of course, we may be in the simplest case, where n = 0 suffices, but the treatment is the same.
Set a M+i = 2 ki for all 0 ≤ i ≤ n − 1.For each such i, the fractional part of a M+i r j = 2 ki r j is less than 1  2 and we have or its negation.But G cannot force ¬Θ, so instead G must force Θ, so that Ψ G = X cannot be UD-random.
This proof cannot be modified in an obvious way to show that no 1-generic tt-computes a UD-random.Were 2-genericity only necessary to force the totality of Ψ X , it would be possible, but the other statements in the problem are Π 0,X 2 as well.
Journal of Logic & Analysis 7:4 (2015) Genericity and UD-random reals 9 4 Future Directions It seems easier to prove negative results about UD-randoms than positive results.In part, this is due to the need to quantify over all intervals; however, this requirement has been minimized by Fact 1.2.Whether this fact has practical significance remains to be seen.However, the main difficulty in relating UD-randomness to other randomness notions is that UD-randomness is intrinsically a notion about tailsets: when we multiply a real by an integer, an initial segment of the real becomes irrelevant.In contrast, every other common randomness notion is defined in terms of initial segments: those that make up the components of a test or serve as inputs for the martingale or Kolmogorov complexity function.From the comment immediately preceding the questions, it is difficult to see how Question 4.1 could have a positive answer-at least, it is difficult to see how such a characterization could resemble that of other randomness notions in any meaningful way.The authors suspect that the answers to the second two questions are negative as well but have weaker grounds for saying so.

Theorem 3 . 1
No 2-generic real can Turing compute a UD-random real.

Question 4 . 1 Question 4 . 2 Question 4 . 3
Is there a test, martingale, or string (e.g.Kolmogorov) complexity characterization of UD-randomness?Do the UD-random reals that are weakly 1-random coincide with the Schnorr random reals?Do the UD-random reals coincide with the Schnorr random reals in the hyperimmune-free degrees?