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TOPICS IN ALGORITHMIC RANDOMNESS AND EFFECTIVE PROBABILITY
A Dissertation
Submitted to the Graduate School
of the University of Notre Dame
in Partial Fulfillment of the Requirements
for the Degree of
Doctor of Philosophy
by
Quinn Culver
Peter Cholak, Director
Graduate Program in Mathematics
Notre Dame, Indiana
April 2015
This document is in the public domain.
TOPICS IN ALGORITHMIC RANDOMNESS AND EFFECTIVE PROBABILITY
Abstract
by
Quinn Culver
This dissertation contains the results from three related projects, each within the
fields of algorithmic randomness and probability theory.
The first project we undertake, which can be found in Chapter 2, contains the
definition a natural, computable Borel probability measure on the space of Borel
probability measures over 2! that allows us to study algorithmically random mea-
sures. The main results here are as follows. Every (algorithmically) random measure
is atomless yet mutually singular with respect to the Lebesgue measure. The random
reals of a random measure are random for the Lebesgue measure, and every random
real for the Lebesgue measure is random for some random measure. However, for a
fixed Lebesgue-random real, the set of random measures for which that real is ran-
dom is small. Relatively random measures, though mutually singular, always share a
random real that is in fact computable from the join of the measures. Random mea-
sures fail Kolmogorov’s 0-1 law. The shift of a random real for a random measure is
no longer random for that measure.
In our second project, which makes up Chapter 3, we study algorithmically ran-
dom closed subsets of 2!, algorithmically random continuous functions from 2! to 2!,
and the algorithmically random Borel probability measures on 2! from Chapter 2,
especially the interplay among these three classes of objects. Our main tools are
preservation of randomness and its converse, the “no randomness ex nihilo princi-
Quinn Culver
ple,” which together say that given an almost-everywhere defined computable map
from 2! to itself, a real is Martin Lof random for the pushforward measure if and
only if its preimage is random with respect to the measure on the domain. These
tools allow us to prove new facts, some of which answer previously open questions,
and reprove some known results more simply.
The main results of Chapter 3 are the following. We answer an open question in
[3] by showing that X ✓ 2! is a random closed set if and only if it is the set of zeros
of a random continuous function on 2!. As a corollary, we obtain the result that the
collection of random continuous functions on 2! is not closed under composition. We
construct a computable measure Q on the space of measures on 2! such that X ✓ 2!
is a random closed set if and only if X is the support of a Q-random measure. We
also establish a correspondence between random closed sets and the random measures
studied in Chapter 2. Lastly, we study the ranges of random continuous functions,
showing that the Lebesgue measure of the range of a random continuous function is
always strictly between 0 and 1.
In Chapter 4 we e↵ectivize a theorem of Erdos and Renyi [11], which says that
for c � 1, if a fair coin is used to generate a length-N string of 1’s and �1’s, which
are interpreted as gain and loss, then the maximal average gain over bc logNc-length
substrings converges almost surely (in N) to the same limit ↵(c). We show that if the
1’s and �1’s are determined by the bits of a Martin Lof random, then the convergence
holds.
CONTENTS
ACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv
CHAPTER 1: INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Summary of Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Summary of Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Summary of Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 A word on notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
CHAPTER 2: ALGORITHMICALLY RANDOM MEASURES . . . . . . . . 72.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2.2 General e↵ective spaces . . . . . . . . . . . . . . . . . . . . . . 82.2.3 The space of probability measures . . . . . . . . . . . . . . . . 92.2.4 Algorithmic randomness . . . . . . . . . . . . . . . . . . . . . 10
2.3 Random measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.4 Random measures and their randoms . . . . . . . . . . . . . . . . . . 152.5 Random measures are atomless . . . . . . . . . . . . . . . . . . . . . 162.6 Random measures are mutually singular (with respect to the Lebesgue
measure) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.7 Relatively random measures . . . . . . . . . . . . . . . . . . . . . . . 23
CHAPTER 3: THE INTERPLAY OF CLASSES OF ALGORITHMICALLYRANDOM OBJECTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.2.1 Some topological and measure-theoretic basics . . . . . . . . . 303.2.2 Some computability theory . . . . . . . . . . . . . . . . . . . . 31
3.3 Algorithmically random objects . . . . . . . . . . . . . . . . . . . . . 323.3.1 Algorithmically random sequences . . . . . . . . . . . . . . . . 323.3.2 Algorithmically random closed subsets of 2! . . . . . . . . . . 333.3.3 Algorithmically random continuous functions on 2! . . . . . . 343.3.4 Algorithmically random measures on 2! . . . . . . . . . . . . 35
3.4 Applications of Randomness Preservation and No Randomness Ex Nihilo 363.5 The support of a random measure . . . . . . . . . . . . . . . . . . . . 40
ii
3.6 The range of a random continuous function . . . . . . . . . . . . . . . 44
CHAPTER 4: A NEW LAW OF LARGE NUMBERS EFFECTIVIZATION . 564.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.2 Stirling’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.3 Maximal average gains over short subgames of a fair game . . . . . . 60
BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
iii
ACKNOWLEDGMENTS
Thanks to Chris Porter, Laurent Bienvenu, Joe Miller, Uri Andrews, Francois
Ledrappier, David Galvin, Greg Igusa, Benoit Monin, Pablo Lessa, Mathieu Hoyrup,
Peter Cholak, Jason Rute, Gerard Misiolek, Julia Knight, Manfred Denker, and
Mushfeq Khan.
Research partially supported by national Science Foundation, EMSW21-RTG-
0838506.
iv
CHAPTER 1
INTRODUCTION
Algorithmic randomness was born out of an attempt to make precise the term
“random” by distinguishing a set of random elements that satisfy every almost-sure
property for the Lebesgue measure, �, on 2!, the space of (one-way) infinite binary
sequences (aka reals). However, being in the complement of a singleton is an almost-
sure property, so satisfying every almost sure property is impossible. Thus, the theory
of computation was brought into the picture, and the only properties considered
were those that were su�ciently computable. This gives rise to a distinguished set
MLR�
✓ 2!, called Martin Lof randoms or just randoms, with the property that
�(MLR) = 1.
One trend in algorithmic randomness has been to code other mathematical objects
by infinite binary sequences, declare an object to be (algorithmically) random if it
has a random code, and then to investigate what those random objects look like and
how they behave. In Chapters 2 and 3, the objects are Borel probability measures
on 2!, continuous functions from 2! to itself, and closed subsets of 2!.
Another trend is the so-called e↵ectivization of classical theorems of probabil-
ity/measure theory. Many theorems of probability are “almost sure” results. To
e↵ectivize such a theorem, one assumes the objects in the hypothesis (e.g. func-
tions/random variables) su�ciently computable and concludes that the result holds
on the algorithmically randoms. In Chapter 4, we e↵ectivize a 1970 result of Erdos
and Renyi [11].
1
1.1 Summary of Chapter 2
In Chapter 2, the objects of study are measures1 over 2!. This project started as
a result of studying the Ergodic Decomposition Theorem (specifically the results in
[14]), which can be viewed as a statement about measures on the space of measures
on 2!. We wondered:
Question 1.1.1. Is there a natural measure on the space of measures over 2!?
Here the word “natural” corresponds to the fact that Lesbesgue measure is consid-
ered the most natural measure on 2!. Just as the randoms for the Lebesgue measure
on 2! constitute the truly random real numbers, if there were a natural measure on
the space of measures, then its random elements should constitute the truly random
measures on 2!.
Definition 2.3.2 defines a measure P on the space of measures on 2! that we
see as answering Question 1.1.1 a�rmatively. Essentially, the measure P says that
the conditional probability of going left from a given node in the full binary tree is
uniformly distributed and independent of other nodes. The measure P is natural in
the sense that every measure on 2! comes from assigning said conditional probabilities
according to some sequence of distributions, and taking that sequence to be IID-
uniform is somehow most natural.
The measure P on determines a collection MLRP
of P -random measures. The
remainder of this section is a synopsis of the results we prove in Chapter 2 about
these P -random measures.
The Lebesgue measure � is the so-called barycenter of P ; that is,Rµ(A) dP (µ) =
�(A) for any Borel set A. By results of Hoyrup [14], this implies that Lebesgue
randoms are exactly those that are random elements for some P -algorithmically-
random measure: MLR�
=S
µ2MLRPMLR
µ
.
1Here the term measure is short for Borel probability measure.
2
Every P -random measure is atomless (i.e., it gives zero probability to singletons)
yet mutually singular with respect to the Lebesgue measure �. (A measure µ is
mutually singular with respect to � if �(A) = 1 and µ(A) = 0 for some Borel set A.)
We conjectured initially that if µ and ⌫ are relatively random measures, then they
share no random reals. Relative randomness is the algorithmic analog of indepen-
dence. So, this conjecture said that if measures µ and ⌫ are generated independently,
then they will not agree on any real’s being random. This conjecture is almost true:
relatively random measures are mutually singular, and, hence, MLRµ
\MLR⌫
has
both µ and ⌫ measure zero. Therefore any agreement between µ and ⌫ on what is
random is rare. Surprisingly, however, there actually is a real that is random for both.
Moreover, there is a uniform construction of such a real using µ and ⌫ as oracles.
Whenever µ is P -random, x 2 MLRµ
, and y di↵ers from that of x at only finitely
many bits, then y /2 MLRµ
; i.e., MLRµ
is an “anti-tailset”. Thus P -random measures
badly fail Kolmogorov’s 0-1 law. We use this fact to prove that if x 2 MLR�
, then
P{µ : x 2 MLRµ
} = 0.
1.2 Summary of Chapter 3
Again, in this project, the focus is on algorithmic randomness in spaces other than
2!. Here, however we focus not just on how the random objects behave, but also on
how they behave with each other. Our main tools are the randomness preservation
principle and its converse, the no randomness ex nihilo principle. This work is joint
with Chris Porter.
The objects in play here are Borel probability measures on 2! (as in Chapter 2),
nonempty closed subsets of 2!, and continuous functions from 2! to 2!. In each of
the three cases, there is a surjective map that assigns to each x 2 2! an object Ox
(a
measure, a nonempty closed set, or a continuous function) and then the object Ox
is
3
said to be (Martin Lof) random if x is (Martin Lof) random. So, we can talk about
random measures, random closed sets (first defined in [2]), and random continuous
functions (first defined in [3]).
In this project, we rely heavily on the preservation of randomness and no ran-
domness ex nihilo principles. Together, they say that given a computable measure
µ on 2! and a µ-a.e. defined computable map � : 2! ! 2!, an element y 2 2! is
Martin Lof random for the pushforward measure µ � ��1 if and only if y = �(x) for
some x 2 2! that is Martin Lof random for µ. These are powerful tools because they
often allow one to draw conclusions about Martin Lof randoms by showing that the
pushforward measure is what was desired and then merely observing that the map at
hand is computable. Using these tools, we reprove, in a much simpler way, the result
(in [2]) that every random closed set contains an element that is Martin Lof random
for the Lebesgue measure and that every element that is Martin Lof random for the
Lebesgue measure is contained in some random closed set.
We also reprove the fact that if F is a random continuous function for which the
zero set F�1{0N} is nonempty, then this set is, in fact, a random closed set. Our tools
then give for free the previously-left-open converse, which says that every random
closed set is realized as the zero set of some random continuous function. The fact
that the composition of random continuous functions need not be random follows as
a corollary.
For a measure µ on 2!, the 1/3-support is defined to be the set of x 2 2! such
that µ([x � n+ 1] | [x � n]) > 1/3, where [x � n] denotes the set of all elements of 2!
that agree with x on the first n bits. We show that a closed subset of 2! is random if
and only if it is the 1/3-support of some random measure. We also show that there is
a di↵erent way of defining a “random measure” (i.e., a di↵erent measure on the space
of measures) so that the (regular) supports of these random measures are exactly the
random closed sets.
4
It was shown in [3] that random continuous functions are not necessarily injective
nor surjective. We extend this by showing that random continuous functions are
never injective and never surjective. Moreover, we show that the Lebesgue measure
of the range of a random continuous function is strictly between 0 and 1.
1.3 Summary of Chapter 4
Erdos and Renyi [11] proved that for any c � 1, if a fair coin is used to generate
a length-N string � = �1
�2
· · · �N
of 1’s and �1’s, then the maximal average
max0nN�bc log
2
Nc
�n+1
+ �n+1
+ · · · �n+bc log
2
Nc
bc log2
Nc
converges almost surely (in N) to the same limit ↵ = ↵(c), which is determined by
the equation1
c= 1� h
✓1 + ↵
2
◆,
where h : [0, 1] ! [0, 1] is the binary entropy function
h(x) = �x log2
x� (1� x) log2
(1� x).
The 1’s and �1’s can be interpreted as the gain or loss of a player in a fair game,
so this result says that the maximal average gain over appropriately-sized subgames
converges almost surely to ↵.
This result is a threshold theorem. As Erdos and Renyi note, if K(N) is an
integer-valued function of N such that K(N)
logN
! 1, then the maximal average
max0nN�K(N)
�n+1
+ �n+2
+ · · · �n+K(N)
K(N)
converges almost surely to 0. If K(N) is an integer-valued function of N such that
5
K(N) c logN for some 0 < c < 1, then the maximal average is almost surely
eventually 1 (and hence converges almost surely to 1). So, the theorem explains
what happens in the only case left to consider, when K(N) grows like c logN for
some c � 1.
Chapter 4 contains an e↵ectivization of this theorem: for any c � 1, not only
does the maximal average converge almost surely to ↵, but, in fact, the convergence
holds on every infinite sequence of 1’s and �1’s that is Martin Lof random.
1.4 A word on notation
Our notation is fairly standard. For various reasons though, we use some di↵erent
notation and conventions in each chapter. In order to make each chapter more self-
contained we also state some definitions more than once (but never more than once
in a given chapter).
6
CHAPTER 2
ALGORITHMICALLY RANDOM MEASURES
2.1 Introduction
Algorithmic randomness attempts to make precise the notion of a random real
number. Coding other objects (e.g., graphs) by real numbers allows for the study
of the algorithmically random versions of these objects (e.g., algorithmically random
graphs). This has been done, for example, in [6], [3], and [1]. Here we undertake a
similar project, where the objects of study are Borel probability measures on 2!.
We define a natural, computable (in the sense of computable analysis) map from
2! to the space P(2!) of Borel probability measures on 2!. This map pushes the
Lebesgue measure forward, yielding a natural, computable Borel probability measure
P on P(2!). The construction of P is a special (and, we think, the most natural)
case of a construction in [18]. We investigate the algorithmically P -random Borel
probability measures and the algorithmically random reals for those measures.
2.2 Preliminaries
2.2.1 Basics
The set of natural numbers is denoted by !. The function h·, ·i : !2 ! ! is any
computable bijection. The computably enumerable (c.e.) subsets of ! are e↵ectively
numbered as hWe
ie2!.
The set of finite binary strings is denoted by 2<!. It is e↵ectively numbered via
�0
= ? (the empty string), �1
= 0, �2
= 1, �3
= 00, �4
= 01, etc. For strings � and
7
⌧ , the notation � � ⌧ means that � is an initial segment of ⌧ and � ? ⌧ means that
neither � � ⌧ nor ⌧ � �.
Cantor space, the space of all one-way infinite binary strings, is denoted by 2!.
For � 2 2<!, let [�] = {x 2 2! : x � �} (where x � � means that � is an initial
segment of x); this is the cylinder set generated by �. The collection of all cylinder
sets forms a clopen basis for a topology on 2!. This topology is metrizable via
d(x, y) = 2�min{n:x(n) 6=y(n)}.
2.2.2 General e↵ective spaces
We assume familiarity with the basics of computability theory and computable
analysis over 2! and R. In order to do computable analysis and probability theory
in spaces other than 2! and R, we, following [13] and [8], work in an e↵ective
Polish space,1 which is a complete metric space (X, d) with a countable dense
subset Q = hqi
i such that d(qi
, qj
) is a computable real number uniformly in i and
j. A representation of X is a partial surjective function ⇢ : 2! ! X. Given
a representation ⇢, a ⇢-name for x 2 X is an element of ⇢�1{x}. Any e↵ective
Polish space is equipped with a representation called its standard fast Cauchy
representation, ⇢C
: 2! ! X, defined by ⇢C
(0n010n110n21 · · · ) = x if d(x, qni) 2�i.
Note that di↵erent elements of X cannot have the same ⇢C
-name. We will simply
say name when ⇢ is clear from the context.
Any e↵ective Polish space admits an e↵ective basis for its topology; BX
k
, with
k = hi, ji, is the ball centered at qi
with radius 2�j. When no confusion will be
caused, BX
k
will be written simply as Bk
. A subset U ✓ X is then e↵ectively open,
or ⌃0
1
, if U =S
i2WeB
i
for some c.e. We
; and C ✓ X is e↵ectively closed, or ⇧0
1
, if
X � C is e↵ectively open. A compact subset K ✓ X is e↵ectively compact if the
set of (indices for) finite covers by the Bi
’s is c.e.
1The authors there use the term computable metric space.
8
A function f : X ! R := R[{1} is left- (resp. right-) computable if f�1(r,1]
(resp. f�1[�1, r)) is e↵ectively open in X for each r 2 Q. Let X and Y be e↵ective
Polish spaces. Then f : X ! Y is computable if f�1(U) is e↵ectively open in X
whenever U is e↵ectively open in Y , uniformly in U . In particular, computability
implies continuity. The following proposition is straightforward
Proposition 2.2.1.
1. A function f : X ! Y , where (Y, ⇢, R) is an e↵ective Polish space, is computableif and only if there is a computable function that outputs a name for f(x) 2 Ywhenever given a name for x 2 X.
2. f : X ! R is left- (resp. right-) computable if and only if there’s a computablefunction that outputs an increasing (resp. decreasing) sequence of rationals con-verging to f(x) whenever given a name for x 2 X.
3. f : X ! R is computable if and only if it is both left- and right-computable.
Where it makes sense, all notions are relativizable. Thus for x 2 2!, we can speak
of an x-computable function, an x-left-computable function, etc.
Proposition 2.2.2 ([14]). Let X and Y be e↵ective Polish spaces.
1. An e↵ectively compact set is e↵ectively closed.
2. If f : X ! Y is computable, and K ✓ X is e↵ectively compact, then f(K) ise↵ectively compact.
3. An e↵ectively closed subset of 2! is e↵ectively compact.
2.2.3 The space of probability measures
For an e↵ective Polish space X, let B(X) be its Borel �-algebra; that is, B(X)
is the smallest class of subsets of X that contains the open sets and is closed under
complementation and countable unions. A Borel probability measure, or just
measure for short, on X is a function µ : B(X) ! [0, 1] such that µ(X) = 1 and
µ(S
i2! Ai
) =P
i
µAi
whenever the Ai
’s are pairwise disjoint elements of B(X). When
X = 2!, Caratheodory’s extension theorem [12] guarantees that the conditions
9
(a) µ([?]) = 1 and
(b) µ([�]) = µ([�0]) + µ([�1]) for all � 2 2<!
uniquely determine a measure on 2!. Thus, a measure on Cantor space is identified
with a function µ : 2<! ! [0, 1] satisfying conditions (a) and (b). We may write µ(�)
instead of µ([�]). The Lebesgue measure � is defined by �(�) = 2�|�| for each
string �.
The space of all Borel probability measures on an e↵ective Polish space X is
denoted by P(X). It is itself an e↵ective Polish space under the (metrizable) weak-⇤
topology [16]. We do not need the details of the e↵ective structure of P(X) but only
the following proposition.
Proposition 2.2.3 ([16]).
1. A measure µ 2 P(X) is computable if and only if µ(U) is uniformly left-computableon e↵ectively open sets U ✓ X.
2. A function f : X ! P(Y ) is computable if and only if a name for x 2 X uniformlyleft-computes the value of f(x)(BY
k
).
2.2.4 Algorithmic randomness
Let X be an e↵ective Polish space endowed with a Borel probability measure µ,
and let y 2 2! be a name for µ. A y-Martin Lof test for µ randomness is a
uniformly ⌃0,y
1
sequence hUn
in2! such that µ(U
n
) 2�n. An element x 2 X passes
the y-Martin Lof test for µ randomness if x /2T
Un
. An element x 2 X is y-Martin
Lof random for µ if it passes all y-Martin Lof tests for randomness.
An element x 2 X is Martin Lof random for µ, or just µ-random, if it is
y-Martin Lof random for µ for some name y 2 2! of µ. We write MLRµ
for the
set of all µ-randoms. Note that because there are only countably many ⌃0,y
1
sets,
µ(MLRµ
) = 1. Because the Lebesgue measure is special, we often write MLR for
MLR�
.
10
As noted in [8], for any name y of µ, there is a single y-Martin Lof test for
µ randomness that su�ces to define y-Martin Lof randomness for µ. Such a y-
Martin Lof test for µ randomness is called a universal y-Martin Lof test for µ
randomness.
The following proposition allows us to work with a single name for µ in the cases
we consider.
Proposition 2.2.4. If µ 2 P(X) has a name y of least Turing degree, then x 2 X
is µ-random if and only if x is y-Martin Lof random for µ.
Proof. Suppose z 2 2! is a name for µ with z �T
y. Then any y-Martin Lof test for
µ randomness is also a z-Martin Lof test for µ randomness. Thus if x is z-Martin Lof
random for µ it is also y-Martin Lof random for µ.
When µ 2 P(X) has a name y 2 2! of least Turing degree, we will call the
universal y-Martin Lof test for µ randomness simply a universal µ-test.
An element x 2 X is µ-Kurtz random if there is a name y of µ such that x 2 U
for every U 2 ⌃0,y
1
with µ(U) = 1. We write KRµ
for the collection of all µ-Kurtz
randoms.
Note that because there are only countably many e↵ectively open sets, µ(KRµ
) =
1. Moreover, the following is true.
Proposition 2.2.5 ([17]). If x 2 MLRµ
then x 2 KRµ
.
2.3 Random measures
Given x 2 2!, the nthcolumn x
n
2 2! of x is defined by xn
(k) = 1 if and only
if x(hn, ki) = 1 (recall that hn, ki is a fixed computable bijection between !2 and !).
We write x = �n2!xn
; this is the infinite join operation in the Turing degrees.
Define the map � : 2! ! P(2!), with �(x) written µx
, by µx
(?) = 1 and
µx
(�n
0) = xn
⇤ µx
(�), where xn
is (the real number represented by) the nth col-
11
umn of x. This map is essentially in [18], but is independently due to Chris Porter.
It is, as the next proposition shows, really just another representation of P(2!).
Proposition 2.3.1. The map � is computable.
Proof. By Proposition 2.2.3, it su�ces to show that given x 2 2! we can uniformly
compute µx
(�). Write µx
(�) =Q
i<|�| µx
(� � i + 1|� � i), where µx
(�|⌧) := µx
([�] \
[⌧ ])/µx
([⌧ ]). Multiplication is computable, so it su�ces to show that µx
(� � i+1|� � i)
is uniformly computable from x. But this is clear since µx
(� � i+ 1|� � i) is either a
column or one minus a column of x, and computably so.
Being computable implies being Borel (indeed continuous), so � pushes � forward
to a Borel probability measure on the space P(2!) of measures.
Definition 2.3.2. The measure P 2 P(P(2!)) is the pushforward via � of the
Lebesgue measure; that is,
P (B) := � � ��1B
for Borel B ✓ P(2!).
Proposition 2.3.3. The measure P is computable.
Proof. By Proposition 2.2.3, it su�ces to show that the measure of an e↵ectively
open set U ✓ P(2!) is uniformly left-computable. Since � is computable, ��1U
is uniformly e↵ectively open. Because � is computable, P (U) = � � (��1(U)) is
left-computable.
The measure P was defined with a goal that Martin Lof random elements of P
are exactly the images under � of random elements of 2!.
Theorem 2.3.4 (Preservation of randomness and no randomness ex nihilo). ⌫ 2
MLRP
if and only if ⌫ = µx
for some (unique) x 2 MLR�
.
12
Proof. (Preservation of randomness) If ⌫ /2 MLRP
, then ⌫ 2T
n2! Un
for some P -
test hUn
in2!. But, � is computable, so ��1U
n
is e↵ectively open in 2! for each n.
Moreover �(��1(Un
)) = P (Un
) 2�n for each n. Thus h��1(Un
)in2! is a �-test,
and, hence, x /2 MLR�
if ⌫ = �(x) = µx
.
(No randomness ex nihilo) Now, we show, following Shen (see [4, Theorem 3.5]),
that if ⌫ 2 MLRP
, then ⌫ = �(x) for some x 2 MLR. Uniqueness follows because
�(x) = �(y) for x 6= y implies each of x and y has a dyadic rational column.
Fix a universal Martin Lof test hUn
in2! for � randomness, and set K
n
:= 2!�Un
.
Define Vn
= P(2!)��(Kn
). Since � is computable, �(Kn
) 2 ⇧0
1
by Proposition 2.2.2,
parts (2) and (3), so Vn
2 ⌃0
1
, and uniformly so. Now
P (Vn
) = 1� P (�(Kn
)) = 1� �(��1(�(Kn
)))
1� �(Kn
)
2�n.
Thus, hVn
in2! is a P -Martin Lof test, so if ⌫ 2 MLR
P
, then ⌫ /2 Vn
for some n; i.e.,
⌫ 2 �(Kn
). The proof is now complete since Kn
✓ MLR�
.
The next proposition shows that � is a measure theoretic isomorphism (see [22])
between (2!,B(2!),�) and (P(2!),B(P(2!)), P ).
Proposition 2.3.5. For any Borel A ✓ 2!, P (�(A)) = �(A).
Proof. Note that �(MLR�
) \ �(A) = �(MLR�
\A) since �(x) = �(x0) and x 2 MLR
13
implies x = x0. Thus,
P (�(A)) = P (MLRP
\�(A))
= P (�(MLR�
) \ �(A))
= P (�(MLR�
\A))
= ���1�(MLR�
\A)
= �(MLR�
\A)
= �(A).
We will need the next result, which is the same as Theorem 2.3.4 for Kurtz
randomness.
Proposition 2.3.6. ⌫ 2 KRP
if and only if ⌫ = µx
for some (unique) x 2 KR�
.
Proof. If ⌫ /2 KRP
, then ⌫ 2 C for some C 2 ⇧0
1
with P (C) = 0. Because �
is computable, ��1C 2 ⇧0
1
and by the definition of P , ���1C = P (C) = 0, so
��1⌫ \ KR�
= ;.
Now, if x /2 KR�
, then x 2 C for some C 2 ⇧0
1
with �(C) = 0. But then
C is e↵ectively compact, so �(C) 2 ⇧0
1
. By Proposition 2.3.5, P (�(C)) = 0, so
�(x) /2 KRP
.
The last preliminaries we need regard relative randomness and a slight variation
of Van Lambalgen’s Theorem.
Definition 2.3.7. A real y is �-random relative to a real x, written y 2 MLRx
�
,
if y /2T
Un
whenever Un
is a uniformly ⌃0,x
1
sequence with �(Un
) 2�n. We write
MLRµ
�
for MLRx
�
, where µ = µx
.
Theorem 2.3.8. In the product space P(2!)⇥ 2!, with the product measure P ⌦�,
the pair (µ, y) is (P ⌦ �)-random if and only if µ 2 MLRy
�
and y 2 MLRµ
�
.
14
Proof. By Theorem 2.3.4, (µ, y) is (P ⌦�)-random if and only if µ = µx
and (x, y) is
(�⌦ �)-random in 2! ⇥ 2!. By Van Lambalgen’s Theorem [21], this happens if and
only if y 2 MLRx
�
= MLRµ
�
and x 2 MLRy
�
= MLRµ
�
.
2.4 Random measures and their randoms
Now, we begin an analysis of MLRP
.
Proposition 2.4.1. If µ 2 MLRP
, then MLRµ
is dense in 2!.
Proof. If µ 2 MLRP
(indeed if µ 2 KRP
), then µ(�) > 0 for any � 2 2<!.
So, MLRµ
is, in some way, topologically large when µ 2 MLRP
. Theorem 2.6.5
below shows, however, that MLRµ
is (Lebesgue) measure theoretically small.
Lemma 2.4.2 below says that � is the barycenter of P . From work of Hoyrup [14],
this gives Theorem 2.4.4, which says, in particular, that the � randoms are exactly
the the P randoms’ randoms.
Lemma 2.4.2. For each Borel A ✓ 2!, �(A) =RP(2
!)
µ(A)dP (µ).
Proof. The function A 7!RP(2
!)
µ(A)dP (µ) is a measure, so it su�ces to consider
sets of the form A = [�], where � 2 2<!.
Z
P(2
!)
µ(�)dP (µ) =
Z
P(2
!)
Y
i<|�|
µ(�(i)|� � i) dP
=Y
i<|�|
Z
P(2
!)
µ(�(i)|� � i) dP (By independence.)
= 2�|�|.
Hoyrup proved the following result, which applies directly to the setting here.
Theorem 2.4.3 ([14, Theorem 3.1, relativized]). Let Q 2 P(P(2!)) be computable
15
with barycenter µ. Then for any z 2 2!,
MLRz
µ
=[
⌫2MLRzQ
MLRz
⌫
.
Since our measure P 2 P(P(2!)) is computable and � is its barycenter, the
following holds.
Corollary 2.4.4. For any z 2 2!,
MLRz
�
=[
µ2MLRzP
MLRz
µ
.
2.5 Random measures are atomless
We show now that every random measure assigns each singleton set measure
zero; i.e., random measures are atomless. An atom of a measure µ 2 P(2!) is x 2 2!
such that µ({x}) > 0. Define A = {µ : µ has an atom} so that A =S
n
An
where
An
:= {µ : µ has an atom with measure � 1/n}.
Lemma 2.5.1. An
is e↵ectively closed.
Proof. Let A(n, �) = {µ : µ(�) � 1/n}. By the proof of Proposition 2.3.1, the
map '�
(µ) = µ(�) is computable, so 2! � A(n, �) = '�1
�
[0, 1/n) 2 ⌃0
1
. Thus,S
�22k A(n, �) 2 ⇧0
1
, and, hence, An
=T
k
S�22k A(n, �) 2 ⇧0
1
.
The notation x =⇤ y for x, y 2 2! means that x and y di↵er on only finitely many
bits; i.e. x =⇤ y if and only if |{i : x(i) 6= y(i)}| < 1.
Lemma 2.5.2 (Kolmogorov’s 0-1 Law [10]). If A 2 B(2!) is closed under =⇤ (i.e. for
all x 2 A, y =⇤ x ) y 2 A), then �(A) = 0 or �(A) = 1.
16
Corollary 2.5.3. If A 2 B(2!) is almost closed under =⇤ (i.e. for �-almost every x,
x 2 A ) y 2 A whenever x =⇤ y), then �(A) = 0 or �(A) = 1.
Proof. The set bA := {x 2 A : 8y[x =⇤ y ) y 2 A]} is closed under =⇤ and
has the same measure as A since A was already almost closed under =⇤. Applying
Lemma 2.5.2 to bA then gives the result.
Lemma 2.5.4. P (A) = 0 or 1.
Proof. By Proposition 2.3.5 and Corollary 2.5.3, it su�ces to show that the set A =
{x : µx
has an atom} is almost closed under =⇤. To that end, we prove that if
x 2 MLR\A and x0 =⇤ x, then x0 2 A. The key here is to notice that y 2 2! is an
atom for µx
if and only ifQ
i2! µx
(y(i)|y � i) > 0. If x0 =⇤ x, then there is N such
that µx
(y(i)|y � i) = µx
0(y(i)|y � i) for all i > N . Thus, unless µx
0(y(i)|y � i) = 0
for some i N , y is also an atom of x0. But µx
0(y(i)|y � i) is either a column of
x0 or one minus a column of x0, and since x 2 MLR, so is x0, which means that
µx
0(y(i)|y � i) = 0 is impossible.
For � 2 2<!, we define a map T�
: P(2!) ! P(2!), and write µ�
for T�
(µ), by
T�
(µ)(⌧) = µ�
(⌧) = µ(⌧ |�) := µ(�⌧)
µ(�)
. This map is like a shift map. It takes a measure
µ, which can be thought of as a tree of conditional probabilities and outputs a new
measure µ�
whose tree of probabilities is the same as that of µ’s above �.
Lemma 2.5.5. For each � 2 2<!, T�
preserves P ; i.e. P (A) = P (T�1
�
A) for every
Borel A ✓ P(2!).
Proof. Since T�
= T�(n�1)
�T�(n�2)
� · · ·�T�(1)
�T�(0)
for � 2 2n, it su�ces to consider
only the maps T0
and T1
. We prove only that T0
preserves P , since the proof that T1
does is essentially the same.
Because � is a measure isomorphism, it su�ces to show that the map T0
:=
17
��1 � T0
� � : 2! ! 2! preserves �. For then
P (B) = �(��1(B)) = �(T�1
0
(��1(B)))
= �(��1(T�1
0
(�(��1(B)))))
= �(��1(T�1
0
(B)))
= P (T�1
0
(B)).
To show that T0
preserves �, we first get a nice description of T0
. With x =
�i2!xi
, we can write T0
(x) = x1
� x3
� x4
� x7
� x8
� x9
� x10
� · · · . There is
a 1-1 (computable) function f : ! ! ! such that T0
(x)(n) = x(f(n)). To show
T0
preserves �, it su�ces to show that �(�) = �(T�1
0
(�)) for each � 2 2<!. But
T�1
0
(�) = {x : 8i < |�|[x(f(i)) = �(i)}, so clearly �(�) = �(T�1
0
(�)).
Lemma 2.5.6. P (A) = 0
Proof. Define m : P(2!) ! [0, 1] by
m(µ) = max{r 2 [0, 1] : 9y 2 2!(µ{y} = r)}.
We want to show that m(µ) = 0 for P -a.e. µ 2 P(2!).
Notice that m(µ) = max{µ(0)m(T0
(µ)), µ(1)m(T1
(µ))}. Let
M = {µ : µ(0)m(T0
(µ)) � µ(1)m(T1
(µ))},
so M is the set where the maximum-mass atom is to the left in the tree. Then
using standard facts about the integral and the fact that the functions µ 7! µ(0) and
18
µ 7! m(T0
(µ)) are P -independent gives
Z
P(2
!)
m(µ) dP (µ) =
Z
M
µ(0)m(T0
(µ)) dP +
Z
P(2
!)�M
µ(1)m(T1
(µ)) dP
Z
P(2
!)
µ(0)m(T0
(µ)) dP +
Z
P(2
!)�M
µ(1)m(T1
(µ)) dP
=
Z
P(2
!)
µ(0) dP
Z
P(2
!)
m(T0
(µ)) dP +
Z
P(2
!)�M
µ(1)m(T1
(µ)) dP
=1
2
Z
P(2
!)
m(T0
(µ)) dP +
Z
P(2
!)�M
µ(1)m(T1
(µ)) dP
=1
2
Z
P(2
!)
m(T0
(µ)) dP +1
2
Z
P(2
!)
m(T1
(µ)) dP
�Z
M
µ(1)m(T1
(µ)) dP
=1
2
Z
P(2
!)
m(µ) dP +1
2
Z
P(2
!)
m(µ) dP �Z
M
µ(1)m(T1
(µ)) dP
=
Z
P(2
!)
m(µ) dP �Z
M
µ(1)m(T1
(µ)) dP.
Thus, either P (M) = 0 or m(T1
(µ)) = 0 for P -a.e. µ 2 M (because µ(1) is P -
a.s. positive). Symmetric computations show that either P (P(2!) � M) = 0 or
m(T0
(µ)) = 0 for P -a.e. µ 2 P(2!)�M .
In the case where P (M) = 0, we have P (P(2!) � M) = 1, so m(T0
(µ)) = 0
for P -a.e. µ 2 P(2!). Similarly, in the case where P (P(2!) � M) = 0, we have
m(T1
(µ)) = 0 for P -a.e. µ 2 P(2!). In either case, because T0
and T1
both preserve
P , m(µ) = 0 for P -a.e. µ 2 P(2!).
In the case where 0 < P (M) < 1, Lemma 2.5.4 implies that P (A) = 0 or
P (A) = 1, so that m(µ), m(T0
(µ)), and m(T1
(µ)) are all P -a.s. 0 or all P -a.s. strictly
positive. Thus in this case m(T1
(µ)) = 0 on a P -positive-measure set, and hence on
P -almost all of P(2!) as well. Therefore m(µ) = 0 for P -a.e. µ 2 P(2!).
Now, we arrive at the main result of this section; random measures are atomless.
Theorem 2.5.7. Every µ 2 KRP
is atomless. In particular every µ 2 MLRP
is
19
atomless.
Proof. By Lemma 2.5.6, A =S
An
is P -null. Hence, An
is P -null for each n. By
Lemma 2.5.1, each An
is ⇧0
1
.
2.6 Random measures are mutually singular (with respect to the Lebesgue measure)
A measure µ is absolutely continuous with respect to another measure ⌫, writ-
ten µ ⌧ ⌫ if ⌫(A) = 0 ) µ(A) = 0. The property µ ⌧ � implies atomlessness
(recall that � is the Lebesgue measure), so it is natural to ask if µ 2 MLRP
implies
µ ⌧ �. This is far from the case. We show now, in fact, that every µ 2 MLRP
is
mutually singular with respect to �, in symbols µ ? �, which means that µ(A) = 1
for some A 2 B(X) with �(A) = 0. We will actually prove a stronger result: that
MLRµ
\MLRµ
�
= ;.2
To show that MLRµ
\MLRµ = ;, we will employ the well-studied notion of selec-
tion functions. A selection function is a partial function f : 2<! ! {select, exclude};
f determines which bits are selected for entry into a subsequence. The next lemma
tells us that a certain selection function we use later will select infinitely often.
Lemma 2.6.1. Let µ 2 MLRP
, let x 2 MLRµ
�
, and let 0 < ↵ < 1. Then there are
infinitely many n such that µ(0|x � n) > ↵.
Proof. In the product space, P(2!)⇥ 2!, the set
EN
= {(µ, x) : 8n � N [µ(0|x � n) ↵]}
is ⇧0
1
, so it su�ces to prove that (P ⌦ �)(EN
) = 0, where P ⌦ � denotes the product
measure, because then for every (µ, x) 2 MLRP⌦�
(actually for every (µ, x) 2 KRP⌦�
)
2The results in this section were proven with Laurent Bienvenu.
20
there are infinitely many n such that µ(0|x � n) > ↵. By Theorem 2.3.8, (µ, x) 2
MLRP⌦�
if and only if µ 2 MLRP
and x 2 MLRµ
�
.
Now, with FN
the complement (in P(2!) ⇥ 2!) of EN
, it is clear that for a fixed
x, P -a.e. µ has the property that (µ, x) 2 FN
. Thus
(P ⌦ �)(FN
) =
Z
P(2
!)⇥2
!
1FN d(P ⌦ �)
=
Z
2
!
Z
P(2
!)
1FN dP d� (By Fubini’s Theorem [12].)
=
Z
2
!
1 d�
= 1.
Lemma 2.6.2. Let µ 2 MLRP
, let x 2 MLRµ
�
, let 0 < ↵ < 1 be rational, and let
n1
< n2
< · · · be the sequence of all ni
such that µ(0|x � ni
) > ↵ for all i, which is
infinite by Lemma 2.6.1. Then yx
2 2! defined by yx
(i) = x(ni
) satisfies the law of
large numbers; i.e.
limn!1
1
n
X
i<n
yx
(i) =1
2.
Proof. This proof is a relativization of the proof of Theorem 7.4.2 in Downey &
Hirschfeldt’s book [10]. The point is that y is the result of a µ-computable selection
strategy that simply selects the bit x(n) from x whenever µ(0|x � n) > ↵. Since
x 2 MLRµ
�
and the MLRµ
�
sequences are among those from which it is impossible to
µ-computably select a subsequence that violates the law of large numbers, yx
must
satisfy the law of large numbers.
The next lemma gives us an e↵ective bound for the proof of Lemma 2.6.4. We
state it in slightly simplified form.
Lemma 2.6.3 (Hoe↵ding’s inequality [23]). Let y(1), . . . , y(n) be independent ran-
21
dom variables on the probability space (2!, µ) taking values in [0, 1].Then
µ
1
n
nX
i=1
y(i)�Z
2
!
1
n
nX
i=1
y(i) dµ � ✏
! e�2n✏
2
for every ✏ > 0.
Lemma 2.6.4. Let µ 2 MLRP
, let x 2 MLRµ
, let 0 < ↵ < 1 be rational, and let
n1
< n2
< · · · be the sequence of all ni
such that µ(0|x � ni
) > ↵ for all i. If hni
ii2!
is infinite, then yx
2 2! defined by yx
(i) = x(ni
) satisfies
lim infn!1
1
n
X
i<n
yx
(i) 1� ↵.
Proof. Let � = 1� ↵ and W ✏
k
= {x : 1
k
Pi<k
yx
(i)� � > ✏} for ✏ 2 Q+. Then W ✏
k
is
uniformly ⌃0,µ
1
and
(x :
1
n
X
i<n
yx
(i)� � > ✏ for infinitely many n
)=\
N
[
k>N
W ✏
k
.
Thus it su�ces to show that µ�S
k>N
W ✏
k
�! 0 e↵ectively in N .
Noticing that
Z
2
!
1
k
X
i<k
yx
(i) dµ =1
k
X
i<k
Z
2
!
yx
(i) dµ 1
k
X
i<k
� = �
and
W ✏
k
✓ {x :1
k
X
i<k
yx
(i)�Z
2
!
1
k
X
i<k
yx
(i) dµ > ✏},
we can apply Lemma 2.6.3 to conclude that µ�S
k>N
W ✏
k
�P
k>N
e�2k✏
2 ! 0 e↵ec-
tively in N .
Theorem 2.6.5. If µ 2 MLRP
, then MLRµ
\MLRµ
�
= ;, so µ ? �.
Proof. Lemmas 2.6.2 and 2.6.4 together imply that MLRµ
\MLRµ
�
= ;. But, since
22
�(MLRµ
�
) = 1, it must be the case that �(MLRµ
) = 0.
2.7 Relatively random measures
Recall that µ 2 MLR⌫
�
means that y 2 MLRx
�
where µ = �(y) and ⌫ = �(x). If
both µ 2 MLR⌫
�
and ⌫ 2 MLRµ
�
, we say that µ and ⌫ are relatively random.
It was conjectured initially that relatively random measures would share no ran-
doms. In fact an immediate consequence of Theorems 2.4.4 and 2.6.5 is the following,
which shows the conjecture almost true.
Theorem 2.7.1. If µ and ⌫ are relatively random, then MLR⌫
µ
\MLR⌫
= ; =
MLRµ
⌫
\MLRµ
. In particular, µ ? ⌫.
Interestingly though, relatively random measures do share a random real, and in
a rather strong way. Before proving this, we need a lemma about a universal test for
µ-randomness.
Lemma 2.7.2. Each µ 2 MLRP
has a name of least Turing degree and hence admits
a universal µ-test.
Proof. The point here is that x := ��1(µ) is essentially a name for µ, and the one of
least Turing degree. The proof of Proposition 2.3.1 shows that x computes a name
for µ. Also any name for µ computes µ(�) for any � 2 2<! and hence also must be
able to compute µ(0|�) for each � 2 2<!; this is the same as computing x.
Theorem 2.7.3. There is a computable function G : P(2!)⇥P(2!) ! 2! such that
if µ and ⌫ are relatively random measures, then G(µ, ⌫) 2 MLRµ
\MLR⌫
.3
Proof. The algorithm we are about to define is a greedy one. It builds the common
random by asking the two measures “Which of you cares the most which direction I
go?” and then acting accordingly.
3This result was proven with Joe Miller and Uri Andrews.
23
Define the function G : P(2!)⇥ P(2!) ! 2! by
G(µ, ⌫)(i) = 0 () µ(0 | G(µ, ⌫) � i) > ⌫(1|G(µ, ⌫) � i).
Note that also
G(µ, ⌫)(i) = 1 () µ(1|G(µ, ⌫) � i) > ⌫(0|G(µ, ⌫) � i).
The key fact here is that P{⌫ : G(µ, ⌫) � �} = µ(�) for each µ 2 P(2!) and
� 2 2<!. Indeed, by induction, if P{⌫ : G(µ, ⌫) � �} = µ(�), then by independence
P{⌫ : G(µ, ⌫) � �ai} = P{⌫ : G(µ, ⌫) � �} · P{⌫ : G(µ, ⌫)(|�|) = i|G(µ, ⌫) � �}
= µ(�) · P{⌫ : ⌫(1� i|�) < µ(i|�)}
= µ(�) · µ(i|�)
= µ(�ai).
Let hUµ
n
in2! be a universal µ-test. The sets V
n
:= {⌫ : G(µ, ⌫) 2 Uµ
n
} are
uniformly ⌃0,µ
1
and, with Sn
a prefix-free set of generators of Un
, the above calculation
gives
P (Vn
) = P
[
�2Sn
{⌫ : G(µ, ⌫) � �}!
=X
�2Sn
P ({⌫ : G(µ, ⌫) � �}) = · · ·
· · · =X
�2Sn
µ(�) = µ(Un
) 2�n.
Thus hVn
in2! is a P -test for Martin Lof randomness relative to µ and so if ⌫ is
relatively random to µ, G(µ, ⌫) 2 MLRµ
. By symmetry, G(µ, ⌫) = G(⌫, µ) 2 MLR⌫
as well.
The random element shared by the relatively random measures in Theorem 2.7.3
24
was derandomized (indeed computed) by the join of those measures. Theorem 2.7.1
shows that in fact any x 2 MLRµ
\MLR⌫
must be derandomized (with respect to
either measure) by µ� ⌫. This raises the following question.
Question 2.7.4. If µ and ⌫ are relatively random and x 2 MLRµ
\MLR⌫
, must x be
computed by µ� ⌫?
Theorem 2.7.1 also leads to Theorem 2.7.7, a special case of which says that the
randoms for a random measure µ form an “anti-tailset”: change even a single bit and
the real loses its µ-randomness.
Recall that for � 2 2<!, µ�
is defined by µ�
(⌧) = µ(⌧ |�).
Lemma 2.7.5. ⌫ and ⇠ are relatively random if and only if ⌫ = µ�
and ⇠ = µ⌧
for
some µ 2 MLRP
and incompatible ⌧, � 2 2<!.
Proof. Clearly µ�
and µ⌧
are relatively random whenever µ 2 MLRP
and � ? ⌧ .
Given two relatively random measure ⌫ and ⇠, by taking p 2 MLR⌫�⇠
�
and defining µ
by µ(0) = p, µ0
= ⌫, and µ1
= ⇠, we have µ 2 MLRP
.
Lemma 2.7.6. If �x 2 MLRµ
, then x 2 MLRµ� .
Proof. If x /2 MLRµ� , then x 2 Uµ�
n
for each n, where hUµ�n
in2! is a universal µ
�
-test.
Then Vn
:= {�⌧ : ⌧ 2 Uµ�n
} is ⌃0,µ
1
since µ �T
µ�
and µ(Vn
) = µ[�]µ�
(Uµ�n
) 2�n.
Therefore, hVn
in2! is a µ-test capturing �x, and, whence, �x /2 MLR
µ
.
The next result shows that µ-random elements in one part of the full binary-
branching tree look much di↵erent from those in another part. Contrast this with the
Lebesgue measure, where randomness does not depend on prefixes (i.e. randomness
is a “tail event”).
Theorem 2.7.7. If �x 2 MLRµ
and ⌧ ? �, then ⌧x /2 MLRµ
.
25
Proof. Suppose �x 2 MLRµ
and ⌧x 2 MLRµ
. Then x 2 MLRµ� \MLR
µ⌧ by Lemma 2.7.6.
Since � ? ⌧ , the measures µ�
and µ⌧
are relatively random and hence by Theo-
rem 2.7.1, x /2 MLRµ�µ⌧. Since µ �
T
µ�
, x /2 MLRµ
µ⌧, so by Lemma 2.7.6, ⌧x /2 MLRµ
µ
=
MLRµ
, a contradiction.
We used Kolmogorov’s 0-1 law earlier (Lemma 2.5.2). It seems that random
measures should fail to satisfy Kolmogorov’s 0-1 law, since changing finitely many
bits of a real puts it in a part of the tree whose conditional probabilities are wildly
di↵erent (for a fixed random µ). We now confirm this intuition.
Corollary 2.7.8. Random measures fail to satisfy Kolmogorov’s 0-1 law.
Proof. Let µ 2 MLRP
. By Theorem 2.7.7, closing the set MLRµ
\[0] under tails
adds no randoms and hence no measure. The result is therefore a tailset of measure
µ(0) 2 (0, 1).
We can also use Theorem 2.7.7 to show that given a random x, the probability of
choosing a measure that thinks x is random is zero.
Corollary 2.7.9. If x 2 MLR�
, then P ({µ : x 2 MLRµ
}) = 0.
Proof. Changing only finitely many bits of (the preimage under � of) any P -random
µ does not a↵ect whether x 2 MLRµ
, so by Kolmogorov’s 0-1 law, either
P ({µ : x 2 MLRµ
}) = 0
or
P ({µ : x 2 MLRµ
}) = 1.
But, P ({µ : x 2 MLRµ
}) = 1 if and only if P ({µ : x0 2 MLRµ
}) = 1, where
x0(i) = x(i) for i > 0 and x0(0) = 1� x(0). So, if P ({µ : x 2 MLRµ
}) = 1, then there
is µ such that x, x0 2 MLRµ
, contrary to Theorem 2.7.7.
26
Let T : 2! ! 2! be the shift map; so Tx(i) = x(i + 1). This map preserves
the Lebesgue measure.Clearly, no element of MLRP
is preserved by T , since that
would introduce dependence amongst the conditional probabilities of µ. Moreover,
the following holds.
Theorem 2.7.10. If µ 2 MLRP
and x 2 MLRµ
, then Tx /2 MLRµ
.
Proof. If x 2 MLRµ
and Tx 2 MLRµ
, then there are incompatible strings � � x and
⌧ � Tx such that x = �y and Tx = ⌧y. But µ�
and µ⌧
are relatively random with
y 2 MLRµ⌧µ�
\MLRµ�µ⌧
contradicting Theorem 2.7.1.
We believe Theorem 2.7.10 can be generalized: Recall (see [19]) that if f : ! ! !
is 1-1 and computable, then x � f 2 MLR�
whenever x 2 MLR�
. Our final conjecture
states that this fails for random elements of random measures.
Conjecture 2.7.11. Suppose µ 2 MLRP
and x 2 MLRµ
. If f : ! ! ! is 1-1,
computable and non-identity, then x � f /2 MLRµ
.
27
CHAPTER 3
THE INTERPLAY OF CLASSES OF ALGORITHMICALLY RANDOM
OBJECTS
Work in this chapter was done jointly with Chris Porter.
3.1 Introduction
In this chapter, we have two primary goals: (1) to study the interplay between
algorithmically random closed sets on 2!, algorithmically random continuous func-
tions on 2!, and algorithmically random measures on 2!; and (2) to apply two central
results, namely the preservation of randomness principle and the no randomness ex
nihilo principle, to the study of the algorithmically random objects listed above.
Barmpalias, Brodhead, Cenzer, Dashti and Weber initiated the study of algorith-
mically random closed subsets of 2! in [2]. Algorithmically random closed sets were
further studied in, for instance, [1], [9], and [7]. In the spirit of their definition of
algorithmically random closed set, Barmpalias, Brodhead, Cenzer, Dashti and Weber
also defined a notion of algorithmically random continuous function on 2! in [3]. The
connection between random closed sets and e↵ective capacities was explored in [6].
Algorithmically random measures on 2! were studied first in Chapter 2.
One of the central results in [3] is that the set of zeroes of a random continuous
function of 2! is a random closed subset of 2!. Inspired by this result, we here
investigate similar “bridge results,” which allow us to transfer information about one
class of algorithmically random objects to another.
28
Two tools that are central to our investigation, mentioned in (2) above, are the
preservation of randomness principle and the no randomness ex nihilo principle. In
2!, the space of infinite binary sequences, the preservation of randomness principle
tells us that if � : 2! ! 2! is an e↵ective map and µ is a computable probability
measure on 2! such that the domain of � has µ measure 1, then � maps µ-random
members of 2! to members of 2! that are random with respect to the measure ⌫
obtained by pushing µ forward via �. Furthermore, the no randomness ex nihilo
principle tells us that any sequence that is random with respect to ⌫ is the image
of some µ-random sequence under �. Used in tandem, these two principles allow
us to conclude that the image of the µ-random sequences under � is precisely the
⌫-random sequences.
With the exception of our work in Chapter 2, the studies listed above do not make
use of these two tools used in tandem. As we will show, they not only allow for the
simplification of a number of proofs in the above-listed studies, but they also allow
us to answer a number of questions that were left open in the above studies.
The outline of the remainder of this chapter is as follows. In Section 3.2, we
provide the requisite background for the rest of the chapter. In Section 3.3, we review
the basics of algorithmic randomness, including preservation and the no randomness
ex nihilo principle. We also provide the definitions of algorithmic randomness for
closed sets in 2!, random continuous functions on 2!, and measures on 2! and we
list some basic properties of these objects. Section 3.4 contains simplified proofs of
some previously obtained results from [2] and [3], as well as a proof of a conjecture in
[3] that every random closed subset of 2! is the set of zeros of a random continuous
function on 2!. We study the support of a certain class of random measures in Section
3.5, and we establish a correspondence between between random closed sets and the
random measures studied in Chapter 2. Lastly, in Section 3.6, we prove that the
Lebesgue measure of the range of a random continuous function on 2! is always non-
29
zero, from which it follows that no random continuous function is injective (which
had not been previously established). We also strengthen a result in [3] (namely,
that not every random continuous function is surjective) by proving that no random
continuous function is surjective, from which it follows that the Lebesgue measure of
the range of a random continuous function is never equal to one.
3.2 Background
3.2.1 Some topological and measure-theoretic basics
For n = {0, 1, . . . n � 1} 2 !, the set of all finite strings over the alphabet n is
denoted n<!. When n = 2, we let �0
, �1
, �2
, . . . be the canonical length-lexicographic
enumeration of 2<!, so that �0
= ✏ (the empty string), �1
= 0, �2
= 1, etc.
The space of all infinite sequences over the alphabet n is denoted n!. The elements
of n! are also called reals. The product topology on n! is generated by the clopen
sets
J�K = {x 2 n! : x � �},
where � 2 n<! and x � � means that � is an initial segment of x. When x is a real
and k 2 !, x � k denotes the initial segment of x of length k.
For �, ⌧ 2 n<!, �_⌧ denotes the concatenation of � and ⌧ . In some cases, we will
write this concatenation as �⌧ .
A tree is a subset of n<! that is closed under initial segments; i.e. T ✓ n<! is
a tree if � 2 T whenever ⌧ 2 T and � � ⌧ . A path through a tree T ✓ n<! is a
real x 2 n! satisfying x � k 2 T for every k. The set of all paths through a tree T is
denoted [T ]. Recall the correspondence between closed sets and trees.
Proposition 3.2.1. A set C ✓ n! is closed if and only if C = [T ] for some tree
T ✓ n<!. Moreover, C is nonempty if and only if T is infinite.
A measure µ on n! is a function that assigns to each Borel subset of n! a number
30
in the unit interval [0, 1] and satisfies µ(S
i2! Bi
) =P
i2! µ(Bi
) whenever the Bi
’s are
pairwise disjoint. Caratheodory’s extension theorem guarantees that the conditions
• µ(J✏K) = 1 and
• µ(J�K) = µ(J�0K) + µ(J�1K) + . . .+ µ(J�_(n� 1)K) for all � 2 n<!
uniquely determine a measure on n!. Thus, a measure is identified with a function
µ : n<! ! [0, 1] satisfying the above conditions, and µ(�) is often written instead of
µ(J�K). The Lebesgue measure � on n! is defined by �(�) = n�|�| for each string
� 2 n<!.
Given a measure µ on n! and �, ⌧ 2 n<!, µ(�⌧ | �) is defined to be
µ(�⌧ | �) = µ(J�⌧K)µ(J�K) .
3.2.2 Some computability theory
A ⌃0
1
class S ✓ n! is an e↵ectively open set, i.e., an e↵ective union of basic clopen
subsets of n!. P ✓ n! is a ⇧0
1
class if 2! \ P is a ⌃0
1
class.
A partial function � : ✓ n! ! m! is computable if the preimage of a ⌃0
1
subset
of m! is a ⌃0
1
subset of the domain of �, uniformly; that is, if for every ⌃0
1
class
U ✓ m!, there is a ⌃0
1
class V ✓ n! such that ��1(U) = V \ dom(�), and an index
for V can be uniformly computed from an index for U . Equivalently, � : ✓ n! ! m!
is computable if there is an oracle Turing machine that when given x 2 n! (as an
oracle) and k 2 ! outputs �(x)(k). We can relativize the notion of a computable
function � : ✓ n! ! m! to any oracle z 2 2! to obtain a z-computable function.
A measure µ on n! is computable if µ(�) is a computable real number, uniformly
in � 2 n<!. Clearly, the Lebesgue measure � is computable.
If µ is a computable measure on n! and � : ✓ n! ! m! is a computable function
31
defined on a set of µ-measure one, then the pushforward measure µ�
defined by
µ�
(�) = µ(��1(�))
for each � 2 m<! is a computable measure.
3.3 Algorithmically random objects
3.3.1 Algorithmically random sequences
Definition 3.3.1. Let µ be a computable measure on n! and z 2 m!. Then MLRz
µ
is the set of all x 2 n! such that x /2T
n
Un
whenever U0
, U1
, . . . is a uniformly
⌃0,z
1
sequence of subsets of n! with µUn
2�n. Such an x is said to be µ-random
relative to z and such a sequence U0
, U1
, . . . is called a µ-test relative to z. When
z is computable, we simply write MLRµ
, say x is µ-random, and call U0
, U1
, . . . a
µ-test.
The following is well-known and straightforward.
Proposition 3.3.2. Let µ be a computable measure on n! and let z 2 m!. If C ✓ n!
is ⇧0,z
1
and µ(C) = 0, then C \MLRz
µ
= ;.
The following is likely folklore, but it was at least observed in [4].
Proposition 3.3.3. Let µ be a computable measure on n!. If T : ✓ n! ! m! is
computable with µ(dom(T )) = 1, then MLRµ
✓ dom(T ).
Lemma 3.3.4 (Folklore). Let T : ✓ 2! ! 2! be computable, and suppose C is a
⇧0
1
subset of dom(T ). Then T (C) 2 ⇧0
1
, uniformly.
The next theorem represents our main tool here.
Theorem 3.3.5 (Preservation of Randomness and No Randomness Ex Nihilo). Let
T : ✓ 2! ! 2! be computable with �(dom(T )) = 1.
32
(i) If x 2 MLR�
then T (x) 2 MLR��T�1 .
(ii) If y 2 MLR��T�1 , then there exists x 2 MLR
�
such that T (x) = y.
Proof.
(i) If T (x) /2 MLR��T�1 , then T (x) 2
Tn
Vn
for some � � T�1 test. Then x 2Tn
T�1Vn
and �(T�1Vn
) 2�n. Moreover, because T is computable (on itsdomain), T�1V
n
= Un
\ dom(T ) for some ⌃0
1
class Un
. Since �(dom(T )) = 1,�(T�1U
n
) 2�n Thus, x /2 MLR�
.
(ii) Let Un
be a universal test for � randomness, and set Kn
= X � Un
. ThenT (K
n
) is uniformly ⇧0
1
by Lemma 3.3.4, so Y �T (Kn
) is uniformly ⌃0
1
. Because��T�1(Y �T (K
n
)) = 1���T�1(T (Kn
)) 1��(Kn
) 2�n, the sets Y �T (Kn
)form a test for � � T�1 randomness. So if y 2 MLR
��T�1 , then y /2 Y � T (Kn
)for some n; i.e. y 2 T (K
n
). The proof is now complete, since Kn
✓ MLR�
.
We will also use a relativization of Theorem 3.3.5.
Corollary 3.3.6. Let T : ✓ 2! ! 2! be computable relative to z 2 2! with
�(dom(T )) = 1.
(i) If x 2 MLRz
�
, then T (x) 2 MLRz
��T�1
.
(ii) If y 2 MLRz
��T�1
, then there is x 2 MLRz
�
such that T (x) = y.
Lastly, the following result, known as van Lambalgen’s Theorem, will be useful
to us.
Theorem 3.3.7 ([21]). Let µ and ⌫ be computable measures on m! and n!, respec-
tively. Then for (x, y) 2 m! ⇥ n!, (x, y) 2 MLRµ⌦⌫
if and only if x 2 MLRy
µ
and
y 2 MLR⌫
.
3.3.2 Algorithmically random closed subsets of 2!
Let C(2!) denote the collection of all nonempty closed subsets of 2!. As noted in
Proposition 3.2.1, these are the sets of paths through infinite binary trees. Thus, to
randomly generate a nonempty closed set, it su�ces to randomly generate an infinite
33
tree. We’ll code infinite trees by reals in 3!, so we can reduce the process of randomly
generating infinite trees to randomly generating reals.
Given x 2 3!, define a tree Tx
✓ 2<! inductively as follows. First, ?, the empty
string is automatically in Tx
. Now, suppose �i
2 Tx
. Then
• �i
_0 2 Tx
and �i
_1 /2 Tx
if x(i) = 0;
• �i
_0 /2 Tx
and �i
_1 2 Tx
if x(i) = 1;
• �i
_0 2 Tx
and �i
_1 2 Tx
if x(i) = 2.
Under this coding Tx
has no dead ends and hence is always infinite. This coding can
be thought of as a labeling of the nodes of 2! by the digits of x; a 0 at a node means
branch only left, a 1 means branch only right, and a 2 means branch both ways. Note
that every tree without dead ends except 2<! itself has infinitely many codes.
Definition 3.3.8. A nonempty closed set C 2 C(2!) is a random closed set if
C = [Tx
] for some x 2 MLR�
.
The main facts about random closed sets that we will use in the sequel are as
follows.
Theorem 3.3.9 ([2]). Every random closed set has Lebesgue measure zero.
Theorem 3.3.10 ([2]). Every random closed set is perfect.
3.3.3 Algorithmically random continuous functions on 2!
Let F(2!) denote the collection of all continuous F : ✓ 2! ! 2!. To define a
random continuous function, we code each element of F(2!) by a real x 2 3!. The
coding is a labeling of the edges of 2! (or equivalently, all nodes in 2<! except ✏) by
the digits of x. Having labeled the edges according to x, the function Fx
coded by x
is defined by Fx
(y) = z if z is the element of 2! left over after following y through
34
the labeled tree and removing the 2’s. (In the case where only finitely many 0’s and
1’s remain after removing the 2’s, Fx
(y) is undefined.)
Formally, define a labeling function `x
: 2<! \ {✏} ! 3 by `x
(�i
) = xi�1
. Now
Fx
2 F(2!) is defined by Fx
(y) = z if and only if z is the result of removing the 2’s
from the sequence `x
(y � 1), `x
(y � 2), `x
(y � 3), . . . .
Definition 3.3.11. A function F 2 F(2!) is a random continuous function if
F = Fx
for some x 2 MLR�
.
Remark 3.3.1. Fx
is continuous (on its domain), because it is computable relative to
some oracle; namely x. Since 2! is compact and Hausdor↵, it follows that Fx
is a
closed map and, hence, that ran(F ) is ⇧0,F
1
.
We will make use of the following facts about random continuous functions.
Theorem 3.3.12 ([3]). If F 2 F(2!) is random and x 2 2! is computable, then
F (x) 2 2! is random.
Theorem 3.3.13 ([3]). If F 2 F(2!) is random, then F is total.
3.3.4 Algorithmically random measures on 2!
Let P (2!) be the space of probability measures on 2!. Given x 2 2!, the nth
column xn
of x is defined by xn
(k) = 1 if and only if x(hn, ki) = 1, where hn, ki
is some fixed computable bijection between !2 and !. We write x = �n2!xn
. Let
(�i
)i2! be the canonical enumeration of 2<! in the length-lexicographical order. We
define a map : 2! ! P (2!) that sends a real x to the measure µx
satisfying (i)
µx
(✏) = 1 and (ii) µx
(�n
0) = xn
· µx
(�n
), where xn
is the real number corresponding
to the nth column of x.
Definition 3.3.14. A measure µ 2 P (2!) is a random measure if µ = µx
for some
x 2 MLR�
.
35
Let P be the pushforward measure on P (2!) induced by � and . Then we have
the following.
Theorem 3.3.15. Let ⌫ 2 P (2!). Then ⌫ 2 MLRP
if and only if ⌫ = µx
for some
x 2 MLR�
.
The support of a measure µ on 2! is defined to be
Supp(µ) = {x 2 2! : (8n)[µ(x�n) > 0]}
It is not hard to see that Supp(µ) = 2! for every random measure µ.
In Chapter 2, it was shown that random measures are atomless and that the reals
that are random with respect to some random measure are precisely the reals in
MLR�
.
3.4 Applications of Randomness Preservation and No Randomness Ex Nihilo
In this section, we demonstrate the usefulness of preservation of randomness and
the no randomness ex nihilo principle in the study of algorithmically random objects
such as closed sets and continuous functions.
The following is a new, simpler proof of a known result from [2].
Theorem 3.4.1. Every random closed set contains an element of MLR�
, and every
element of MLR�
is contained in some random closed set.
Proof. We define a computable map T : C(2!) ⇥ 2! ! 2! that pushes forward the
product measure �C ⌦ � to � and satisfies T (C, x) 2 C for every (C, x) 2 C(2!)⇥ 2!.
Once we have done this, preservation of randomness and no randomness ex nihilo
imply that the image of a �C ⌦ �-random pair is �-random and any �-random is
the image of some �C ⌦ �-random pair. The result then follows because by Van
36
Lambalgen’s Theorem (Theorem 3.3.7), a pair (C, x) is �C ⌦ �-random if and only if
C is �C-random and x is �-random relative to C.
The map works by using x to tell us which way to go through C (viewed as a
tree) when we have a choice to make. Specifically, having T (C, x) � n = � such
that J�K \ C 6= ;, we define T (C, x)(n) = 0 if J�1K \ C = ; and T (C, x)(n) = 1 if
J�0K \ C = ;. If neither J�0K \ C = ; nor J�1K \ C = ;, then T (C, x)(n) := x(n).
The map T is clearly computable. It pushes �C ⌦� forward to � because if T has
output � 2 2n, then T outputs a next bit of 0 if and only if either J�1K \ C = ; or
both J�1K \ C 6= ; 6= J�0K \ C and x(n) = 0. The former happens with probability
1
3
, and the latter happens with probability 1
3
· 1
2
by independence. The proof is now
complete since 1
3
+ 1
6
= 1
2
.
Let F 2 F(2!). We define ZF
= {x : F (x) = 0}. This is clearly a closed subset
of 2!. In [3], the following was shown.
Theorem 3.4.2 ([3]). Let F 2 F(2!) be random. Then ZF
is a random closed set
provided it is nonempty.
In [3], it was conjectured that the converse also holds, but this was left open. We
prove this conjecture. To do so, we provide a new proof of Theorem 3.4.2, from which
the converse follows immediately. We also make use of an alternative characterization
of random closed sets, due to Diamondstone and Kjøs-Hanssen [9].
Just as a binary tree with no dead ends is coded by a sequence in 3! (see the
paragraph preceding Definition 3.3.8), an arbitrary binary tree is coded by a sequence
in 4!, except now a 3 at a node indicates that the tree is dead above that node. That
is, given x 2 4!, we define a tree Sx
✓ 2<! inductively as follows. First ✏, the empty
string, is included in Sx
by default. Now suppose that �i
2 Sx
. Then
• �i
_0 2 Sx
and �i
_1 /2 Sx
if x(i) = 0;
• �i
_0 /2 Sx
and �i
_1 2 Sx
if x(i) = 1;
37
• �i
_0 2 Sx
and �i
_1 2 Sx
if x(i) = 2;
• �i
_0 /2 Sx
and �i
_1 /2 Sx
if x(i) = 3.
This coding can be thought of as a labeling of the nodes of 2! by the digits of x; a 0
at a node means that only the left branch is included, a 1 means that only the right
branch is included, a 2 means that both branches are included, and a 3 means that
neither branch is included. Note that every tree except 2<! itself has infinitely many
codes.
Let µGW
be the measure on 4! induced by setting, for each � 2 4<!,
µGW
(�0 | �) = µGW
(�1 | �) = 2/9, µGW
(�2 | �) = 4/9, and µGW
(�3 | �) = 1/9
Via this coding, we can also think of µGW
as a measure on Tree, the space of
binary trees. Then the probability of extending a string in a tree by only 0 is 2/9,
by only 1 is 2/9, by both 0 and 1 is 4/9, and by neither is 1/9. We call a tree T
GW-random if it has a random code; i.e., there is x 2 MLRµGW such that T = S
x
.
Lemma 3.4.3 (Diamondstone and Kjøs-Hanssen [9]). A closed set C is random if
and only if C is the set of paths through an infinite GW-random tree.
Theorem 3.4.4. (i) For every random F 2 F(2!), ZF
is a random closed setprovided that it is nonempty.
(ii) For every random C 2 C(2!), there is some random F 2 F(2!) such thatC = Z
F
.
Proof. We define a computable map : F(2!) ! Tree that pushes forward �F to
µGW
such that the set of paths through (F ) \ dom(F ) is exactly ZF
. Given our
representation of functions as members of 3! and binary trees as members of 4!, we
are really defining a computable map b : 3! ! 4! that pushes forward � to µGW
.
Given F 2 F(2!), which we think of as a {0, 1, 2}-labeling of the edges of the
full binary tree, we build the desired tree by declaring that � 2 (F ) if and only
38
if the labels by F of the edges of � consists only of 0’s and 2’s. More formally,
as in the paragraph preceding Definition 3.3.11, F comes with a labeling function
`F
: 2<! \ {✏} ! 3 defined by `F
(�i
) = j if and only if x(i) = j where x is the given
code for F . So, � 2 (�) if and only if `F
(��k) 2 {0, 2}<! for every 0 < k |�|.
Clearly, this map is computable.
Now we show that the map pushes �F forward to µGW
. Suppose � 2 (F ),
which, as stated above, means that `F
(��k) 2 {0, 2}<! for every 0 < k |�|. Then
�0 2 (F ) & �1 /2 (F ) , `F
(�0) 2 {0, 2} & `F
(�1) = 1.
The right-hand side of the equivalence occurs with probability (2/3)(1/3) = 2/9.
Similarly,
�0 /2 (F ) & �1 2 (F ) , `F
(�0) = 1 & `F
(�1) 2 {0, 2},
where this latter event also occurs with probability 2/9. Next,
�0 2 (F ) & �1 2 (F ) , `F
(�0) 2 {0, 2} & `F
(�1) 2 {0, 2},
with the latter event occurring with probability (2/3)(2/3) = 4/9. Lastly,
�0 /2 (F ) & �1 /2 (F ) , `F
(�0) = `F
(�1) = 1,
where the event on the right-hand side occurs with probability (1/3)(1/3) = 1/9.
Now, by construction, it follows immediately that any path through the tree (F ) is
a sequence X such that either F (X) = 0! (in the case that `(X�n) = 0 for infinitely
many n) or F (X)" (in the case that `(X�n) = 0 for only finitely many n).
By preservation of randomness and no randomness ex nihilo, a tree is GW-random
39
if and only if it is the image of some random continuous function F . The conclusion
then follows, by Lemma 3.4.3.
One consequence of Theorem 3.3.12 and Theorem 3.4.4(ii), not noted in [3], is
that the composition of two random continuous functions need not be random.
Corollary 3.4.5. For every random F 2 F(2!), there is some random G 2 F(2!)
such that G � F is not random.
Proof. By Theorem 3.3.12, there is some R 2 MLR such that F (0!) = R. By
Theorem 3.4.1, there is some random C 2 C(2!) containing R. By Theorem 3.4.4(ii),
there is a G 2 F(2!) such that G�1({0!}) = C. It follows that G(F (0!)) = 0!. This,
together with Theorem 3.3.12, implies that G � F is not random.
Another consequence of Theorem 3.4.4 lets us answer an open question from [3]
involving random pseudo-distance functions. Given a closed set C 2 C(2!), a function
� : 2! ! 2! is a pseudo-distance function for C if C is the set of zeroes of �. In
[3] it was shown that if � is a random pseudo-distance function for some C 2 C(2!),
then C is a random closed set, but the converse was left open. By Theorem 3.4.4,
the converse immediately follows.
Corollary 3.4.6. Let C 2 C(2!). Then C has a random pseudo-distance function if
and only if C is a random closed set.
3.5 The support of a random measure
In the previous section, we established a correspondence between random closed
sets and and random continuous functions: a closed set C is random if and only if it
is the set of zeroes of some random continuous function. In this section, we establish
similar correspondences between random closed sets and random measures.
40
Since the support of a measure µ, i.e., the set Supp(µ) = {x 2 2! : 8n µ(x�n) > 0}
is a closed set, one might hope to establish such a correspondence by considering the
supports of random measures. However, it is not hard to see that for each random
measure µ, Supp(µ) = 2!.
If we consider a computable measure on P (2!) di↵erent from the measure P
defined in Section 3.3.4, then such a correspondence can be given. In the first place,
we want a measure Q on P (2!) with the property that no Q-random measure has
full support. In fact, we can choose a measure Q such that each Q-random measure
is supported on a random closed set.
Theorem 3.5.1. There is a computable measure Q on P (2!) such that
(i) every Q-random measure is supported on a random closed set, and
(ii) for every random closed set C ✓ 2!, there is a Q-random measure µ such thatSupp(µ) = C.
Proof. We will define the measure Q so that each Q-random measure is obtained
by restricting Lebesgue measure to a random closed set. That is, each Q-random
measure will be uniform on all of the branching nodes of its support.
We define Q in terms of an almost total functional � : 3! ! 2!. On input x 2 3!,
� will treat x as the code for a closed set and will output the sequence y = �i2!yi
defined as follows. For each i 2 !, we set
yi
=
8>>>><
>>>>:
11 if xi
= 0
01 if xi
= 1
101 if xi
= 2
.
If we think of the columns of y as encoding the conditional probabilities of a measure
µy
, then if (�i
)i2! is the standard enumeration of 2<!, these conditional probabilities
41
are given by
p�i =
8>>>><
>>>>:
1 if xi
= 0
0 if xi
= 1
1/2 if xi
= 2
.
That is, �(x) = y, where y represents the unique measure µy
such that µy
(�0 | �) =
p�
for each � 2 2<!. Let Q be the measure on P (2!) induced by the composition of
� and the representation map : 2! ! P (2!) defined in Section 3.3.4.
We now verify (i) by showing that � maps each x 2 MLR to a Q-random measure
supported on a random closed set. Let x 2 MLR and set �(x) = y. By preservation
of randomness, (�(x)) = µy
is Q-random.
Next, since x 2 MLR, [Tx
] is a random closed set. We claim that Supp(µy
) = [Tx
].
Suppose that � 2 2<! is the (n + 1)-st extendible node of Tx
. Then one of the
following holds:
(a) �0 2 Tx
and �1 /2 Tx
;
(b) �0 /2 Tx
and �1 2 Tx
; or
(c) �0 2 Tx
and �1 2 Tx
.
Moreover, we have
• Condition (a) holds i↵ x = 0 i↵ µy
(�0 | �) = 1 and µy
(�1 | �) = 0.
• Condition (b) holds i↵ x = 1 i↵ µy
(�0 | �) = 0 and µy
(�1 | �) = 1.
• Condition (c) holds i↵ x = 2 i↵ µy
(�0 | �) = µy
(�1 | �) = 1/2.
One can readily verify that µy
(�_i | �) > 0 if and only if �_i 2 Tx
. Thus,
Z 2 Supp(µy
) , µy
(Z�n) > 0 for every n
, µy
(Z�(n+ 1) | Z�n) > 0 for every n
, Z�(n+ 1) 2 Tx
for every n
, Z 2 [Tx
].
42
We have thus established that µy
is supported on a random closed set.
To show (ii), let C ✓ 2! be a random closed set. By no randomness ex nihilo,
there is some Martin-Lof random z 2 3! such that C = [Tz
]. Hence, (�(z)) is a Q-
random measure ⌫. By the definition of �, ⌫ has support [Tz
] = C, which establishes
the claim.
Instead of changing the measure on P (2!) we can also establish a correspondence
between random closed sets and random measures by considering, not the support of
a random measure, but what we refer to as its 1/3-support.
Definition 3.5.2. Let µ 2 P (2!) and set
Tµ
= {� : (8i < |�|) [ µ���(i+ i) | ��i
�> 1/3 ]} [ {✏}.
Then the 1/3-support of the measure µ is the closed set [Tµ
].
Theorem 3.5.3. A closed set C 2 C(2!) is random if and only it is the 1/3-support
of some random measure µ 2 P (2!).
Proof. We define an almost-total, computable, and Lebesgue-measure-preserving map
� : 2! ! 3! that induces a map � : P (2!) ! C(2!) such that �(µ) = [Tµ
]. Suppose
x = �xi
2 2! such that µ(�i
_0 | �i
) = xi
for each i. Then for � 2 Tµ
(which must
exist since ✏ 2 Tµ
),
• if µ(�0) 2 [0, 1/3), then �1 2 Tµ
and �0 /2 Tµ
;
• if µ(�0) 2 (2/3, 1], then �0 2 Tµ
and �1 /2 Tµ
;
• if µ(�0) 2 (1/3, 2/3), then �0 2 Tµ
and �1 2 Tµ
; and
• if µ(�0) = 1/3 or µ(�0) = 2/3, then �(x) is undefined.
Clearly � is defined on a set of measure one, since it is defined on all sequences x
such that xi
6= 1/3 and xi
6= 2/3. Observe that each � 2 Tµ
extends to an infinite
path in [Tµ
]. Thus, if � is the (n+1)-st extendible node in Tµ
, then each of the events
43
• �0 2 Tµ
and �1 /2 Tµ
,
• �0 /2 Tµ
and �1 2 Tµ
, and
• �0 2 Tµ
and �1 2 Tµ
,
occurs with probability 1/3, since each event corresponds to whether µ(�0) 2 [0, 1/3),
µ(�0) 2 (2/3, 1], or µ(�0) 2 (1/3, 2/3), respectively. It thus follows that the pushfor-
ward measure induced by � and � is the Lebesgue measure on 3!. By preservation
of randomness, each random measure µ is mapped to a random closed set, and by
no randomness ex nihilo, each random closed set is the image of a random measure
under �. This establishes the theorem.
3.6 The range of a random continuous function
In [3], it was shown that for each y 2 2!
�({x 2 2! : y 2 ran(Fx
)}) = 3/4.
From this, it follows that every y 2 2! is in the range of some random F 2 F(2!).
In this section, we prove that �(ran(F )) 2 (0, 1) for every random function F . First,
we will prove that �(ran(F )) > 0 for each random function. This implies that no
random function is injective and that the range of a random function is never a
random closed set. These improve two results of [3] according to which (i) not every
random function is injective and (ii) the range of a random function is not necessarily
a random closed set. Our proof requires us to prove some auxiliary facts about the
measure induced by a random function.
To prove that �(ran(F )) < 1 for every F 2 F(2!), we will show that no random
function is surjective, from which the result immediately follows. Our result on
surjectivity also improves a result of [3] according to which not every random function
is surjective.
44
We begin by proving the following, which is similar to Lemma 2.4.2 in Chapter 2
for random measures.
Lemma 3.6.1. Let �F be the natural measure on F(2!). Then the measure PF on
P (2!) induced by the map F 7! � � F�1 has barycenter �; i.e.
�(�) =
Z
P(2!)µ(�) dPF(µ)
for each � 2 2<!.
Proof. By change of variables, it su�ces to show that
2�|�| =
Z
F(2
!)
�(F�1J�K) d�F (3.1)
for each � 2 2<!. Without loss of generality, we assume � = 0n. We proceed then by
induction on n.
Equation (3.1) holds when � = ✏, since each random F is total by Theorem 3.3.13.
Now supposing that equation (3.1) holds for 0n, we show it also holds for 0n+1.
Suppose thatRF(2
!)
�(F�1J0nK) d�F = 2�n. To computeRF(2
!)
�(F�1J0n+1K) d�F , we
note that by symmetryRF(2
!)
�(F�1J0n+1K) d�F = 2 ·RF(2
!)
�(J0K \ F�1J0n+1K) d�F
and we proceed to compute sn+1
:=RF(2
!)
�(J0K \ F�1J0n+1K) d�F .
Recall that any F 2 F(2!) can be viewed as a labeling by 0’s, 1’s, and 2’s of
the nodes of the full binary branching tree (where the root node is unlabeled). We
computeRF(2
!)
�(J0K \ F�1J0n+1K) d�F by considering the three equiprobable cases
for the label of the node 0 for an arbitrary F 2 F(2!). The point is that the label
0 contributes to producing an output beginning with 0n+1, the label 1 rules out the
possibility of producing an output beginning with 0n+1, and the label 2 neither con-
tributes to nor rules out the possibility of producing an output beginning with 0n+1.
45
Case 1: If the node 0 is labeled with a 0, then the measure of all sequences extendingthe node 0 that (after removing 2’s) yield an output extending 0n+1 is equalto the measure of all sequences that yield an output extending 0n times 1/2(the measure determined by the initial label 0), i.e., 1/2 · 2�n.
Case 2: If the node 0 is labeled with a 1, then the measure of all sequences extendingthe node 0 that (after removing 2’s) yield an output extending 0n+1 is equalto 0.
Case 3: If the node 0 is labeled with a 2, then the measure of all sequences extendingthe node 0 that (after removing 2’s) yield an output extending 0n+1 is equalto the measure of all sequences that yield an output extending 0n+1 times1/2 (the measure determined by the initial label 2), i.e., 1/2 · s
n+1
.
Putting this all together gives
sn+1
=1
3· 12· 2�n +
1
3· 0 + 1
3· 12· 2s
n+1
,
which yields sn+1
= 2�n/4, as desired.
Lemma 3.6.2 (Hoyrup [15], relativized). Let Q be a computable measure on P (2!)
with barycenter µ. Then for any z 2 2!,
MLRz
µ
=[
⌫2MLRzQ
MLRz
⌫
.
Theorem 3.6.3. If F 2 F(2!) is random, then �(ran(F )) > 0.
Proof. Fix a random F 2 F(2!). We show that ran(F ) always contains an element
of MLRF
�
. Since ran(F ) is ⇧0,F
1
by Remark 3.3.1, it follows by Proposition 3.3.2 that
�(ran(F )) > 0.
46
By preservation of randomness relative to F , if x 2 MLRF
�
, then F (x) 2 MLRF
��F�1
.
By Lemmas 3.6.1 and 3.6.2, MLRF
��F�1
✓ MLRF
�
, so F (x) 2 MLRF
�
, as desired.
Corollary 3.6.4. If F 2 F(2!) is random, then F is not injective.
Proof. For any y 2 2!, a relativization of Theorem 3.4.4(i) shows that F�1({y}), if
nonempty, is a random closed set relative to y provided that F is random relative to
y. Since ran(F ) has positive Lebesgue measure, there is y 2 ran(F ) that is random
relative to F . Then by Van Lambalgen’s Theorem, F is also random relative to y.
So, F�1({y}) is a nonempty random closed set and, hence, has size continuum by
Theorem 3.3.10. Thus, F is not injective.
Corollary 3.6.5. If F 2 F(2!) is random, then ran(F ) is not a random closed set.
Proof. By Theorem 3.3.9, every random closed set has Lebesgue measure 0. But by
Theorem 3.6.3, the range of a random F 2 F(2!) has positive Lebesgue measure.
This gives the conclusion.
From the proof of Corollary 3.6.4, we can also obtain the following.
Corollary 3.6.6. The measures induced by random functions are atomless.
Proof. Let F 2 F(2!) be random and suppose that z 2 2! is an atom of �F
, i.e.,
�F
({z}) > 0. It follows that z 2 MLRF
�F, since z is not contained in any �
F
-nullsets.
As we argued in the proof of Corollary 3.6.4, F�1({z}) is a nonempty random closed
set and, thus, has Lesbesgue measure zero, by Theorem 3.3.9. This contradicts our
assumption.
We now turn to showing that �(ran(F )) < 1 for every random F 2 F(2!). Instead
of proving this directly, we will first prove the following.
Theorem 3.6.7. If F 2 F(2!) is surjective, then F is not random.
47
To prove Theorem 3.6.7, we provide a careful analysis of the result from [3] stated
at the beginning of this section; namely, that for each y 2 2!,
�({x 2 2! : y 2 ran(Fx
)}) = 3/4.
This result is obtained by showing that the strictly decreasing sequence (qn
)n2! de-
fined by
qn
= �({x 2 2! : ran(Fx
) \ J0nK})
converges to 3/4 and using the fact that
�({x 2 2! : ran(Fx
) \ J0nK}) = �({x 2 2! : ran(Fx
) \ J�K})
for each � 2 2<! of length n. The sequence (qn
)n2! is obtained by using a case
analysis to derive the following recursive formula:
qn+1
=3
2
p1 + 4q
n
� 3
2� q
n
. (3.2)
For details, see [3, Theorem 2.12].
For F 2 F(2!) and � 2 2<!, let us say that F hits J�K if ran(F )\ J�K 6= ;. Thus,
qn
is the probability that a random F 2 F(2!) hits J�K for some fixed � 2 2<! such
that |�| = n. It is worth noting that the function T (J�K) = qn
for each � of length n
induces an e↵ective capacity on C(2!); see [6] for details on e↵ective capacities.
We will proceed by proving a series of lemmas. First, for each n 2 !, let ✏n
satisfy
qn
= 3/4 + ✏n
. Since
(i) qn
> qn+1
for every n, and
(ii) limn!1 q
n
= 3/4,
we know that each ✏n
is non-negative and limn!1 ✏
n
= 0. Moreover, we have the
48
following.
Lemma 3.6.8. For each n � 1,
(a) ✏n+1
1
2
✏n
,
(b) ✏n
2�(n+2),
(c) ✏n+1
� 1
2
✏n
� 2�(2n+5), and
(d) ✏n
� 1
2
n+5�1
.
Proof. First, let n � 1. If we substitute 3/4 + ✏n+1
and 3/4 + ✏n
for qn+1
and qn
,
respectively, into Equation (3.2), we obtain (after simplification)
✏n+1
= 3p1 + ✏
n
� 3� ✏n
. (3.3)
Sincep1 + x 1 + x
2
on [0, 1], from (3.3) we can conclude
✏n+1
3�1 +
✏n
2
�� 3� ✏
n
=1
2✏n
,
thereby establishing (a). To show (b), we proceed by induction. Using the fact from
[3] that q1
=p45�5
2
, it follows by direct calculation that
✏1
=
p45� 5
2� 3
4 2�3.
Next, assuming that ✏n
2�(n+2), it follows from (a) that
✏n+1
1
2✏n
1
22�(n+2) = 2�(n+3).
To show (c), for each fixed n � 1, we use a di↵erent approximation ofp1 + x from
below. By (b), since ✏n
2�(n+2), we use the Taylor series approximation 1 + x
2
of
49
p1 + x on [0, 2�(n+2)], with error term
maxc2[0,2�(n+2)
]
1
4(1 + c)3/2x2
2=
x2
8.
Thus,p1 + x � 1 +
x
2��2�(n+2)
�2
/8 = 1 +x
2� 2�(2n+7)
on [0, 2�(n+2)]. Combining this with Equation (3.3) yields
✏n+1
� 3(1 +✏n
2� 2�(2n+7))� 3� ✏
n
� 1
2✏n
� 2�(2n+5).
Lastly, to prove (d), first observe that
✏1
=
p45� 5
2� 3
4� 2�4 (3.4)
and thus it certainly follows that
✏1
� 1
26 � 1.
Next, using (c), we verify by induction that for n � 2,
✏n
� 1
2n�1
✏1
��2�(n+5) + . . .+ 2�(2n+3)
�. (3.5)
For n = 2, by part (c) we have
✏2
� 1
2✏1
� 2�7.
Supposing that
✏n
� 1
2n�1
✏1
��2�(n+5) + . . .+ 2�(2n+3)
�,
50
again, by part (c), we have
✏n+1
� 1
2✏n
� 2�(2n+5) � 1
2
⇣ 1
2n�1
✏1
��2�(n+5) + . . .+ 2�(2n+3)
�⌘� 2�(2n+5)
=1
2n✏1
��2�(n+6) + . . .+ 2�(2n+4)
�� 2�(2n+5)
=1
2n✏1
��2�(n+6) + . . .+ 2�(2n+5)
�,
which establishes Equation (3.5). Combining Equations (3.4) and (3.5) yields
✏n
� 1
2n�1
2�4 ��2�(n+5) + . . .+ 2�(2n+3)
�.
=1
2(n+3)
� 2�(n+4)
�2�1 + . . .+ 2�(n�1)
�
=1
2(n+3)
� 2�(n+4)(1� 2(n�1))
� 2�(n+3) � 2�(n+4)
� 2�(n+4)
� 1
2n+5 � 1.
Lemma 3.6.9. For n � 1, we have
qn+1
qn
1� 2�(n+6).
Proof. By Lemma 3.6.8(d),
✏n
� 1
2n+5 � 1=
2�(n+5)
1� 2�(n+5)
,
which implies
�1� 2�(n+5)
�✏n
� 2�(n+5) = 4 · 2�(n+7) � 3 · 2�(n+7).
51
Multiplying both sides by 1/2 yields
1
2
�1� 2�(n+5)
�✏n
� 3
42�(n+6).
Expanding the left hand side and using the fact from Lemma 3.6.8(a) that 1
2
✏n
� ✏n+1
,
we have1
2✏n
+⇣14+ . . .+ 2�(n+6)
⌘✏n
� 3
42�(n+6) + ✏
n+1
,
which is equivalent to
(1� 2�(n+6))✏n
+3
4(1� 2�(n+6)) � 3
4+ ✏
n+1
.
This yields the inequality�1� 2�(n+6)
�qn
� qn+1
,
from which the conclusion follows.
Lemma 3.6.10. For n � 1, we have
2⇣q
n+1
qn
⌘� 1
!2
n
132
pe< 1.
Proof. First, it follows from Lemma 3.6.9 that
2⇣q
n+1
qn
⌘� 1 1� 2�(n+5)
and hence 2⇣q
n+1
qn
⌘� 1
!2
n
⇣1� 2�(n+5)
⌘2
n
. (3.6)
Next, it is straightforward to verify by cross-multiplication that
2n+5 � 1
2n+5
2n+6 � 1
2n+6
52
and2n+5 � 1
2n+5
2n+6 � 1
2n+6
!2
,
from which it follows that
2n+5 � 1
2n+5
!2
n
2n+6 � 1
2n+6
!2
n+1
.
Lastly, we have
limn!1
⇣1� 2�(n+5)
⌘2
n
=1
32
pe.
From Equation (3.6) and the fact that the sequence�(1 � 2�(n+5))2
n�n2! is non-
decreasing and converges to 1/ 32
pe, the claim immediately follows.
The proof of following result is essentially the proof of the e↵ective Choquet
Capacity Theorem in [6]. We reproduce the proof here for the sake of completeness.
Lemma 3.6.11. The probability that a random continuous function F hits both J0K
and J1K is 2q1
� 1, and the probability that F hits both J�0K and J�1K for a fixed
� 2 2<! of length n � 1, given that F hits J�K, is equal to 2⇣q
n+1
qn
⌘� 1.
Proof. We write the probability that F hits J�K for some fixed � as P(F 2 H�
). Now
since P(F 2 H0
) = q1
, it follows that P(F 2 H1
\H0
) = 1� q1
(here we use the fact
that every random function is total). By symmetry, we have P(F 2 H0
\H1
) = 1�q1
.
Since F is total with probability one, it follows that
P(F 2 H0
\H1
) = 1��P(F 2 H
0
\H1
) + P(F 2 H1
\H0
)�
and thus
P(F 2 H0
\H1
) = 1� ((1� q1
) + (1� q1
)) = 2q1
� 1.
Next, let � be a string of length n � 1 and let i 2 {0, 1}. Since P(F 2 H�
) = qn
and
53
P(F 2 H�
_i
) = qn+1
, it follows that
P(F 2 H�
_i
| F 2 H�
) =P(F 2 H
�
_i
& F 2 H�
)
P(F 2 H�
)=
P(F 2 H�
_i
)
P(F 2 H�
)=
qn+1
qn
.
Consequently,
P(F 2 H�1
\H�0
| F 2 H�
) = P(F 2 H�0
\H�1
| F 2 H�
) = 1� qn+1
qn
Thus,
P(F 2 H�0
\H�1
| F 2 H�
) = 1��P(F 2 H
�0
\H�1
| F 2 H�
)
+ P(F 2 H�1
\H�0
| F 2 H�
)�
= 1�⇣�
1� qn+1
qn
�+�1� q
n+1
qn
�⌘
= 2⇣q
n+1
qn
⌘� 1.
To complete the proof of Theorem 3.6.7, we now define a Martin-Lof test on F(2!)
that covers all surjective functions. We say that a function F 2 F(2!) is onto up
to level n if F 2 H�
for every � 2 2n. By Lemma 3.6.11, the probability that a
function is onto up to level n is
(2q1
� 1)n�1Y
i=1
2⇣q
i+1
qi
⌘� 1
!2
i
132
pe
!n
.
Thus, if we set
Un
= {F 2 F(2!) : F is onto up to level n},
and
f(n) = min{k : ( 32
pe)�k 2�n},
54
which is clearly computable, then (Uf(n)
)n2! is a Martin-Lof test with the property
that F 2 F(2!) is onto if and only if F 2T
n2! Uf(n)
. This completes the proof.
Corollary 3.6.12. If F 2 F(2!) is random, then �(ran(F )) < 1.
Proof. Suppose �(ran(F )) = 1. Since ran(F ) is closed, it follows that ran(F ) = 2!.
Then F is onto, so it cannot be random.
We also have the following corollary.
Theorem 3.6.13. No measure induced by a random function is a random measure
in the sense of Definition 3.3.14.
Proof. Let F 2 F(2!) be random. Then by Corollary 3.6.12, �(ran(F )) < 1. It
follows that 2! \ ran(F ) is non-empty and open, so J�K ✓ 2! \ ran(F ) for some
� 2 2<!. Thus, �F
(�) = 0. By contrast, for every random measure µ, we have
µ(�) > 0, and the result follows.
55
CHAPTER 4
A NEW LAW OF LARGE NUMBERS EFFECTIVIZATION
4.1 Introduction
To e↵ectivize a classical theorem of mathematics roughly means to make all the
objects mentioned in the theorem (su�ciently) computable and gauge the e↵ectivity
of the conclusion. When the classical theorem comes from probability, the e↵ectivity
of the conclusion can usually be gauged by whether the conclusion holds of algo-
rithmically randoms. For example, Birkho↵’s Ergodic Theorem (a generalization of
the Strong Law of Large Numbers) says that if f is an integrable function on the
probability space (X,µ) and T : X ! X is measure-preserving, then
1
n
X
i<n
f(T i(x)) !Z
f dµ
for µ-almost-every x 2 X. This theorem was e↵ectivized in [5], where it was con-
cluded that if X = 2! and µ, f , and T are all computable, then the convergence
holds on every Martin Lof random.
This chapter contains an e↵ectivization of a theorem of Erdos and Renyi [11],
which says that the maximal average winnings of short subgames of a fair game
converges almost-surely to a certain constant depending on the length of the subgame.
We basically follow their proof, injecting e↵ectivity wherever necessary.
56
4.2 Stirling’s Formula
We recall Stirling’s formula, which is a crucial part of the proof the main theorem
of this chapter.
Theorem 4.2.1 ([20]).
n! = (1 + o(1))p2⇡nnne�n. (4.1)
In other words, there is a function R(n) such that R(n) ! 0 as n ! 1 and
n! = (1 +R(n))p2⇡nnne�n. (4.2)
The next lemma is a consequence of Stirling’s formula and gives an e↵ective bound
for the probability that a Bernoulli-1/2 random variable has at least �n successes
where � is some parameter between 1/2 and 1.
Lemma 4.2.2. For all � 2 (1/2, 1) there are positive reals A and B, uniformly
computable from �, such that for all n
An�1/22n(h(�)�1) 2�n
X
n�kn
✓n
k
◆ Bn�1/22n(h(�)�1), (4.3)
where h(�) := �� log2
� � (1� �) log2
(1� �).
Proof. First we prove the upper bound in (4.3). Letm = dn�e. For everym k n,
✓n
k
◆=
✓n
m
◆k�1Y
i=m
n� i
i+ 1✓n
m
◆✓n�m
m+ 1
◆k�m
.
Note that n�m
m+1
< 1 since n 2m. This allows for a first bound using a geometric
57
series:
2�n
X
mkn
✓n
k
◆ 2�n
✓n
m
◆ X
mkn
✓n�m
m+ 1
◆k�m
< 2�n
✓n
m
◆X
mk
✓n�m
m+ 1
◆k�m
= 2�n
✓n
m
◆m+ 1
2m+ 1� n
Now we use Stirling’s formula to bound�n
m
�from above. Since R(n) ! 0, there is
a natural number R such that R � R(n) for all n. We will use the fact that � m/n,
together with the following claim.
Claim 1. For any � 2 (�, 1), there is an N , computable from � and �, such that
m/n < � whenever n � N .
Proof of claim. Since n� m n� + 1, � m/n � + 1/n. So, with � and � as
oracles, we can compute N such that � + 1/n < � whenever n � N . This proves the
claim.
Fix such a � computable from � (e.g. � = (� + 1)/2) and the corresponding N .
Then for n � N ,
✓n
m
◆=
(1 +R(n))p2⇡nnne�n
(1 +R(m))p2⇡mmme�m(1 +R(n�m))
p2⇡(n�m)(n�m)n�me�(n�m)
=1pn
1p2⇡
1 +R(n)
(1 +R(m))(1 +R(n�m))
1p(m/n)(1�m/n)
2nh(m/n) (4.4)
1pn
1p2⇡
(1 +R)1p
(m/n)(1�m/n)2nh(m/n)
1pn
1p2⇡
(1 +R)1p
(m/n)(1�m/n)2nh(�)
1pn
1p2⇡
(1 +R)1p�
1p(1�m/n)
2nh(�)
1pn
1p2⇡
(1 +R)1p�
1p(1� �)
2nh(�) (4.5)
58
It remains only to bound m+1
2m+1�n
. We again use the fact that � m/n < � for
n � N :
m+ 1
2m+ 1� n=
m/n+ 1/n
2m/n+ 1/n� 1
� + 1
2� � 1.
Thus B := 1p2⇡
(1 + R) 1p�
1p(1��)
�+1
2��1
works, where � = (1 + �)/2 and R � R(n)
for all n. (Actually B only works for n � N , but we can compute B1
, B2
, . . . , BN�1
that work for n < N , and then compute max{B,B1
, B2
, . . . , BN�1
}.)
Now to finding A. This is a bit easier since we’ll actually show that�
n
dn�e
��
An�1/22nh(�) for some positive constant A. We’ll use (4.4), still with m = dn�e, and
the following facts:
• there is N such that 1+R(n)
(1+R(m))(1+R(n�m))
� 1/2 whenever n � N ,
• 1
m/n
� 1
�+1
for n � 1,
• 1
1�m/n
� 1
1��
for all n.
These facts, together with (4.4), yield
✓n
dn�e
◆� 1
2p2⇡(� + 1)(1� �)
n�1/22nh(dn�e/n).
It is su�cient then to show that 2nh(dn�e/n) � C2nh(�) for some positive constant C
computable from �. To do this, we start by noting that 2nh(dn�e/n) � 2nh(�+1/n), since
dn�e/n � + 1/n and h is decreasing on [1/2, 1]. Using the Taylor expansion of h
centered at �, evaluated at � + 1/n, we get
h(� + 1/n) = h(�) +h0(�)
n+
h00(�)
2n2
+ · · · ,
59
so that
nh(� + 1/n) = nh(�) + h0(�) +h00(�)
2n+ · · · .
The tail of that series,h00(�)
2n+ · · ·
goes to 0 as n ! 1; and, moreover, we can use Taylor’s inequality to compute N
after whichh00(�)
2n+ · · · � �1.
Putting this all together, we have
2nh(dn�e/n) � 2nh(�+1/n) = 2nh(�)2h0(�)2h
00(�)/(2n)+··· � 2h
0(�)�12nh(�).
So, in summary, A = 2
h0(�)�1
2
p2⇡(�+1)(1��)
works. (Again, A actually only works for su�-
ciently large n (computable from �), but we can check the first finitely many n’s to
find an A that works for all.)
4.3 Maximal average gains over short subgames of a fair game
LetX1
, X2
, . . . be a sequence of IID random variables taking on the values ±1 with
probability 1/2. Each Xn
represents the winnings in a fair game. Let Sn
=P
in
Xi
and
#(N, k) = max0nN�k
Sn+k
� Sn
k= max
0nN�k
Xn+1
+Xn+2
+ · · ·+Xn+k
k
So #(N, k) represents the maximal average gain over length-k subgames of the length-
N game.
For those who prefer the non-probabilistic point-of-view, we are working in the
Cantor space {�1, 1}! with the Lebesgue (fair coin) measure. Sn
and #(N, k) are
60
then both functions from {�1, 1}! to R; thus we will write Sn
(x) and #(N, k)(x),
where x 2 {�1, 1}!.
Lemma 4.3.1. Let c � 1 and define ↵ 2 (0, 1] via 1/c = 1� h�1+↵
2
�. Then for any
✏ > 0, with ↵0 = ↵+ ✏, there are positive constants B and �, depending only on and
uniformly computable from c and ✏, such that
P (#(N, bc log2
Nc) � ↵0) BN��,
where P (. . .) denotes the probability of the event in parenthesis.
Proof. We begin by noticing that #(N, bc log2
Nc) � ↵0 if and only ifSn+bc log
2
Nc�Sn
bc log2
Nc �
↵0 for some n 2 {0, 1, . . . , N � bc log2
Nc}. When there are exactly d 1’s among
Xn+1
, Xn+2
, . . . , Xn+bc log
2
Nc,
Sn+bc log
2
Nc � Sn
bc log2
Nc =d� (bc log
2
Nc � d)
bc log2
Nc .
This is � ↵0 if and only if d � bc log2
Nc↵
0+1
2
. The probability that d � bc log2
Nc↵
0+1
2
can then be bounded above using (4.3) with � = ↵
0+1
2
:
P (#(N, bc log2
Nc) � ↵0) = P
0
@[
0nN�bc log2
Nc
Sn+bc log
2
Nc � Sn
bc log2
Nc � ↵0
1
A
X
0nN�bc log2
Nc
P
✓Sn+bc log
2
Nc � Sn
bc log2
Nc � ↵0◆
X
0nN�bc log2
Nc
Bbc log2
Nc�1/22bc log2 Nc(h(�)�1)
NBbc log2
Nc�1/22bc log2 Nc(h(�)�1)
NB2bc log2 Nc(h(�)�1)
NB2(c log2 N)(h(�)�1)
= BN c(h(�)�1)+1
61
Since h is decreasing on [1/2, 1], h((1+↵+ ✏)/2) < h((1+↵)/2), which is equivalent
to c(h(�) � 1) + 1 < 0, so � = �(c(h(�) � 1) + 1) and the B from Lemma 4.2.2
work.
In Chapters 2 and 3, we primarily used the Martin Lof test definition of Martin
Lof randomness. Here, however, we use the Solovay test definition. Recall also from
those chapters that S ✓ {�1, 1}! is e↵ectively open if it is a c.e. union of cylinder
sets. Also, the preimage of an e↵ectively open set via a computable map is e↵ectively
open.
Definition 4.3.2 ([10]). A sequence x 2 {�1, 1}N is Martin Lof random relative
to c if x is not in infinitely many Ai
’s when hAi
ii2N is a sequence of subsets of
{�1, 1}N, uniformly e↵ectively open relative to c, such thatP
P (Ai
) < 1.
Lemma 4.3.3. Let c � 1 be Martin Lof random (relative to ;) and define ↵ 2 (0, 1]
via 1/c = 1� h�1+↵
2
�. Then
lim supN!1
#(N, bc log2
Nc)(x) ↵
for every x 2 {�1, 1}N that is Martin Lof random relative to c.
Proof. Let ✏ > 0 be a (small) computable number, and set ↵0 = ↵+ ✏. It is straight-
forward to show that bc log2
(2(j+1)/c � 1)c = j, so by Lemma 4.3.1, the series
1X
j=1
P (#(2(j+1)/c � 1, j) > ↵0)
converges.
The hypothesis that c is random means that the random variables #(2(j+1)/c�1, j)
are uniformly computable relative to c (note that if c were not random, determining
b2(j+1)/c � 1c could be noncomputable relative to c, since the floor function is not
computable). Thus, the events {#(2(j+1)/c � 1, j) > ↵0} are e↵ectively open relative
62
to c. It follows that for every x that is Martin Lof random relative to c, the inequality
#(2(j+1)/c � 1, j)(x) ↵0 holds for all but finitely many j.
Since bc log2
Nc = j for 2j/c N 2(j+1)/c�1, the random variables #(N, bc log2
Nc)
and #(2(j+1)/c� 1, j) are looking at the same length of windows to take the max, and
since the latter has more windows,
#(N, bc log2
Nc) #(2(j+1)/c � 1, j)
when 2j/c N 2(j+1)/c � 1. Thus, we know now that for any x that is Martin Lof
random relative to c,
#(N, bc log2
Nc)(x) ↵0
for all but finitely many N . Since ✏ is arbitrary, the proof is complete.
Lemma 4.3.4. Let c � 1 be Martin Lof random (relative to ;) and define ↵ 2 (0, 1]
via 1/c = 1� h�1+↵
2
�. Then
lim infN!1
#(N, bc log2
Nc)(x) � ↵
for every x 2 {�1, 1}N that is Martin Lof random relative to c.
Proof. Let 0 < ✏ < ↵ be computable and set ↵00 = ↵� ✏.
If #(N, k) ↵00, thenS
(r+1)k�Srk
k
↵00 for each 0 r n/k � 1. The random
63
variables S(r+1)k
� Srk
are IID for di↵erent r’s, so
P (#(N, bc log2
Nc) ↵00) P
✓S(r+1)bc log
2
Nc � Srbc log
2
Nc
bc log2
Nc ↵00,
0 r N
bc log2
Nc � 1
◆
= P
✓Sbc log
2
Nc
bc log2
Nc ↵00◆bN/bc log
2
Nc�1c+1
= P
✓Sbc log
2
Nc
bc log2
Nc ↵00◆bN/bc log
2
Ncc
P
✓Sbc log
2
Nc
bc log2
Nc ↵00◆
N/bc log2
Nc�1
If there are d many 1’s among X1
, . . . , Xbc log2
Nc, then
↵00 �Sbc log
2
Nc
bc log2
Nc =2d� bc log
2
Ncbc log
2
Nc () d bc log2
Nc1 + ↵00
2.
Now, we use the lower bound in (4.3) to estimate P�d bc log
2
Nc1+↵
00
2
�, with
� := 1+↵
00
2
.
P
✓d bc log
2
Nc1 + ↵00
2
◆= 1� P
✓d > bc log
2
Nc1 + ↵00
2
◆
= 1� P
✓d � bc log
2
Nc1 + ↵00
2
◆
1� Abc log2
Nc�1/22bc log2 Nc(h(�)�1)
1� Abc log2
Nc�1/22(c log2 N�1)(h(�)�1)
= 1� Abc log2
Nc�1/22(c log2 N)(h(�)�1)21�h(�)
= 1� Abc log2
Nc�1/2N c(h(�)�1)21�h(�)
1� Abc log2
Nc�1/2N c(h(�)�1)
Because the function h is decreasing on [1/2, 1], h(�) > h((1+↵)/2), so c(h(�)�1) >
64
c(h((1 + ↵)/2)� 1) = �1, say c(h(�)� 1) = �1 + �. Thus,
1� Abc log2
Nc�1/2N c(h(�)�1) = 1� Abc log2
Nc�1/2N�1+�
1� AN�1+�/2.
The last inequality holds because bc log2
Nc N � for su�ciently large N , so that
bc log2
Nc�1/2 � N��/2.
Putting these inequalities together, with �1
:= �/2, and then using the inequality
1� x e�x, we get
P (#(N, bc log2
Nc) ↵00) ✓1� AN �
1
N
◆N/bc log
2
Nc�1
⇣e�
AN�1
N
⌘N/bc log
2
Nc�1
= e�AN�
1
bc log
2
Nc�1
e�AN�
1
c log
2
N
= e�AN�
1
c log
2
N
e�N
�1
/2(eventually)
N�2 (eventually).
Thus,1X
N=1
P (#(N, bc log2
Nc) < ↵00)
converges.
Because c is random, the random variables #(N, bc log2
Nc) are uniformly com-
putable relative to c (note that if c weren’t random, the floor function could be
problematic), so the sets {x 2 {�1, 1}! : #(N, bc log2
Nc)(x) < ↵00} are uniformly
e↵ectively open. Thus, if x is Martin Lof random relative to c, it is in only finitely
many of those sets; i.e., there is M such that #(N, bc log2
Nc)(x) � ↵00 = ↵ � ✏
65
whenever N � M . Since ✏ is arbitrary, we are done.
Putting Lemmas 4.3.3 and 4.3.4 together,we have so far shown:
Lemma 4.3.5. For any Martin Lof random c � 1,
#(N, bc log2
Nc)(x) ! ↵
for every x that is Martin Lof random relative to c.
We strengthen this now by showing that in fact the convergence holds for any
c � 1 and any Martin Lof random x, even when x is not random relative to c.
Theorem 4.3.6. Let c � 1, and define ↵ = ↵(c) 2 (0, 1] via 1/c = 1�h�1+↵
2
�. Then
limN!1
#(N, bc log2
Nc)(x) = ↵
for every x 2 {�1, 1}N that is Martin Lof random (relative to ;).
Proof of Theorem 4.3.6. Fix c � 1, a Martin Lof random x 2 {�1, 1}!, and let
✏ > 0. By Van Lambalgen’s Theorem, Lemma 4.3.5, and the continuity of ↵(c), there
is c1
� c that is Martin Lof random relative to x, and there is M1
2 ! such that
#(N, bc1
log2
Nc)(x) > ↵(c) � ✏ whenever N � M1
. Further, we assume M1
is large
enough to guarantee that for every integer n � bc1
log2
M1
c, there is N � M1
such
that bc1
log2
Nc = n.
Let M be such that bc log2
Mc � bc1
log2
M1
c and let N � M . By the hy-
pothesis on M1
, there is N1
2 [M1
, N ] such that bc log2
Nc = bc1
log2
N1
c. Then
#(N, bc log2
Nc) � #(N1
, bc1
log2
N1
c) > ↵(c)� ✏. Since ✏ is arbitrary, we have that
lim infN!1
#(N, bc log2
Nc)(x) � ↵(c).
66
For the lim sup direction, let c2
c be random relative to x and M2
2 ! be such
that #(N, bc2
log2
Nc)(x) < ↵(c)+✏ whenever N � M1
. Again we assume M2
is large
enough to guarantee that for every integer n � bc2
log2
M2
c, there is N � M2
such
that bc2
log2
Nc = n.
If N � M2
, then bc log2
Nc = [c2
log2
K] for some K � N . But then
#(N, bc log2
Nc) #(K, bc2
log2
Kc) < ↵(c) + ✏.
Since ✏ is arbitrary, we have that
lim supN!1
#(N, bc log2
Nc)(x) ↵(c),
and the proof is now complete.
We do not know if this Theorem 4.3.6 holds for other notions of randomness. It
would be interesting to know, for example, if the theorem is satisfied by all Kurtz
randoms (recall from Chapter 2 that a real is Kurtz random if it avoids all null e↵ec-
tively closed sets). It would be even more interesting to know whether Theorem 4.3.6
characterizes a certain notion of randomness, so that the theorem holds on a real x
if and only if x is that type of random.
67
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