# Mass Problems and Almost Everywhere Domination

Stephen G. Simpson

Pennsylvania State University

First draft: September 15, 2006
This draft: April 28, 2008

http://www.personal.psu.edu/t20/

AMS Subject Classifications: 03D28, 03D30, 03D80, 03D25, 68Q30.

This research was partially supported by NSF grant DMS-0600823.

Accepted April 24, 2007 for Mathematical Logic Quarterly.

Mathematical Logic Quarterly, 53, 2007, pp. 483-492.

### Abstract:

We examine the concept of almost everywhere domination from the viewpoint of mass problems. Let and be the set of reals which are almost everywhere dominating and Martin-Löf random, respectively. Let , , be the degrees of unsolvability of the mass problems associated with the sets , , respectively. Let be the lattice of degrees of unsolvability of mass problems associated with nonempty subsets of . Let and be the top and bottom elements of . We show that and and belong to and that . Under the natural embedding of the recursively enumerable Turing degrees into , we show that and but not are comparable with some recursively enumerable Turing degrees other than and . In order to make this paper more self-contained, we exposit the proofs of some recent theorems due to Hirschfeldt, Miller, Nies, and Stephan.

# Introduction

In our previous papers [3,31,34,32,7] we studied the lattice of degrees of unsolvability of mass problems associated with nonempty subsets of . We showed that contains many specific, natural degrees in addition to and , the top and bottom elements of . We showed that many specific, natural degrees in arise from foundationally interesting topics such as reverse mathematics, algorithmic randomness, computational complexity, hyperarithmeticity, and subrecursive hierarchies from Gentzen-style proof theory.

The purpose of the present paper is to exhibit and discuss some relatively new examples of specific, natural degrees in . The new examples arise from almost everywhere domination, a concept which was originally introduced by Dobrinen/Simpson [9]. Let be a Turing oracle. We say that is almost everywhere dominating if, for all except a set of measure , each function computable from is dominated by some function computable from . It is known [9] that almost everywhere domination is closely related to the reverse mathematics of measure theory.

In order to succinctly state our results, let is Martin-Löf random and is almost everywhere dominating. For we write and and the intersection of and . With these conventions, let , , be the respective degrees of unsolvability of the mass problems associated with , , . Trivially . Our main results may be summarized by saying that the degrees , , belong to and

.
The proof of this chain of inequalities uses virtually everything that is currently known about the relationship between Martin-Löf randomness and almost everywhere domination. See Theorems 2.2 and 2.3 below.

Historically, there has been a great deal of interest in the semilattice of recursively enumerable Turing degrees. Therefore, it seems desirable to examine the relationships between recursively enumerable Turing degrees and the various specific, natural degrees in . In order to state our results, let us temporarily identify each recursively enumerable Turing degree with its image in under the natural one-to-one embedding given in [34, Theorem 5.5]. In particular, we identify and , the top and bottom recursively enumerable Turing degrees, with and , the top and bottom degrees in . In our papers [31,34] written in 2004, we remarked that all of the specific, natural degrees in which were known at that time are incomparable with all recursively enumerable Turing degrees except and . In this respect it turns out that our new examples of specific, natural degrees in behave somewhat differently from the old ones. Namely, although is again incomparable with all recursively enumerable Turing degrees except and , this turns out not to be the case for and . See Theorems 3.1 and 3.2 below.

Our work in this paper owes much to conversations with Bjørn Kjos-Hanssen, Antonín Kucera, and Joseph S. Miller. In particular, the fact that belongs to was already implicit in Kjos-Hanssen [18], and Miller corrected an error in one of our early proofs of the inequality .

The reader who is familiar with the basics of recursion theory will find that this paper is largely self-contained. If is an expression which may or may not denote a natural number, we write to mean that is defined (i.e., denotes a natural number), otherwise . If and are two such expressions, we write to mean that and are both defined and equal, or both undefined. Throughout this paper, a convenient background reference is our recent paper [33], which includes a fairly thorough exposition of almost everywhere domination and Martin-Löf randomness.

# Some mass problem inequalities

The purpose of this section is to prove our mass problem inequalities in . The proofs use some recent theorems of Cholak/Greenberg/Miller, Kjos-Hanssen, Hirschfeldt/Miller, Nies, and Stephan concerning almost everywhere domination and Martin-Löf randomness. In order to make this paper more self-contained, we exposit the proofs of the theorems of Hirschfeldt/Miller, Nies, and Stephan respectively in Sections 4, 5, 6 below.

Our notion of reducibility for decision problems is standard. Given , we say that is Turing reducible to , abbreviated , if is computable using as an oracle. A Turing degree is an equivalence class of elements of under mutual Turing reducibility. The Turing degree of is denoted .

Our notion of reducibility for mass problems is as follows. Given , we say that is weakly reducible to , abbreviated , if for all there exists such that is Turing reducible to . A weak degree is an equivalence class of subsets of under mutual weak reducibility. The weak degree of is denoted . Weak degrees have sometimes been known as Muchnik degrees [23].

Note that for and we have and . Note also that for we have if and only if . Here denotes the singleton set whose only element is . Therefore, the Turing degree is sometimes identified with the weak degree .

Definition 2.1
1. Let where
is almost everywhere dominating.
2. Let where
is Martin-Löf random.
3. Let .

Theorem 2.2   The weak degrees , , belong to .

Proof.An important tool in the study of is the Embedding Lemma [34, Lemma 3.3], [32, Lemma 4]. The Embedding Lemma says: For any where is , we have . Therefore, in order to prove Theorem 2.2, it suffices to show that and are .

After Dobrinen/Simpson [9], the concept of almost everywhere domination was subsequently explored by Binns/Kjos-Hanssen/Lerman/Solomon [2], Cholak/Greenberg/Miller [5], Kjos-Hanssen [18], Kjos-Hanssen/Miller/Solomon [19], and Simpson [33]. We now know [18,19] that is almost everywhere dominating if and only if . Here denotes the Halting Problem, and denotes -reducibility: if and only if if is random relative to , then is random relative to . Moreover, as shown in [19], -reducibility is equivalent to -reducibility: if and only if for all . Here denotes the prefix-free Kolmogorov complexity of relative to a Turing oracle . The concepts of -reducibility and -reducibility were originally introduced by Nies [24, Section 8]. A convenient reference for these results is Simpson [33].

One way to see that is is to use the characterization in terms of -reducibility. We know that if and only if , i.e., , i.e., . But if and only if and . Here is a universal prefix-free oracle machine. The last satement is , so is . The fact that is has previously been noted by Kjos-Hanssen [18] and Kjos-Hanssen/Miller/Solomon [19]. See also [33, Corollary 5.9].

It remains to show that is . In fact, is in view of the existence of a universal Martin-Löf test. See for instance [20] or [31, Theorem 8.3] or [33, Theorem 3.2]. This completes the proof of Theorem 2.2.

Theorem 2.3   In we have .

Proof.Trivially , hence . Recall from [31,34] that where

is a complete extension of Peano Arithmetic.
We have because by [9] no member of is recursive. In order to prove , consider
is diagonally nonrecursive
where is said to be diagonally nonrecursive if .

Proof.This follows from items 1 and 2 in Lemma 2.7 below. The next two lemmas are well known.

Proof.Consider the recursive functional given by for all . Let . Note that is uniformly and . If is random, it follows by Solovay's Lemma (see for instance [33, Lemma 3.7]) that for only finitely many . Therefore, with finitely many exceptions, is diagonally nonrecursive. This proves the lemma. For refinements, see Jockusch [14, Proposition 3] and Ambos-Spies/Kjos-Hanssen/Lempp/Slaman [1] and Simpson [31].

Proof.Since is and nonempty, we can find a nonempty set . Since is a nonempty subset of , we have in view of Scott/Tennenbaum [30] and Scott [29]. The lemma follows. By Lemmas 2.4 and 2.5 and 2.6 we have . From this it follows trivially that . In other words, .

Next we prove . We use the forcing construction of Cholak/Greenberg/Miller [5] referring to -genericity.

Lemma 2.7 (Cholak/Greenberg/Miller)
1. If is sufficiently -generic, then is almost everywhere dominating.
2. If is sufficiently -generic, then .
3. For each nonrecursive , if is sufficiently -generic then .

Proof.See Cholak/Greenberg/Miller [5, Section 4]. We also use the following result due to Hirschfeldt and Miller, 2006.

Lemma 2.8 (Hirschfeldt/Miller)   There is a nonrecursive, recursively enumerable set such that .

Proof.See Nies [25, Theorem 5.6]. Alternatively, see Section 4 below. The next lemma is well known.

Lemma 2.9
1. If , then .
2. If , then .

Proof.The first statement is a relativized form of Sacks [28, Theorem 1, page 154]. The second statement is a relativived form of Jockusch/Soare [16, Theorem 5.3]. Both statements follow from the general non-helping'' result in [31, Lemma 7.3]. By Lemma 2.8 let be nonrecursive such that . Since is nonrecursive, by Lemma 2.7. Hence by Lemmas 2.5 and 2.6. But then, by Lemma 2.9, , since . It now follows by our choice of that . From this it follows trivially that . In other words, .

It remains to prove . We use the following results of Nies 2006 and Stephan 2002.

Lemma 2.10 (Nies)   There exists such that .

Proof.See Nies [26, Theorem VI.18]. Alternatively, see Section 5 below.

Lemma 2.11 (Stephan)   If and , then .

Proof.See Stephan [35]. Alternatively, see Section 6 below. By Lemma 2.10 let be such that . By Lemma 2.11 we have . Thus . It follows trivially that . In other words, .

This completes the proof of Theorem 2.3.

# Comparison with r.e. Turing degrees

In this section we discuss the relationship between the weak degrees , , and the recursively enumerable Turing degrees. Recall from [34, Theorem 5.5] that there is a natural one-to-one embedding of the recursively enumerable Turing degrees into given by .

Theorem 3.1   There is no recursively enumerable Turing degree such that is comparable with .

Proof.Let where is recursively enumerable and . Since is nonrecursive, we have by Lemma 2.7, hence by Lemmas 2.5 and 2.6, hence by Lemma 2.9. From this it follows trivially that . In other words, . On the other hand, since is recursively enumerable and not Turing complete, we have by the Arslanov Completeness Criterion [14], hence by Lemmas 2.5 and 2.6. From this it follows trivially that . In other words, . This completes the proof.

Theorem 3.2   There are recursively enumerable Turing degrees and such that and .

Proof.By Lemma 2.8 due to Hirschfeldt and Miller, let be recursively enumerable such that . By Simpson [33, Example 6.8] or Cholak/Greenberg/Miller [5, Theorem 2.1], let be recursively enumerable and almost everywhere dominating. Let and . Clearly and and and . By Theorems 2.3 and 3.1 it follows that and .

# A theorem of Hirschfeldt and Miller

In this section we exposit the proof of the following theorem of Hirschfeldt and Miller 2006, generalizing a much earlier theorem of Kucera [21].

Theorem 4.1 (Hirschfeldt/Miller)   Let be of measure . Then we can find a nonrecursive, recursively enumerable set such that for all random .

Proof.We follow the writeup of Nies [25, Theorem 5.6].

We first prove the theorem for sets. Let be of measure . Write where is uniformly and .

The construction of is as follows. At stage , for each such that , look for such that and and put the least such into .

Clearly is recursively enumerable. Moreover, for each , at most one gets into for the sake of , and for this we have . Hence has at most members . Thus is infinite.

We claim that if is infinite then . To see this, fix so large that and . Let be such that . We have , so by construction , Q.E.D.

It follows from the previous claim that is nonrecursive. Indeed, is simple in the sense of Post (compare Rogers [27, Section 8.1]).

We claim that for all random . To see this, note that by construction

Since is random, it follows by Solovay's Lemma (see [33, Lemma 3.7]) that for all but finitely many pairs such that . Let be so large that . Given , since we can effectively find such that . We then have . Thus , Q.E.D. This proves the theorem for sets.

Suppose now that is of measure . Write where is uniformly . Write where is uniformly and for each . This implies that , so we can build as before replacing by . The construction insures that

hence for each

hence for any random we have by Solovay's Lemma for all but finitely many such that . It follows as before that for all random . This proves the theorem.

Remark 4.2   Hirschfeldt, Miller and Nies describe the proof of Theorem 4.1 as a cost function'' construction. In the case of a set, the cost of putting into at stage is . In the case of a set, the cost of putting into at stage is . The construction insures that the total cost of building is finite, so that Solovay's Lemma can be applied.

The following corollary is originally due to Kucera [21].

Corollary 4.3 (Kucera)   Let be random and . Then we can find a nonrecursive, recursively enumerable set such that for all .

Proof.Let and apply Theorem 4.1.

The following lemma is well known.

Proof.It suffices to show that whenever is random relative to . (Generalizations of this result are in Kautz's thesis [17, Theorem III.2.1].) Consider the sets . Obviously these sets are uniformly . Let least such that . Note that . Thus is uniformly and these sets form a Solovay test relative to . Now assume that is random relative to . By Solovay's Lemma relative to , we have for all but finitely many . In other words, for all but finitely many , if and only if . Since , it follows that . This completes the proof.

Theorem 4.5 (Hirschfeldt/Miller)   We can find a nonrecursive, recursively enumerable set such that for all such that is random and almost everywhere dominating.

Proof.By Theorem 4.1 with , it suffices to show that is and of measure . We have seen in the proof of Theorem 2.2 that is . To show that , recall from [19] and [33, Section 8] that . Thus is disjoint from the intersection of two sets: and . The first set is of measure by Lemma 4.4, and the second set is of measure by Lemma 2.9. This completes the proof.

Remark 4.6   According to a theorem of Nies (see Section 5 below), there are uncountably many 's as in Theorem 4.5 which are . On the other hand, a result of Hirschfeldt/Nies/Stephan [13, Corollary 3.6] says that if is random and then any recursively enumerable is -trivial [24], i.e., low-for-random [22,24,33]. In particular, any as in Theorem 4.5 is low-for-random.

# A theorem of Nies

In this section we present a new proof of a theorem of Nies [26, Theorem VI.18] refining the Jockusch/Shore Pseudojump Inversion Theorem [15, Theorem 2.1]. By a pseudojump operator we mean an operator given by for all .

Theorem 5.1 (Nies)   Let be of positive measure. For any pseudojump operator and any Turing oracle , we can find such that .

Proof.Our proof is based on a sketch given by Kucera at an American Institute of Mathematics workshop on algorithmic randomness, August 7-11, 2006. The idea is to combine the proofs of the Pseudojump Inversion Theorem and the Kucera/Gács Theorem.

Let us write . Note that , is a standard, recursive enumeration of all subsets of . Given a set , an index of is any integer such that . Let us say that a set is rich if and there exists a recursive function such that for all , if then .

We claim that every set of positive measure includes a set which is rich. To see this, let be such that . Write and note that , is a uniformly recursive descending sequence of clopen sets such that . Define a recursive ascending sequence of clopen sets , as follows. Begin with . Given define where the union is taken over all such that . Finally let where . Clearly is . Moreover , hence . If then for all we have , hence . Thus is rich via . This proves our claim.

To prove the theorem, let be of positive measure. By our claim, we may safely assume that is rich. Under this assumption we shall carry out the proof of the Pseudojump Inversion Theorem within '' to produce with the desired properties.

For strings let us write . For each string we define a string and a nonempty set . Begin with and . Assume inductively that and have already been defined.

In order to control the pseudojump , define a string and a nonempty set as follows. Let . If and , let and let . Otherwise, consider the least such that and , and let and . Thus decides'' whether or not.

Because is rich, given an index of we can effectively find such that . But then there are at least two strings of length such that . Let and be the lexicographically leftmost and rightmost such . Let and .

We have now defined , , , for all . It is straightforward to check that and and the indices of and are uniformly computable from .

Given let . Then if and only if . Moreover, it is straightforward to show that and and the indices of and are uniformly computable from each of the Turing oracles and and . From this, the desired conclusions follow easily.

Theorem 5.2 (Nies)   For any pseudojump operator and any Turing oracle , we can find a random such that .

Proof.In Theorem 5.1 let be a nonempty set such that is random.

Remark 5.3   Theorem 5.2 is a common generalization of several known theorems. If we omit the conclusion that is random, we get the Jockusch/Shore Pseudojump Inversion Theorem [15, Theorem 2.1]. If we keep the conclusion that is random but let be the identity operator, we get the Kucera/Gács Theorem [33, Theorem 3.5]. If we let be the Turing jump operator, we get the Friedberg Jump Inversion Theorem (see Rogers [27, Section 13.3]) with the additional conclusion that is random.

Corollary 5.4 (Nies)   For any we can find a random such that and is low-for-random relative to .

Proof.By Theorem 5.2, it suffices to produce a psuedojump operator such that for all , and is low-for-random relative to . Such an operator is obtained by uniformly relativizing the Kucera/Terwijn [22] construction of a nonrecursive, recursively enumerable set which is low-for-random. See also the exposition in Simpson [33, Section 6].

Corollary 5.5 (Nies)   We can find a random such that is low-for-random relative to .

Proof.This is the special case of Corollary 5.4 in which we let .

Corollary 5.6 (Nies)   There are uncountably many random such that is low-for-random relative to .

Proof.This follows from Corollary 5.4 by considering uncountably many .

Corollary 5.7 (Nies)   We can find a random which is almost everywhere dominating.

Corollary 5.8 (Nies)   There are uncountably many random such that is almost everywhere dominating.

Proof.Corollaries 5.7 and 5.8 are immediate from Corollaries 5.5 and 5.6 plus the following fact: If is low-for-random relative to then is almost everywhere dominating. This fact is due to Kjos-Hanssen/Miller/Solomon [19]. See also the exposition in Simpson [33, Section 5].

# A theorem of Stephan

In this section we exposit the proof of the following theorem of Stephan [35].

Theorem 6.1 (Stephan)   If is random and , then .

Proof.We shall define a recursively bounded, partial recursive function with the following property: For all random , if some total function extending then . This suffices to prove the theorem, because clearly every computes a total extension of every recursively bounded, partial recursive function (see for instance [31, Theorem 4.10]).

Recall that is a recursively enumerable set. If let be undefined for all . If , say , then for each compute the rational numbers

and define chosen so as to minimize . Note that is partial recursive, and if and only if . Moreover, if then , so is recursively bounded. In addition where
and .
Assume now that is random and computes a total extension of . Let be such that is total and extends . Define least such that . Since is total, is total and . Since is random, it follows by Solovay's Lemma that for all but finitely many . In other words, for all but finitely many , if then . But then, since and extends , it follows that , i.e., . We now see that, for all but finitely many , if then . Thus . This completes the proof.

Remark 6.2   A similar proof yields the following more detailed result. Given a recursively enumerable set , we can find recursively enumerable sets and such that and and, for all random , if and then .

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Mass Problems and
Almost Everywhere Domination

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