1

Arnold W Miller

University of Wisconsin

Madison WI

millermathwiscedu

2

Set theoretic prop erties of Lo eb measure

Abstract

In this pap er we ask the question to what extent do basic set

theoretic prop erties of Lo eb measure dep end on the nonstandard

universe and on prop erties of the mo del of in which

it lies We show that assuming Martins axiom and saturation

the smallest cover by Lo eb measure zero sets must have cardinal

ity less than In contrast to this we show that the additivity

of Lo eb measure cannot b e greater than Dene cof H as

1

the smallest cardinality of a family of Lo eb measure zero sets

which cover every other Lo eb measure zero set We show that

H

car dblog H c cof H car d where car d is the external

2

cardinality We answer a question of Paris and Mills concerning

cuts in nonstandard mo dels of theory We also present a

pair of nonstandard universes M and N and hypernite

H

H M such that H is not enlarged by N contains new el

ements but every new subset of H has Lo eb measure zero We

show that it is consistent that there exists a Sierpi nskiset in the

reals but no Lo ebSierpi nskiset in any nonstandard universe We

also show that is consistent with the failure of the continuum hy

p othesis that Lo ebSierpi nskisets can exist in some nonstandard

universes and even in an ultrap ower of a standard universe

Let H b e a hypernite set in an saturated nonstandard universe Let

1

b e the counting measure on H ie for any internal subset A of H let A

jAj

b e the nonstandard rational where jAj is the internal cardinality of A a

jH j

Lo eb showed that the standard part of has a nat

ural extension to a countably additive measure on the algebra generated

1

Partially supp orted by NSF grant AMS classication H A A

2

in Journal of Symbolic Logic

by the internal subsets of H For more background on nonstandard measure

theory and its applications see the survey article Cutland Refer

ences on general prop erties of Lo eb measure and the algebra generated by

the internal sets include Henson and Keisler et al

Some prop erties of Lo eb measure are the following For every Lo eb mea

surable set X there exists an A H such that X A A X

has Lo eb measure zero Also every Lo eb measure zero set can b e covered by

one of the form A where each A is an internal set of measure less than

n n n

1

It is related to Leb esgue measure on the unit interval Identify H

n+1

1

Let st T with the time line T fnt n jH jg where t

jH j

b e the standard part map Then for any Leb esgue measurable Z the

1

set st Z is Lo eb measurable with the same measure as Z

Theorem KeislerLeth If F is a family of internal Loeb mea

sure zero subsets of an innite hypernite set H the nonstandard universe

is saturated and F has external cardinality less than then there exists

an internal set A H of Loeb measure zero which covers every element of

F

pro ofConsider the sentences A

jK j

fB A B F g f n g

jH j n

where A is a variable Clearly this set of sentences is nitely satisable and

has cardinality the same as F It follows by saturation that some internal

A satises them all simultaneously

2

Note that this implies that in a saturated universe any external X H

of cardinality less that has Lo eb measure zero

Theorem Suppose the universe is saturated and every set of reals of

1

cardinality has Lebesgue measure zero then for any innite hypernite

set H and X H of an external cardinality X has Loeb measure zero

pro ofThe standard part map shows this

2

In this case it may b e imp ossible to cover X with an internal set of Lo eb

measure zero

Theorem For any M an saturated universe and H an innite hy

1

pernite set in M there exists any elementary saturated extension N of

1

M which has the property that there exists a family fA H g of

1

internal Loeb measure zero subsets of N such that for every internal Loeb

measure zero B H in N there exists with B A

1

pro ofBuild an elementary chain of saturated universes M for

1 1

Use the pro of of Theorem to get A M such that every Lo eb measure

+1

zero set B M is covered by A

2

By using a simple diagonal argument in a universe given by the last

theorem there will b e an X H of cardinality which is not covered by

1

any internal Lo eb measure zero set In fact the set X will have the prop erty

that for every internal Lo eb measure zero A H X A is countable We

say that X H is a Lo ebSierpi nski set i it is uncountable and meets

every Lo eb measure zero set in a Thus what we have here

is a weak kind of Lo ebSierpi nski set If MACH is true then every set

of reals of cardinality has measure zero So by Theorem this weak

1

Lo ebSierpi nski set would not b e a Lo ebSierpi nski set

Now we ask for the smallest cardinality of a cover of H by Lo eb measure

1

zero sets Note that since monads st fpg for some p have measure

zero H can always b e covered by continuum many Lo eb measure zero sets

MA stands for the version of Martins Axiom which says that for any par

tially ordered set P which has the countable chain condition and any family

D of dense subsets of P of cardinality less than there exists a Plter G

which meets all the dense sets in D

Theorem Suppose MA and the nonstandard universe is saturated

then H cannot be covered by fewer than sets of Loeb measure zero

pro ofLet P b e the partial order of all internal subsets of H of p ositive Lo eb

measure ordered by inclusion where stronger conditions are smaller Forc

ing with P is the same as using the measure algebra formed by taking the

algebra generated by the internal subsets of H and dividing out by the

Lo eb measure zero sets It has the countable chain condition b ecause it is

1

imp ossible to nd n sets of measure greater than which have pairwise

n

intersections of measure zero

T

For let f K g b e a family of many Lo eb measure

n

n

zero sets where each K is an internal subset of H of Lo eb measure less than

n

1

For each dene a dense D P by

n+1

D fb P n b K g

n

Using MA let G b e a Plter meeting every D for For each let

n

G Consider the family of sentences x n b e such that H K

n

g fx H K

Since G is a Plter this family of sentences is nitely satisable Hence by

saturation some internal x H satises them all But this shows that the

family of Lo eb measure zero sets did not cover H

2

This partial order was also used in Kaufmann and Schmerl

Next we consider the additivity of Lo eb measure Here we show that it is

always as small as p ossible

Theorem For any innite hypernite H in an saturated universe

1

there exists a family of Loeb measure zero sets whose union does not have

1

measure zero

K

pro ofWithout loss of generality we may assume H for some internal

hyperinteger K since H may b e replaced by its internal cardinality and for

K K +1 K

some hyperinteger K we have H and has Lo eb measure

at least in H Supp ose K where K is the innite countable

subsets of K and let f n g For each n dene

n

K

H fh h h h g

n 0 1 n1

1

Then each H is an internal set of measure

n

n

2

T

Lemma If A is an internal set and H A then there exists

n

n

n such that H A

n

This follows from the fact that the H are a descending and the

n

universe is saturated

1

2

Let K for b e a family of disjoint countable subsets of K

1

Let H b e dened as H was but using instead of Dene A H

n n

n n

So each A has Lo eb measure zero We show that the union A cannot

1

have Lo eb measure zero Supp ose A is an internal set and

A A

1

By the Lemma there exists n and f m g such that

m 1

m m

for every m H A Let K H Note that each K for

m m

n n

1

m has Lo eb measure and they are indep endent ie for any

n

2

m m m the Lo eb measure of

1 2 k

K K K

m m m

1 2

k

k

is Consequently K has Lo eb measure one since the measure of

m m

N N

H K is and as N Consequently the

mN m

Lo eb measure of A is one and the theorem is proved

2

The last cardinal asso ciated with Lo eb measure we will consider is the

conality cof H This is the smallest cardinality of a family F of Lo eb

measure zero sets such every Lo eb measure zero set is covered by a member

of the family F

Theorem Let car d be the external cardinality and H an innite

hyperinteger in some saturated universe then

1

H

car dblog H c cof H car d

2

pro ofThe rst inequality is proved by a similar argument to the pro of of the

K K +1

last theorem Let the hypernite integer K blog H c so H

2

Let f car dK g b e disjoint countable subsets of K and dene

A H as ab ove The argument ab ove shows that no Lo eb measure zero

set covers uncountably many of the A hence car dK cof H

The second inequality follows from the fact that every Lo eb measure zero

set is covered by a Lo eb measure zero set in the algebra generated by the

internal subsets of H and the following result of Shelah if L is

an innite hypernite set in an saturated universe and car dL then

1

H

take L the number of internal subsets of H

2

In Keisler it is shown under GCH that for any set of innite

successor cardinals C there exists a nonstandard universe M in which

C fcar dH H is a hypernite set in M g

For nite C the nonstandard universe M can b e an ultrap ower of a standard

universe

In Shelah it is shown that for any countable theory T with

distinguished unary predicates Q and P if for every n T has a mo del M

n M n M M

such that n jQ j jP j then T has a mo del M where jQ j

M

and jP j Taking T to b e any theory containing arithmetic and Q

2

2 n

to b e n and P to b e we see that there is a nonstandard universe with

H

a hyperinteger H where car dH and car d If follows from

Changs two cardinal theorem and its pro of that assuming the continuum

hypothesis there is an saturated nonstandard universe with a hyperinteger

1

H

H where car dH and car d

1 2

This result was also proved by Paris and Mills using a metho d

similar to the MacDowellSpecker theorem For any innite cardinal and

nonstandard universe M let

M

fH car dH and H an integer of Mg I

2

Since H has the same internal cardinality as the cartesian pro duct H H

M

which has the same external cardinality as H for innite H it is clear that I

must b e closed under multiplication Paris and Mills show this is sucient

by showing If M is any countable nonstandard universe and I is a prop er

initial segment of the of M closed under multiplication then there

N

exists an elementary extension N of M in which I I and every integer H

of N not in I has cardinality The following theorem answers a question

raised by Paris and Mills concerning cuts in nonstandard mo dels of number

theory

Theorem Assume the continuum hypothesis Then there exists an

1

saturated universe N with a hyperinteger H which has external cardinality

n

but every integer greater than H for al l n has cardinality

1 2

pro ofLet M b e any saturated universe of cardinality and let H b e any

1 1

innite hyperinteger of M Let

n

I fK a hyperinteger of M n K H g

and let J b e the complement of I with resp ect to the of M

We construct a sequence X of elements of M such each X is

1

a hypernite set of hyperintegers of M such that

the internal cardinality of X is an element of J

if then X X

for any set X in M there exists an such that either X X or X

is disjoint from X

for any and function f in M whose domain includes X there exists

a such that f  X is either onetoone or constant

for any d J some X has internal cardinality less than d

Note that since M is saturated and the conality of I is the coinitiality

1

of J is ie for any countable set B J there exists c J such that for

1

every b B we have c b To obtain X for a ordinal rst

1

nd b J such that for every the internal cardinality of X is greater

than b Now use saturation to nd X of internal cardinality greater than

1

b and for all X X To make functions onetoone or constant use

the following lemma

Lemma Suppose f X Y is an onto function in M such that the

internal cardinality of X is in J then there exists an internal set X X

with internal cardinality in J and f  X is either onetoone or constant

S

1 1

pro ofNote that X f y If some f y has internal cardinality

y Y

1

b J then let X f y and hence f  X is constant Otherwise since

internally f is a nite function and the maximum of a hypernite set of

integers is always achieved there exists c I such that for every y Y

1

the internal cardinality of f y is less than c Since I is closed under

multiplication it follows that the internal cardinality of Y is in J In M

choose a set X X such that f maps X onetoone onto Y and hence the

internal cardinality of X is in J

2

This ends the construction of the sequence X Let c b e a

1

new constant symbol and let T b e the theory which consists of the elementary

diagram of M plus all statements of the form c X for Let M

1 1

b e any mo del of T which is an elementary sup erstructure of M and let M

0

M

1

b e the set of all f c such that f M is a function whose domain contains

some X

Claim For any formula x x x and f hf f f i a se

1 2 n 1 2 n

quence of functions from M

M j f c i b X M j f b

1 1

This is proved just like Loss theorem

Claim M is an elementary substructure of M

0

We have that M M b ecause of the constant functions in M So it is

0

enough to note that M is an elementary substructure of M This follows

0 1

from the TarskiVaught criterion Supp ose

M j x x f c f c

1 1 n

then there must b e some X contained in the domain of each f such that

i

M j b X x x f b f b

1 n

In M nd g with domain X such that

M j b X g b f b f b

1 n

and hence

M j g c f c f c

1 1 n

It follows that M is an elementary substructure of M

0

Claim If a I and b M with b a then b M ie the initial

0

segment of the hyperintegers of M determined by I is not enlarged

0

To see this let f c b for some f M and let X b e contained in the

domain of f have the prop erty that for every x X we have f x a

By our construction we may assume that f  X is onetoone or constant

However it cannot b e onetoone since the internal cardinality of X is in J

and a is in I so it must b e constant Hence this constant must b e in M and

therefore in I

Claim for every d J there exists b M M with b d ie there

0

are arbitrarily small new elements of J

Some X has internal cardinality less than d Let f M b e a onetoone

M M

0 0

function from X into d Clearly f c d f c must b e new b ecause

1

for any b M it must b e true that some X X f b It is also true

that no new element of M is b eneath every element of J

0

Claim M is an saturated mo del of cardinality

0 1 1

Clearly M has cardinality so it is enough to check that it is

0 1 1

saturated So let x f x f c n g b e a nitely realizable

n n

type in M We may assume parameters are singletons since ntuples are

0

elements of our universe Let b e suciently large so that X is a subset

of the domain of f for every n Since x is nitely realizable we can

n

also choose large enough so that for every n

M j b X x x f b

m m

mn

Consider the following type g with variable g over M g contains the

sentence g is a function with domain X and for each n the sentence

b X g b f b Since the type g is nitely realizable in M and

m n

M is saturated some function g in M realizes it and so g c realizes x

1

in M Kotlarski shows that every simple conal extension of an

0

saturated mo del is saturated

1 1

Now we prove Theorem This is proved similarly to the standard pro of

of Changs two cardinal Theorem see Chang and Keisler Theorem

p If M is an elementary substructure of N let

N

I fb a hyperinteger of N a I b ag

Construct an elementary chain of mo dels N for such that N M

2 0

N

each N is isomorphic M and I I

For successor steps to obtain N just use the pair M M So the only

+1 0

thing to do is the step for a limit ordinal Letting N b e the union of N

N

for and I b e the initial segment of N determined by I we need to

N

see that N can b e embedded into M in such a way that I is mapp ed onto

I This is done by showing that N while not necessarily saturated is

1

N

saturated for types realizable in I

1

Claim Any countable nitely realizable type x over N which con

N

tains for some b I the formula x b is realized in N

Let x f x a n g Cho ose a sequence hb n i from N

n n n

N

such that for each n b b and for m n N j b a Since I

n m n m

N

0

I and N is saturated there exists an internal sequence hb n K i for

0 1 n

an innite hyperinteger K in N which is an extension of hb n i Since

0 n

N is substructure of N we have that hb n K i N Working in N for

0 n

each n let K b e the least m K if any such that N j b a

n n m n

N N

0

By construction each K is an innite integer of N Since the I I the

n

coinitiality of the nonstandard integers of N must b e and so there exists

1

some hypernite L in N less than all K for n It follows that b realizes

n L

x

N is now obtained by a backandforth argument starting with taking

n

H to itself where I fb n b H g This concludes the pro of of

Theorem

2

H

I do not know what either cof H or cof are in this mo del This

argument needs only that the coinitiality of J is I do not know how to

1

do it if the coinitiality of J is as for example in the pro of of Theorem

+

The result easily generalizes to in place of if

1 2

This technique can also b e used to prove the following theorem

Theorem Let M be a countable nonstandard universe and H an in

nite hyperinteger in M Then there exists an elementary extension N of M

such that there exists new subsets of H ie X H with X N M

jX j

but every new X H has nonzero Loeb measure ie if X N and is

H

innitesimal then X M ie the internal Loeb measure zero sets are the

same in M and N

H

n

pro ofLet J fK a hyperinteger of M n K g and let I b e

the complement of J in the hyperintegers of M Similar to the last pro of

construct an sequence hX n i of hypernite sets of integers in M

n

such that

the internal cardinality of each X is an element of J

n

if n m then X X

m n

for any set X in M there exists an n such that either X X or X

n n

is disjoint from X

for any n and function f in M whose domain includes X there exists

n

an m n such that f  X is either onetoone or constant

m

for any d J some X has internal cardinality less than d of course

n

H

this will automatically b e true if we take X

0

Let c b e a new constant symbol and let T b e the theory which consists of

the elementary diagram of M plus all statements of the form c X for

n

n Let M b e any mo del of T which is an elementary sup erstructure of

1

M

1

M and let N b e the set of all f c such that f M is a function whose

domain contains some X for n The following set of claims also go

n

thru

Claim For any formula x x x and f hf f f i a se

1 2 n 1 2 n

quence of functions from M

M j f c i n b X M j f b

1 n

Claim M is an elementary substructure of N

Claim If a I and b N with b a then b M ie the initial

segment of N determined by I is not enlarged

Claim for every d J there exists b N M with b d ie there are

arbitrarily small new elements of J

jX j

It remains only to show that if X H in N has the prop erty that

H

is innitesimal then X M To do that we need the following claim from

nonstandard

K

Claim Supp ose K H are innite hyperintegers and then

H

1

H

H

K

K

th

ie if is innitesimal then the H ro ot of the number of subsets of H of

H

size K is innitesimally close to

We will use Stirlings approximation for n

p

n

n

n n

n

e

where if n is innite Since H K and H K are all innite we get

n

that

1

H

2

H H H

H

x

K H K

K

H K K K H K K H K

1

p

where x For any p ositive nite but not innitesimal x

2

1

H

Consequently we need only show we have that x

1

1

H

H

2

H H

A

K H K

K H K K H K

K 1

Calculating this let so that K H and

H H

1

1

H

H

2

H H

A

K H K

K H K K H K

1

(2H )

H H

KH 1KH

K H K K H K

1

(2H )

H H

(1)

H H H H H H

1(2H )

(1)

H

12

1H

(1) 1H 1H

H

(1) 1H

Since we have and Using LHopitols

rule it easy to check that for any p ositive innitesimal that hence

1 1

1H

and since we have that

H H

1H 1H

H

1H

and so This proves the Claim

Now we prove the theorem Supp ose X H is in N but not M and

M

1

supp ose X f c where f is a function from M whose domain includes

X Clearly f cannot b e constant so we may assume it is a onetoone map

n

from X to the internal set of all subsets of H Working in M dene the

n

function g X H by letting g x b e the internal cardinality of f x

n

As H is in I the function g cannot b e made onetoone and hence for some

m n g  X is constantly equal to some hyperinteger K Since the internal

m

K

is not cardinality of X is K and X is new clearly K must b e innite If

H

K

innitesimal then we are done So assume that By our last claim

H

we have that for any n

1

H

1

H

n

K

so that for any n

H

H

n

K

H

It follows that is in I This contradicts the fact that the range of f

K

on X must have internal cardinality the same as X an element of J

m m

2

A set of real X is a Sierpi nskiset i it is uncountable but meets

every measure zero set in a countable set KeislerLeth have in

tro duced the analogous notion of a Lo ebSierpi nski set A set X H is a

Lo ebSierpi nski set i it is uncountable but meets every Lo eb measure zero

set in a countable set

Keisler and Leth have proved that in an saturated universe in which

1

H

the external cardinality of is there exists a Lo ebSierpi nskiset X H

1

Such a universe can exist i the continuum hypothesis holds They also note

that the standard part map takes a Lo ebSierpi nski set to a Sierpi nski set

in the reals Note that by Theorem if there is a Lo ebSierpi nski set in a

nonstandard universe that universe cannot b e saturated

2

For any cardinal let b e the pro duct measure on This measure is

determined by for any F and s F

fx s xg

jF j

We dene X is a Sierpi nskiset i X is uncountable but meets every

measure zero set in a countable set We b egin by establishing a relation

b etween Lo ebSierpi nski sets and Sierpi nskisets in

Theorem If there exists a LoebSierpinski set X H in some

1

saturated universe and innite cardinal car ddlog H e where card is

2

the external cardinality then there exists an Sierpi nski set in

K 1

pro ofLet K dlog H e so that K is a hyperinteger satisfying

2

K K

H Since H has measure at least half in a Lo ebSierpi nski set

K

in H is also Lo ebSierpi nski set in Without loss of generality we may

K

assume H Let fx g K b e distinct and dene H

K

by h hx where we identify with internal maps from K into

f g Note that for any and i or

1 K

fx x ig fh hx ig

which is an internal set of Lo eb measure

1

Claim If Y has measure zero then Y has Lo eb measure

zero

T S

Supp ose Y Y where Y t each t X is a nite

n n n n

n tX

n

jtj

partial function from to t fx t xg with t

1 1

each X countable and t n But then Y Y

n tX n

n

S

1 1 1

t Since each t is an internal set with Lo eb and Y

n

tX

n

1

measure t it follows that the Lo eb measure of Y n This

n

proves the claim

Hence if X H is a Lo ebSierpi nskiset then X is a Sierpi nski

set

2

1

Theorem Suppose car ddlog H e then there cannot be a

2

LoebSierpinski set in H

+

1

pro ofIt suces to show there cannot b e a Sierpi nski set for

Supp ose for contradiction that X fx g is a Sierpi nskiset

1

+

1 1

g b e dened by y x Since and let fy

1

such that for all A we have y y there exists A and y

1

Cho ose B and i f g such that y  B is constantly i It follows

1

that for all A that x  B is constantly equal to i however the set

fx x  B ig

has measure zero

2

Theorem It is consistent with ZFC that there exists a Sierpi nski set

in but no LoebSierpinski set in any saturated nonstandard universe

1

It suces to nd a mo del of set theory which contains a Sierpi nski set

but do es not contain a Sierpi nski set where c is the cardinality of the

c

continuum The mo del is obtained as follows Let M b e a countable standard

M

mo del of ZFCGCH and let and let B b e the measure algebra on

ie the complete b o olean algebra of Borel subsets of mo dulo the

measure zero sets Let G b e B generic over M Then

c

M G j There is a Sierpi nskiset in but none in

c

We need only show there is no Sierpi nskiset in The argument will b e

similar to one found in Miller We will use the following Lemma

of Kunen from that pap er

Lemma Kunen Suppose B X for i n are measurable sets

i

with B then

i

r X jfi n r B gj n

i

For any sentence in the forcing language let j j b e the b o olean value

of ie j j fb B b j g

M [G]

1

then n g M G such that Lemma Suppose f

n

M

1

domaing and domaing jf g j

1

pro ofWorking in M for each let C b e a clop en set such that

1

jf j C where denotes symmetric dierence Since C

S

F

such that C is clop en there exists F and T t ie

tT

S

a nite union of cylinders over nitely many co ordinates Since

n

n

1

such that for every Y we have F there exists n and Y

n 1

Dene g Y by g i C G Then g M G since C G i

n

C G Also jf g jf j C which has measure less

n

than This proves the Lemma

2

+

+

In M G we have that c Now supp ose fx g

+1 1

c +

1

is a Sierpi nski set in in M G Let fy g b e dened by

y x By the Lemma and the fact that

1

M G j

n n

+ +

there exists Q M in fact one of size and g Y with

M

1

g M G and Y such that Q Y

n 1

jy g j

equivalently

jx g j

Q

Now fx  Q Y g must b e a Sierpi nskiset in Consequently letting

Q f n g by the strong law of large numbers for all but countably

n

many Y

jfi n x gj

i

lim

n

n

Cho ose Y and n so that if

jfi n x gj

i

j b j

n

2

Let B jx g j and note that B so then b

i i i

3

by Kunens Lemma

r jfi n r B gj n

i

Let

n jfi n x g gj nj c r jfi n r B gj

i i

I claim that b c which contradicts the fact that b and

c This is b ecause it is imp ossible that

n or jfi n x gj n jfi n x gj

i i

and

jfi n x gj

i

n

This proves the theorem

2

In a mo del due to Bartoszynski and Iho da there exists a subset

of of cardinality which is a Sierpi nskiset but also every Sierpi nskiset

1

in has cardinality This do es not mean that there is no Sierpi nskiset

1

c

in To construct their mo del start with a mo del M of ZFCMACH

1

then add random reals ie force with the measure algebra of In

1

2 1

this mo del there is a Sierpi nski set in To see this let D for

1

b e almost disjoint sets uncountable sets with pairwise intersection

2

2

countable and dene x for by x GD where

1

th

G is the generic map and D is the element of D It is not

1

2

hard to show that fx g is a Sierpi nski set We do not know

1

of mo del for Theorem where the continuum is

2

Theorem It is relatively consistent with ZFC that the continuum hy

pothesis is false but in some saturated universe there is a LoebSierpinski

1

set

pro ofSupp ose M is a countable standard mo del of ZFCCH And let W

M b e an saturated nonstandard universe and H W some hypernite

1

set Let B b e the b o olean algebra obtained by taking the algebra generated

by the internal subsets of H and dividing out by the Lo eb measure zero sets

Then B has the countable chain condition Let G b e B g eneric over M

Working in M G nd an elementary extension of W say W such that the

type x fx A A Gg is realized in W It is easy to check that

for any hA n i M such that each A H is internal and the Lo eb

n n

measure of A less than n then there exists n with H A G It

n n

T

follows that x A Iterate this forcing times using nite supp ort

n 1

n

at limits and obtain W M and x H W then using the countable

S

W is chain condition in the nal mo del M the universe W

1 1

1

saturated and fx g is a Lo ebSierpi nski set

1 1

2

We say that a nonstandard universe W is an p ower i there exists

an innite ordinal a typical example is and a nonprincipal

ultralter U on such that W is the ultrap ower V U where V is the set

of all sets of rank less than All p owers are saturated

1

Theorem It is relatively consistent with ZFC that continuum hypoth

esis is false and we have an power in which there exists a LoebSierpinski

set

pro ofWe will use the following lemmas

Lemma If M is a countable standard model of ZFC and U M a

nonprincipal ultralter on then there exists a generic extension N of M

which satises the countable chain condition and

N j Z X U Z X

where A B means inclusion modulo nite

pro ofThis is a wellknown forcing fs X s and X U g see

Mathias

2

Supp ose F P is a eld of sets and F is a nitely additive

measure with and n for each n Let P fb F b

g ordered by inclusion

Lemma P has the countable chain condition If G is P generic over a

model M of ZFC then for every ha n i M F with lim a

n n n

there exists n with a G

n

pro ofThe countable chain condition holds b ecause there cannot b e n sets

of measure greater than n and pairwise intersection measure zero The

second sentence is an easy density argument

2

Given Z and h with hn all n Z then dene

X hn and give X the pro duct measure determined as follows

nZ

Let A fx X xn ig for i hn then A hn and for

in in

n n n distinct

1 2 k

A A

i n i n

1 1

k k

hn hn hn

1 2 k

Let B b e the measure algebra determined by X and ie the Borel subsets

of X mo dulo the measure zero sets Then B is isomorphic to the usual

random real forcing and a generic lter G determines and is determined by

a random real r X

Lemma Suppose h and Z are elements of M a countable standard

model of ZFC r X is a random real over M g Z with g M

jg (n)j

g n hn al l n and lim p Then the fol lowing limit exists

n nZ

h(n)

and equals p

jfm n m Z and r m g mgj

p lim

n

jfm n m Z gj

pro ofThis follows from the strong law of large numbers for variable dis

tributions see Feller X page For each n let X

n

b e a random variable which is with probability p and with probabil

n

jg (z )j

n

ity p where p and fz n g is an enumeration of Z

n n n

h(z )

n

The sequence is mutually indep endent and satises Kolmogorovs Criterion

variances are less than so that the strong law of large numbers holds

S m

n n

This implies that with probability the sequence tends to zero where

n

S X X X and m p p Since any random

n n n 0 n1

real r must b e in any measure one set co ded in the ground mo del we must

have that for

a jfm n r k g k gj

n m m

a m

n n

To nish the pro of it is enough to see that that lim

n

n

a

n

lim p

n

n

m

n

This is true b ecause lim p since

n

n

p p p np m

0 1 n1 n

p

n n

p p p p p p p p

0 m m+1 n1

n n

and choosing m large makes the second quotient small and n large drives the

rst down

2

Now supp ose H is an innite hyperinteger in an p ower via a nonprin

cipal ultralter U from M a countable standard mo del of ZFC and h

represents H in this ultrap ower ie h H We combine the last three

U

lemmas

Lemma There exists a countable chain condition generic extension N



of M and an ultralter U N extending U and f such that f

U

 

is not in any Loeb measure zero set coded in M ie and f H h

U U

hg n i M such that g H and such that Loeb measure of g is

n n U n

 

g less than n for each n there exists n with f

n U U

pro ofApply Lemma to obtain Z so that X U we have Z X

Let r X hn b e a random real over M Z Working in M Z r

nZ

dene the eld of sets

F fX g M g and X Z fm r m g mgg

Dene X for any X F by

jX Z nj

X lim

n

jZ nj

By Lemma this limit exists and in fact equals the Lo eb measure of the

g that put X into F Use Lemma to obtain G Pgeneric over M Z r

U

where P fb F b g and let U b e any ultralter extending G

Let f b e any map extending r Since Z we have U U If

hg n i M is the co de for a Lo eb measure zero subset of H then

n

letting X fm f m g mg f r on Z we have that the Lo eb

n n

measure of g is X which go es to zero as n So by Lemma

n U n

there exists n such that X G U This means that

n

fm f m g mg U

n

 

g so f

n U U

2

Now we prove Theorem Start with any M countable standard

0

mo del of ZFCCH nonprincipal ultralter U M on and H h an

0 U

innite hyperinteger with h M Iterate Lemma with nite supp ort

times to obtain hf i and hU i with f h H

1 1 1 U U

and f not in any Lo eb measure zero set co ded in M Since the

+1 U

+1

S

entire iteration has the countable chain condition at the end U U

1

is an ultralter in M and ff g will b e a Lo ebSierpi nski set

1

1

2

I do not know if a Lo ebSierpi nski set can have cardinality

2

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