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\begin{document}

\begin{flushright}
  Arnold W. Miller\footnote{Partially supported by
  NSF grant 8801139, AMS classification 03H15, 04A15, 28A05}\\
  University of Wisconsin\\
  Madison, WI  53706\\
  miller@math.wisc.edu
\end{flushright}


\begin{center}
 {\large Set theoretic properties of Loeb measure}
 \footnote{in Journal
 of Symbolic Logic, 55(1990), 1022-1036.}
\end{center}

\bigskip
\begin{quote}
\centerline{Abstract}
In this paper we ask the question: to what extent do basic set
theoretic properties of Loeb measure depend on the nonstandard universe
and on properties of the model of set theory in which it lies?
We show that assuming Martin's axiom and $\kappa$-saturation the
smallest cover by Loeb measure zero sets must have cardinality less
than $\kappa$.  In contrast to this we show that the additivity of
Loeb measure cannot be greater than $\omega_1$.
Define $cof(H)$ as the smallest cardinality of a family
of Loeb measure zero sets which cover every other
Loeb measure zero set.  We show that
 $card(\lfloor \log_2(H)\rfloor)\leq cof(H)\leq card(2^H)$
where $card$ is the external cardinality.
We answer a question of Paris and Mills concerning cuts in
nonstandard models of number theory.
 We also present a pair of nonstandard universes
$M$ and $N$ and hyperfinite integer $H\in M$ such that
$H$ is not enlarged by $N$, $2^H$  contains new elements, but
every new subset of $H$ has Loeb measure zero.
We show that it is consistent that there exists a
Sierpi\'{n}ski set in the reals but no Loeb-Sierpi\'{n}ski set in any
nonstandard universe.  We also show that is consistent with the failure
of the continuum hypothesis that
Loeb-Sierpi\'{n}ski sets can exist
in some nonstandard universes and even in an ultrapower of a standard
universe.
\end{quote}


Let $H$ be a hyperfinite set in an $\omega_1-$saturated nonstandard universe.
Let $\mu$ be the counting measure
on $H$, i.e. for any internal subset $A$ of $H$ let $\mu(A)$ be the
nonstandard
rational: $|A|\over|H|$ where $|A|$ is the internal cardinality of
$A$, a hyperinteger.
Loeb (1975)\cite{L} showed that the standard part of $\mu$ has
a natural extension to a countably additive
measure on the $\sigma$-algebra generated by the internal
subsets of H.
For more background on nonstandard measure theory and its applications, see
the survey article Cutland (1983)\cite{C}.
References on general properties of Loeb measure and the $\sigma$-algebra
generated by the internal sets include Henson (1979)\cite{H1}\cite{H2}
and Keisler et. al. (1989)\cite{KK}.

Some properties of Loeb measure
are the following. For every Loeb measurable set $X$ there exists an
internal set $A\subseteq H$
such that $(X -  A)\cup (A -  X)$ has
Loeb measure zero.
Also every Loeb measure zero set can be covered by one of the form
$\cap_{n\in\omega}A_{n}$ where each $A_n$ is an internal set
of measure less than $1\over n+1$.
It is related to Lebesgue measure on the unit interval $[0,1]$.
Identify $H$ with the time line $T=\{n\triangle t:n<|H|\}$
where $\triangle t={1\over |H|}$.  Let $st:T\to [0,1]$ be the standard
part map. Then for any Lebesgue measurable $Z\subseteq [0,1]$ the
set $st^{-1}(Z)$ is Loeb measurable with the same measure as $Z$.


\begin{theorem} Keisler-Leth (19..)\cite{KL} \label{KL}
If $F$ is a family of internal Loeb measure
zero subsets
of an infinite hyperfinite set $H$,
the nonstandard universe is
$\kappa-$saturated,  and $F$ has external cardinality less than $\kappa$,
then there exists an internal set $A\subseteq H$ of Loeb measure zero
which covers every element of $F$.
\end{theorem}
\proof
Consider the sentences $\Sigma(A)$:
$$\{B\subseteq A :  B\in F\}\cup\{{|K|\over |H|}<
{1\over n+1} : n\in\omega\}$$
where $A$ is a variable.  Clearly this set of sentences
is finitely satisfiable and has cardinality the same as $F$. It follows
by $\kappa-$saturation that some internal $A$ satisfies them all
simultaneously.
\qed

Note that this implies that in a $\kappa$-saturated universe
any external $X\subseteq H$ of cardinality less that $\kappa$
has Loeb measure zero.

\begin{theorem} \label{lebzero}
   Suppose the universe is $\omega_1$-saturated and every set
   of reals of cardinality $<\kappa$ has Lebesgue measure zero,
   then for any infinite hyperfinite set $H$ and
   $X\subseteq H$ of an
   external cardinality $<\kappa$, $X$ has Loeb measure zero.
\end{theorem}
\proof
The standard part map shows this.
\qed

In this case it
may be impossible to cover $X$ with an internal set of Loeb
measure zero.

\begin{theorem}
For any $M$ an $\omega_1$-saturated universe and $H$ an infinite
hyperfinite set in $M$, there exists any elementary $\omega_1$-saturated
extension $N$ of $M$ which has the property that there exists a
family $\{A_{\alpha}\subseteq H:\alpha<\omega_1\}$ of internal Loeb
measure zero subsets of $N$ such that for every internal Loeb measure zero
$B\subseteq H$ in $N$ there exists $\alpha<\omega_1$ with
$B\subseteq A_{\alpha}$.
\end{theorem}
\proof
Build an elementary chain of $\omega_1$-saturated universes $M_{\alpha}$
for $\alpha<\omega_1$.  Use the proof of Theorem \ref{KL} to get
$A_\alpha\in M_{\alpha+1}$ such that every
Loeb measure zero set $B\in M_{\alpha}$ is covered
by $A_\alpha$.
\qed

By using a simple diagonal argument in a
universe given by the last theorem there will be an $X\subseteq H$
of cardinality $\omega_1$  which is
not covered by any internal Loeb measure zero set.
In fact, the set $X$ will have the property
that for every internal Loeb measure zero  $A\subseteq H$,
$X\cap A$ is countable.
We say that $X\subseteq H$ is a Loeb-Sierpi\'{n}ski set
iff it is uncountable and meets every Loeb measure zero set in a
countable set.  Thus what we have here is a
weak kind of Loeb-Sierpi\'{n}ski set.
If MA$+\neg$CH is true, then every set of
reals of cardinality $\omega_1$ has measure zero.  So by
Theorem~\ref{lebzero} this weak Loeb-Sierpi\'{n}ski set would not be
a Loeb-Sierpi\'{n}ski set.



Now we ask for the smallest cardinality of a cover of $H$ by Loeb measure
zero sets.  Note that since monads ($st^{-1}\{p\}$ for some $p\in [0,1]$)
have measure zero, $H$ can always
be covered by continuum many Loeb measure zero sets.
MA$_{\kappa}$ stands for the version of Martin's Axiom which says that
for any partially ordered set $\Bbb P$ which has the countable chain
condition and any family $\cal D$ of dense subsets of $\Bbb P$  of
cardinality less than $\kappa$ there exists a $\Bbb P$-filter $G$
which meets all the dense sets in $\cal D$.

\begin{theorem} \label{nextthm}
Suppose MA$_{\kappa}$ and the nonstandard
universe is $\kappa$-saturated, then $H$ cannot be covered by fewer
than $\kappa$ sets of Loeb measure zero.
\end{theorem}
\proof
Let ${\Bbb P}$ be the partial order of all internal subsets of $H$
of positive Loeb measure ordered by inclusion (where stronger
conditions are smaller).  Forcing with $\Bbb P$ is the same
as using the
measure algebra formed by taking
the $\sigma-$algebra generated by the internal subsets of $H$ and dividing
out by the Loeb measure zero sets.
It has the countable chain condition because it is impossible to find
$n+1$ sets of measure greater than $1\over n$ which have
pairwise intersections of measure zero.

For $\lambda<\kappa$ let
$\{\bigcap_{n\in\omega}K_{n}^{\alpha}:\alpha<\lambda\}$ be
a family of $\lambda$ many Loeb measure zero sets
where each $K_{n}^{\alpha}$
is an internal subset of $H$ of Loeb measure less than $1\over n+1$.
For each $\alpha$ define a dense $D_{\alpha}\subseteq {\Bbb P}$ by
$$D_{\alpha}=\{b\in{\Bbb P}:\exists n\;
b\cap K_{n}^{\alpha}=\emptyset\}$$
Using MA$_{\kappa}$ let
$G$ be a $\Bbb P$-filter meeting every $D_{\alpha}$ for $\alpha<\lambda$.
For each $\alpha$ let $n_{\alpha}$ be such that
$(H -  K_{\alpha}^{n_{\alpha}})\in G$.  Consider the family
of sentences $\Sigma(x)$:
$$\{x\in (H -  K_{\alpha}^{n_{\alpha}}): \alpha<\lambda\}$$
Since $G$ is a $\Bbb P$-filter this family of sentences is finitely
satisfiable.  Hence by $\kappa$-saturation some internal $x\in H$
satisfies them all.  But this shows that the family of Loeb measure zero
sets did not cover $H$.
\qed

This partial order was also used in Kaufmann and Schmerl (1987)\cite{KS}.
Next we consider the additivity of Loeb measure.  Here we show that
it is always as small as possible.

\begin{theorem} For any infinite hyperfinite $H$ in an
$\omega_1-$saturated universe
there exists a family of $\omega_1$ Loeb measure zero sets whose union does
not have measure zero.
\end{theorem}
\proof
Without loss of generality we may assume $H=2^{K}$ for some internal
hyperinteger $K$, since $H$ may be replaced by its internal cardinality
and for some hyperinteger $K$ we have $2^K\leq H\leq 2^{K+1}$ and
$2^K$ has Loeb measure at least $1/2$ in $H$.
  Suppose $\Sigma\in [K]^{\omega}$ where $[K]^{\omega}$ is
the infinite countable subsets of $K$ and let
$\Sigma=\{\alpha_n: n\in\omega\}$. For each $n\in\omega$ define
$$H_{n}=\{h\in 2^K: h(\alpha_0)=h(\alpha_1)=\ldots=h(\alpha_{n-1})=0\}$$
Then each $H_n$ is an internal set of measure $1\over {2^n}$.

\begin{lemma} If $A$ is an internal set and
$\bigcap_{n\in\omega}H_n\subseteq A$, then there exists $n\in\omega$
such that $H_n\subseteq A$.
\end{lemma}
This follows from the fact that the $H_n$ are a descending sequence and
the universe is $\omega_1-$saturated.
\qed

Let $\Sigma_{\beta}\in [K]^{\omega}$ for $\beta<\omega_1$ be a
family of disjoint countable subsets of $K$. Let $H_n^{\beta}$
be defined as $H_n$ was but using $\Sigma_{\beta}$ instead of
$\Sigma$.
Define $A_{\beta}=\cap_{n\in\omega}H_n^{\beta}$.
So each $A_{\beta}$ has Loeb measure zero.  We show that
the union $\cup_{\beta<\omega_1}A_{\beta}$ cannot have Loeb
measure zero.   Suppose $A$ is an internal set and
$$\cup_{\beta<\omega_1}A_{\beta}\subseteq A$$
By the Lemma  there exists $n\in\omega$ and $\{\beta_m:m\in\omega\}\in
[\omega_1]^{\omega}$ such that for every $m\in\omega$
$H_n^{\beta_m}\subseteq A$. Let $K_m=H_n^{\beta_m}$.  Note that each
$K_{m}$ for $m\in\omega$ has Loeb measure $\epsilon={1\over 2^n}$ and
they are independent, i.e. for any $m_1<m_2<\ldots<m_k<\omega$
the Loeb measure of
$$K_{m_1}\cap K_{m_2}\cap\ldots\cap K_{m_k}$$
is $\epsilon^k$.  Consequently $\cup_{m\in\omega}K_m$ has Loeb measure
one (since the measure of
$\cap_{m<N}(H -  K_m)$ is $(1-\epsilon)^N$ and
$(1-\epsilon)^N\rightarrow 0$ as $N\rightarrow\infty$).
Consequently the Loeb measure of $A$ is one and the theorem is proved.
\qed

The last cardinal associated with Loeb measure we will consider is the
cofinality, $cof(H)$.  This is the smallest cardinality of a
family $F$ of Loeb measure zero sets such every Loeb measure zero set
is covered by a member of the family $F$.

\begin{theorem}
Let $card$ be the external cardinality function and $H$ an infinite
hyperinteger in some $\omega_1-$saturated universe,
then
$$ card(\lfloor \log_2(H)\rfloor)\leq cof(H)\leq card(2^H)$$
\end{theorem}
\proof
The first inequality is proved by a similar argument to the proof of
the last theorem.
Let the hyperfinite integer $K=\lfloor \log_2(H)\rfloor$
so $2^K\leq H\leq 2^{K+1}$.
Let
$\{\Sigma_{\alpha}:\alpha<card(K)\}$ be disjoint countable subsets of $K$
and define $A_{\alpha}\subseteq H$ as above.
The argument above shows that no Loeb measure zero set covers uncountably
many of the $A_{\alpha}$, hence $card(K)\leq cof(H)$.

The second inequality follows from the fact that every Loeb measure zero set
is covered by a Loeb measure zero set in the $\sigma$-algebra generated
by the internal subsets of $H$ and the following
result of Shelah (1970)\cite{S2}:
if $L$ is an infinite hyperfinite set in an $\omega_1$-saturated
universe and $\kappa=card(L)$, then $\kappa^{\omega}=\kappa$
(take $L=2^H$ the number of internal subsets of $H$).
\qed

In Keisler (1967)\cite{K} it is shown under GCH that for any set of
infinite successor cardinals $C$
there exists a nonstandard universe $M$ in which
$$C=\{card(H):\mbox{$H$ is a hyperfinite set in $M$}\}$$
For finite $C$ the nonstandard universe $M$
can be an ultrapower of a standard universe.

In Shelah (1975)\cite{S1} it is shown that for any countable theory $T$ with
distinguished unary predicates $Q$ and $P$, if for every $n<\omega\;T$
has a model $M$ such that $n^n\leq|Q^M|^n\leq|P^M|<\omega$, then
$T$ has a model $M$ where $|Q^M|=\omega$ and $|P^M|=2^{\omega}$.
Taking $T$ to be any theory containing arithmetic and $Q$ to
be $n^2$ and $P$ to be $2^{n^2}$ we see that there is a nonstandard
universe with a hyperinteger $H$ where $card(H)=\omega$ and
$card(2^H)=2^\omega$.  If follows from Chang's two cardinal
theorem and its proof that assuming the continuum hypothesis there is an
$\omega_1$-saturated nonstandard
universe
 with a hyperinteger $H$ where $card(H)=\omega_1$ and
$card(2^H)=\omega_2$.

This result was also proved by Paris and Mills (1979)\cite{P}
 using a method similar
to the MacDowell-Specker theorem.
For any infinite cardinal $\kappa$ and nonstandard universe $M$ let
$$I^M_{\kappa}=\{H:\mbox{$card(H)\leq\kappa$ and $H$ an integer of M}\}$$
Since $H^2$ has the same internal cardinality as the cartesian product
$H\times H$ which has the same external cardinality as $H$ for infinite
$H$ it is clear that $I^M_{\kappa}$ must be closed under multiplication.
Paris and Mills show this is sufficient by showing: If $M$ is any countable
nonstandard universe and $I$ is a proper initial segment of the integers
of $M$ closed under multiplication, then there exists an elementary
extension $N$ of $M$ in which $I^N_{\omega}=I$ and every integer $H$ of
$N$ not in $I$ has cardinality $2^\omega$.   The following theorem answers
a question raised by Paris and Mills concerning cuts in
nonstandard models of number theory.

\begin{theorem} \label{PM}
   Assume the continuum hypothesis. Then there exists an
   $\omega_1$-saturated universe $N$
   with a hyperinteger
   $H$ which has external cardinality $\omega_1$ but every
   integer greater than $H^n$ for all $n<\omega$ has cardinality
   $\omega_2$.
\end{theorem}
\proof
Let M be any $\omega_1$-saturated universe of cardinality
$\omega_1$ and let $H$ be any infinite hyperinteger of $M$.
Let
$$I=\{K\mbox{ a hyperinteger of M }:\exists n\in\omega\;K<H^n\}$$
and let $J$ be the complement
of $I$ (with respect to the hyperintegers of $M$).
We construct a sequence $<X_{\alpha}:\alpha<\omega_1>$ of elements of
$M$ such each $X_{\alpha}$ is a hyperfinite set of hyperintegers of $M$
 such that
\begin{itemize}
  \item the internal cardinality of $X_{\alpha}$ is an element of $J$
  \item if $\alpha<\beta$ then $X_{\beta}\subseteq X_{\alpha}$
  \item for any set $X$ in $M$ there exists an $\alpha$ such that
     either $X_{\alpha}\subseteq X$ or $X_{\alpha}$ is disjoint from $X$
  \item for any $\alpha$ and function $f$ in $M$ whose domain includes
     $X_{\alpha}$ there exists a $\beta\geq\alpha$ such that
     $f\res X_{\beta}$ is either one-to-one or constant
  \item for any $d\in J$ some
     $X_{\alpha}$ has internal cardinality less than $d$
\end{itemize}
Note that since $M$ is $\omega_1$-saturated and the cofinality of
$I$ is $\omega$ the coinitiality of $J$ is $\omega_1$, i.e. for any
countable set $B\subseteq J$ there exists $c\in J$ such that for every
$b\in B$ we have $c<b$.
To obtain $X_{\lambda}$
for $\lambda<\omega_1$ a limit ordinal,
first find $b\in J$ such that for every $\alpha<\lambda$ the internal
cardinality of $X_{\alpha}$ is greater than $b$. Now use
$\omega_1$-saturation to find $X_{\lambda}$ of internal cardinality
greater than $b$ and for all $\alpha<\lambda\;\;
X_{\lambda}\subseteq X_{\beta}$.
To make functions one-to-one or constant use the following lemma.

\begin{lemma}
Suppose $f:X\to Y$ is an onto function in $M$ such that the internal
cardinality of $X$ is in $J$, then there exists an internal set
$X^\prime\subseteq X$ with internal cardinality in $J$ and
$f\res X^\prime$ is either one-to-one or constant.
\end{lemma}
\proof
 Note that $X=\bigcup_{y\in Y}f^{-1}(y)$.
If some $f^{-1}(y)$ has internal cardinality $b\in J$, then
let $X^\prime=f^{-1}(y)$ and hence $f\res X^\prime$ is constant.
Otherwise (since internally $f$ is a finite function
and the maximum of a hyperfinite set of integers is always achieved)
 there exists $c\in I$ such that for every
$y\in Y$ the internal cardinality of $f^{-1}(y)$ 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
$X^\prime\subseteq X$ such that $f$ maps $X^\prime$ one-to-one onto $Y$
and hence the internal cardinality of $X^\prime$ is in $J$.
\qed

This ends the construction of the sequence $<X_{\alpha}:\alpha<\omega_1>$.
Let $c$ be a new constant symbol and let $T$ be the theory which consists
of the elementary diagram of $M$ plus all statements of the form
``$c\in X_{\alpha}$'' for $\alpha<\omega_1$.  Let $M_1$ be any model of
$T$ which is an elementary superstructure of $M$ and let
$M_0$ be the set of all $f^{M_1}(c)$ such that
$f\in M$ is a function whose domain
contains some $X_{\alpha}$.

{\bf Claim:} For any formula $\psi(x_1,x_2,\ldots,x_n)$
and $\vec{f}=\langle f_1,f_2,\ldots,f_n\rangle$ a sequence of
functions from $M$
$$M_1\models \psi(\vec{f}(c))\;\;\mbox{iff}\;\;
\exists \alpha<\omega_1\;\forall b\in X_{\alpha}\;
M\models \psi(\vec{f}(b))$$

This is proved just like {\L}os's theorem.

{\bf Claim:} $M$ is an elementary substructure of $M_0$.

We have that $M\subseteq M_0$ because of the constant functions in $M$.
So it is enough to note that $M_0$ is an elementary substructure
of $M_1$.
This follows from the Tarski-Vaught criterion.
 Suppose
$$M_1\models \exists x \;\theta(x,f_1(c),\ldots,f_n(c))$$
then there must
be some $X_{\alpha}$ contained in the domain of each $f_i$ such that
$$M\models \forall b\in X_{\alpha}
 \exists x \;\theta(x,f_1(b),\ldots,f_n(b))$$
In $M$ find $g$ with domain $X_{\alpha}$ such
that
$$M\models \forall b\in X_{\alpha}\;
 \theta(g(b),f_1(b),\ldots,f_n(b))$$
and hence
$$M_1\models  \theta(g(c),f_1(c),\ldots,f_n(c))$$
It follows that $M$ is an elementary substructure of $M_0$.

{\bf Claim:} If $a\in I$ and $b\in M_0$ with $b<a$ then $b\in M$, i.e.
the initial segment of the hyperintegers of $M_0$ determined by $I$
is not enlarged.

To see this let $f(c)=b$ for some $f\in M$
and let $X_{\alpha}$ be contained in the domain
of $f$ have the property that for every $x\in X_{\alpha}$
we have $f(x)<a$. By our construction we may assume that
$f\res X_{\alpha}$ is one-to-one or constant.
However it cannot be one-to-one since the internal cardinality
of $X_{\alpha}$ is in J and $a$ is in $I$, so it must be constant.
Hence this constant must be in $M$ and therefore in $I$.

{\bf Claim:} for every $d\in J$ there exists $b\in M_0- M$ with
$b<d$, i.e. there are arbitrarily small new elements of $J$.

Some $X_{\alpha}$ has internal cardinality less than $d$. Let $f\in M$
be a one-to-one function from $X_{\alpha}$ into $d$.  Clearly
$f^{M_0}(c)<d$.  $f^{M_0}(c)$ must be new because for
any $b\in M$ it must be true that some
$X_{\beta}\subseteq (X_{\alpha}- f^{-1}(b))$.
It is also true that no new element of $M_0$ is beneath every element of
$J$.

{\bf Claim:} $M_0$ is an $\omega_1$-saturated model of cardinality $\omega_1$.

Clearly $M_0$ has cardinality $\omega_1$, so it is enough to check that
it is $\omega_1$-saturated.  So let
$\Sigma(x)=\{\theta_n(x,{f_n(c)}):n\in\omega\}$ be a finitely
realizable type
in $M_0$. We may assume parameters are singletons since $n$-tuples
are elements of our universe.
Let $\alpha$ be sufficiently large so that $X_{\alpha}$
is a subset of the domain of ${f_n}$ for every $n<\omega$.
Since $\Sigma(x)$ is finitely realizable we can also choose $\alpha$
large enough so that for every $n\in\omega$
$$M\models\forall b\in X_{\alpha}\exists x\conj_{m<n}
\theta_m(x,{f_m(b)})$$
Consider the following type $\Gamma(g)$ with variable $g$ over $M$.
$\Gamma(g)$ contains the sentence  ``g is a function with domain
$X_{\alpha}$''and for each $n\in\omega$ the sentence
$\forall b\in X_{\alpha}\;\theta_m(g(b),{f_n(b)})$.
Since the type $\Gamma(g)$ is finitely realizable in $M$ and $M$ is
$\omega_1$-saturated some function $g$ in $M$ realizes it and so
$g(c)$ realizes $\Sigma(x)$ in $M_0$.
(Kotlarski (1983)\cite{Ko} shows that every simple cofinal extension of
an $\omega_1$-saturated model is $\omega_1$-saturated.)

Now we prove Theorem \ref{PM}.  This is proved similarly to the standard
proof of Chang's two cardinal Theorem (see Chang and Keisler (1973)\cite{CK}
Theorem 7.2.7 p.438). If $M$ is an elementary substructure of $N$ let
$$I^N=\{b\mbox{ a hyperinteger of $N$} :\exists a\in I\;b<a\}$$
Construct an elementary chain of models
$N_{\alpha}$ for $\alpha<\omega_2$ such that $N_0=M$,
each $N_{\alpha}$ is isomorphic $M$,  and $I^{N_{\alpha}}=I$.

For successor steps to obtain $N_{\alpha+1}$ just use the pair
$M,M_0$. So the only thing to do is the step for $\lambda$ a
limit ordinal.
Letting $N$ be the union of $N_{\alpha}$ for $\alpha<\lambda$
and  $I^N$ be the initial segment of $N$ determined by $I$
we need to see that $N$ can be embedded into $M$ in such a
way that $I^N$ is mapped onto $I$.
This is done by showing that $N$ (while not necessarily
$\omega_1$-saturated) is  $\omega_1$-saturated
for types realizable in $I^N$:

\medskip
{\bf Claim:} Any countable
finitely realizable type $\Sigma(x)$ over $N$, which contains for some
$b\in I^N$ the formula $x<b$, is realized in $N$.

\medskip
Let $\Sigma(x)=\{\theta_n(x,{a_n}):n\in\omega\}$. Choose
a sequence $\langle b_n:n\in\omega\rangle$ from $N$ such
that for each $n\in\omega$  $b_n<b$ and for $m<n$
$N\models \theta_m(b_n,{a_m})$.  Since $I^N=I^{N_0}$ and
$N_0$ is $\omega_1$-saturated, there exists an internal sequence
$\langle b_n:n\in K\rangle$ for an infinite hyperinteger $K$
in $N_0$ which is an extension of $\langle b_n:n\in\omega\rangle$.
Since $N_0$ is substructure of $N$ we have that
$\langle b_n:n\in K\rangle\in N$.  Working in $N$ for each $n\in\omega$ let
$K_n$ be the least $m<K$ (if any) such that
$N\models\neg\theta_n(b_m,{a_n})$.  By construction each
$K_n$ is an infinite integer of $N$.  Since the $I^N=I^{N_0}$
the coinitiality of the nonstandard integers of $N$ must be
$\omega_1$ and so there exists some hyperfinite $L$ in $N$ less
than all $K_n$ for $n\in\omega$.  It follows that $b_L$ realizes
$\Sigma(x)$.

$N_{\lambda}$ is now obtained by a back-and-forth argument
starting with taking $H$ to itself where
$I=\{b:\exists n\in\omega\;\;b<H^n\}$.
This concludes the proof of Theorem \ref{PM}.
\qed

I do not know what either $cof(H)$ or $cof(2^H)$ are in this model.
This argument needs only that the coinitiality of $J$ is $\omega_1$.
I do not know how to do it if the coinitiality of $J$ is $\omega$, as
for example in the proof of Theorem~\ref{inLoeb}.
The result easily generalizes to $(\kappa,\kappa^+)$ in place
of $(\omega_1,\omega_2)$ if $\kappa^{<\kappa}=\kappa$.

This technique can also be used to prove
the following theorem.
\begin{theorem}\label{inLoeb}
   Let $M$ be a countable nonstandard universe and $H$ an infinite
   hyperinteger in $M$.  Then there exists an elementary extension $N$ of
   $M$ such that there exists new subsets of $H$, i.e.
   $X\subseteq H$ with $X\in(N- M)$ but every new
   $X\subseteq H$ has nonzero Loeb measure, i.e.
   if $X\in N$ and $|X|\over H$ is infinitesimal, then $X\in M$, i.e.
   the internal Loeb measure zero sets are the same in $M$ and $N$.
\end{theorem}
\proof
Let $J=\{K\mbox{ a hyperinteger of M }:\exists n\in\omega\;
K > 2^{H\over n}\}$ and let
$I$ be the complement of $J$ in the hyperintegers of $M$.
Similar to the last proof construct an $\omega$-sequence
$\langle X_n:n\in\omega\rangle$
of hyperfinite sets of integers in $M$ such that
\begin{itemize}
  \item the internal cardinality of each $X_n$ is an element of $J$
  \item if $n<m$, then $X_m\subseteq X_n$
  \item for any set $X$ in $M$ there exists an $n$ such that
     either $X_{n}\subseteq X$ or $X_{n}$ is disjoint from $X$
  \item for any $n$ and function $f$ in $M$ whose domain includes
     $X_{n}$  there exists an $m>n$ such that $f\res X_{m}$
     is either one-to-one or constant
  \item for any $d\in J$ some
     $X_{n}$ has internal cardinality less than $d$ (of course
     this will automatically be true if we take $X_0=2^H$)
\end{itemize}

Let $c$ be a new constant symbol and let $T$ be the theory which consists
of the elementary diagram of $M$ plus all statements of the form
``$c\in X_{n}$'' for $n<\omega$.  Let $M_1$ be any model of
$T$ which is an elementary superstructure of $M$ and let
$N$ be the set of all $f^{M_1}(c)$ such that $f\in M$ is a function
whose domain
contains some $X_{n}$ for $n<\omega$.
The following set of claims also go thru:

{\bf Claim:} For any formula $\psi(x_1,x_2,\ldots,x_n)$
and $\vec{f}=\langle f_1,f_2,\ldots,f_n\rangle $ a sequence
of functions  from $M$
$$M_1\models \psi(\vec{f}(c))\;\;\mbox{iff}\;\;
\exists n<\omega\;\forall b\in X_{n}\;
M\models \psi(\vec{f}(b))$$

{\bf Claim:} $M$ is an elementary substructure of $N$.

{\bf Claim:} If $a\in I$ and $b\in N$ with $b<a$ then $b\in M$, i.e.
the initial segment of $N$ determined by $I$ is not enlarged.

{\bf Claim:} for every $d\in J$ there exists $b\in N-M$ with
$b<d$, i.e. there are arbitrarily small new elements of $J$.

It remains only to show that if $X\subseteq H$ in $N$ has the property
that $|X|\over H$ is infinitesimal, then $X\in M$. To do that we need
the following claim from nonstandard calculus:

{\bf Claim:} Suppose $K<H$ are infinite hyperintegers and
${K\over H}\approx 0$, then
$${\left(\begin{array}{c} H\\ K\end{array}\right)}^{1\over H}\approx 1$$
i.e. if $K\over H$ is infinitesimal, then the $H^{\mbox{th}}$ root of the
number of subsets of $H$ of size $K$ is infinitesimally close to $1$.

We will use Stirling's approximation for $n!$
$$n!=({n\over e})^n\sqrt{2\pi n}(1+\epsilon_n)$$
where $\epsilon_n\approx 0$ if $n$ is infinite.
Since $H,K$, and $H-K$ are all infinite
we get that
$$
{\left(\begin{array}{c} H\\ K\end{array}\right)}=
{H!\over (H-K)!K!}=
{{H^{H}}\over{K^{K}(H-K)^{H-K}}}{\left(H\over K(H-K)\right)^{1\over 2}}x
$$
where $x\approx {1\over\sqrt{2\pi}}$.
For any positive finite but not
infinitesimal real number $x$ we have that $x^{1\over H}\approx 1$.
Consequently we need only show:
$$
\left( {{H^{H}}\over{K^{K}(H-K)^{H-K}}}
          {\left(H\over K(H-K)\right)^{1\over 2}}\right)^{1\over H}
\approx 1$$
Calculating this let $\epsilon={K\over H}$ so that $K=\epsilon H$ and
${1\over H}<\epsilon\approx 0$:

\begin{eqnarray*}
%
  \lefteqn{\left( {{H^{H}}\over{K^{K}(H-K)^{H-K}}}
          {\left(H\over K(H-K)\right)^{1\over 2}}\right)^{1\over H}}
          \hspace{.75in} \\
%
   & = &{ H \over K^{K/H} (H-K)^{1-{K/H}}}
        \left({H \over K(H-K)}\right)^{1\over (2H)}\\
%
   & = &{H  \over (\epsilon H)^{\epsilon}(H-\epsilon H)^{(1-\epsilon)}}
        \left({H \over\epsilon H(H-\epsilon H)}\right)^{1\over (2H)}\\
%
   & = &{1\over\epsilon^{\epsilon}(1-\epsilon)^{(1-\epsilon)}}
        \left({1\over\epsilon(1-\epsilon)H}\right)^{1/(2H)}\\
%
   & = &{1\over\epsilon^{\epsilon}(1-\epsilon)^{(1-\epsilon)}}
        \left({1\over\epsilon^{1/H}(1-\epsilon)^{1/H}}({1\over H})^{1/H}
        \right)^{1/2}
%
\end{eqnarray*}

Since $\epsilon\approx 0$ we have
$(1-\epsilon)^{(1-\epsilon)}\approx 1$,
 and $(1-\epsilon)^{1/H}\approx 1$.
Using L'Hopitol's rule it easy to check that for any positive infinitesimal
$\delta$ that $\delta^\delta\approx 1$,
hence
$\epsilon^{\epsilon}\approx 1$, $({1\over H})^{1/H}\approx 1$, and
since ${1\over H}<\epsilon<1$ we have that
$$1\approx (1/H)^{1/H}<\epsilon^{1/H}<1$$
and so $\epsilon^{1/H}\approx 1$.  This proves the Claim.

   Now we prove the theorem.  Suppose $X\subseteq H$ is in $N$ but not $M$,
and suppose $X=f^{M_1}(c)$ where $f$ is a function from $M$ whose domain
includes
$X_n$. Clearly $f$ cannot be constant so we may assume it
is a one-to-one map from $X_n$ to the (internal) set of all subsets of $H$.
Working in $M$ define the function $g:X_n\to H$ by letting
$g(x)$ be the internal cardinality of $f(x)$.  As $H$ is in $I$ the function
$g$ cannot be made one-to-one and hence for some $m\geq n$,
$g\res X_m$ is constantly equal to some hyperinteger $K$.
Since the internal cardinality
of $X$ is $K$ and $X$ is new, clearly $K$ must be infinite.
If ${K\over H}$ is not infinitesimal, then we are done. So assume that
${K\over H}\approx 0$.  By our last claim we have that
for any $n<\omega$
$${\left(\begin{array}{c} H\\ K\end{array}\right)}^{1\over H}<2^{1\over n}$$
so that for any $n<\omega$
$$\left(\begin{array}{c} H\\ K\end{array}\right)<2^{H\over n}$$
It follows that
$\left(\begin{array}{c} H\\ K\end{array}\right)$ is in $I$.
This contradicts the fact that the range of $f$ on $X_m$
must have internal cardinality the same as $X_m$, an element of $J$.
\qed





A set of real numbers $X$ is a Sierpi\'{n}ski set iff it is uncountable
but meets every measure zero set in a countable set.
Keisler-Leth (19..)\cite{KL} have introduced the analogous notion of
a Loeb-Sierpi\'{n}ski set.
A set $X\subseteq H$ is a Loeb-Sierpi\'{n}ski set iff it is uncountable
but meets every Loeb measure zero set in a countable set.

Keisler and Leth have proved that in an $\omega_1$-saturated universe
in which the external cardinality of $2^H$ is $\omega_1$ there exists
a Loeb-Sierpi\'{n}ski set $X\subseteq H$. Such a universe can exist
iff the continuum hypothesis holds.
 They also note
that the standard part map takes a Loeb-Sierpi\'{n}ski set to
a Sierpi\'{n}ski set in the reals.
Note that by Theorem~\ref{KL} if there is a Loeb-Sierpi\'{n}ski set
in a nonstandard universe that universe cannot be $\omega_2$-saturated.

For any cardinal $\kappa$ let $\mu_{\kappa}$ be the product measure on
$2^{\kappa}$.  This measure is determined by: for any
$F\in [\kappa]^{<\omega}$ and $s:F\to 2$
$$\mu_{\kappa}\left(\{x\in 2^{\kappa}:s\subseteq x\}\right)={1\over 2^{|F|}}$$
We define $X\subseteq 2^{\kappa}$ is a $\mu_{\kappa}$-Sierpi\'{n}ski set
iff $X$ is uncountable but meets every $\mu_{\kappa}$ measure zero
set in a countable set.  We begin by establishing a relation between
Loeb-Sierpi\'{n}ski sets and Sierpi\'{n}ski sets in $2^\kappa$.

\begin{theorem}
   If there exists a Loeb-Sierpi\'{n}ski set $X\subseteq H$ in some
   $\omega_1$-saturated universe and infinite cardinal
   $\kappa\leq card(\lceil\log_2(H)\rceil)$
   (where card is the external cardinality), then there exists
   an $\mu_{\kappa}$-Sierpi\'{n}ski set in $2^{\kappa}$.
\end{theorem}
\proof
Let $K=\lceil\log_2(H)\rceil)$, so that $K$ is a hyperinteger
satisfying
$2^{K-1}\leq H\leq 2^{K}$.
Since $H$ has measure at least half in $2^K$ a Loeb-Sierpi\'{n}ski set
in $H$ is also Loeb-Sierpi\'{n}ski set in $2^K$.
Without loss of generality we may assume $H=2^K$.
Let $\{x_{\alpha}:\alpha<\kappa\}\subseteq K$ be distinct and
define $\rho:H\to 2^{\kappa}$ by $\rho(h)(\alpha)=h(x_{\alpha})$,
where we identify $2^K$ with internal maps from $K$ into $2=\{0,1\}$.
Note that for any $\alpha<\kappa$ and $i=0$ or $1$
$$\rho^{-1}(\{x\in 2^{\kappa}:x(\alpha)=i\})=\{h\in 2^K: h(x_{\alpha})=i\}$$
which is an internal set of Loeb measure $1/2$.

{\bf Claim:}  If $Y\subseteq 2^{\kappa}$ has $\mu_{\kappa}$ measure zero,
then $\rho^{-1}(Y)$ has Loeb measure zero.

Suppose $Y\subseteq \bigcap_{n<\omega}Y_n$ where
$Y_n=\bigcup_{t\in X_n}[t_n]$,
each $t\in X_n$ is a finite partial function from $\kappa$ to $2$,
 $[t]=\{x\in 2^{\kappa}:t\subseteq x\}$ with
$\mu_{\kappa}([t])=1/{2^{|t|}}$, each $X_n$ countable, and
$\Sigma_{t\in X_n}\mu_{\kappa}([t])< 1/n$.
But then $\rho^{-1}(Y)\subseteq\rho^{-1}(Y_n)$ and
$\rho^{-1}(Y_n)=\bigcup_{t\in X_n}\rho^{-1}([t])$.
Since each $\rho^{-1}([t])$ is an internal set with Loeb measure
$\mu_{\kappa}([t])$, it follows that the Loeb measure of
$\rho^{-1}(Y_n)<1/n$.  This proves the claim.

Hence if $X\subseteq H$ is a Loeb-Sierpi\'{n}ski set, then
$\rho^{\prime\prime}X\subseteq 2^{\kappa}$ is  a
$\mu_{\kappa}$-Sierpi\'{n}ski set.\qed

\begin{theorem}
   Suppose $card(\lceil\log_2(H)\rceil)> 2^{\omega_1}$, then there
   cannot be a Loeb-Sierpi\'{n}ski set in $H$.
\end{theorem}
\proof
It suffices to show there cannot be a $\mu_{\kappa}$-Sierpi\'{n}ski set
for $\kappa=(2^{\omega_1})^+$.  Suppose for contradiction that
$X=\{x_{\alpha}\in 2^{\kappa}:\alpha<\omega_1\}$ is a
$\mu_{\kappa}$-Sierpi\'{n}ski set and let
$\{y_{\beta}\in 2^{\omega_1}:\beta<\kappa\}$
be defined by $y_{\beta}(\alpha)=x_{\alpha}(\beta)$.  Since
$\kappa=(2^{\omega_1})^+$ there exists $A\in [\kappa]^{\omega}$
and $y\in 2^{\omega_1}$ such
that for all $\alpha\in A$ we have $y_{\alpha}=y$.  Choose
$B\in[\omega_1]^{\omega_1}$ and $i\in\{0,1\}$ such that
$y\res B$ is constantly $i$.  It follows that
for all $\alpha\in A$ that $x_{\alpha}\res B$ is constantly equal to
$i$,  however the set $$\{x\in 2^{\kappa}:x\res B=i\}$$ has measure zero.
\qed

\begin{theorem} \label{consierp}
   It is consistent with ZFC that there exists a
   Sierpi\'{n}ski set in $2^{\omega}$, but no
   Loeb-Sierpi\'{n}ski set in any $\omega_1$-saturated
   nonstandard universe.
\end{theorem}

It suffices to find a model of set theory which contains
a Sierpi\'{n}ski set, but does not contain a
$\mu_{c}$-Sierpi\'{n}ski set where $c$ is the cardinality of
the continuum.
The model is obtained as follows. Let $M$ be a countable
standard model of ZFC+GCH and let $\kappa=\aleph_{\omega}^M$ and
 let ${\Bbb B}_{\kappa}$ be the
measure algebra on $2^{\kappa}$, i.e.
the complete boolean algebra of Borel subsets
of $2^{\kappa}$ modulo the $\mu_{\kappa}$-measure zero sets.
 Let $G$ be ${\Bbb B}_{\kappa}$ generic
over $M$.  Then
$$M[G]\models \mbox{There is a Sierpi\'{n}ski set in $2^\omega$ but none in
$2^c$}$$

We need only show there is no Sierpi\'{n}ski set in $2^c$.
The argument will be similar to one found
in Miller (1982)\cite{Mi}.
 We will use the following Lemma of Kunen from that paper:
\begin{lemma}
   (Kunen) Suppose $B_i\subseteq X$ for $i<n$ are $\mu$-measurable sets
   with $\mu(B_i)\geq 3/4$, then
   $$\mu\left(\left\{r\in X: |\{i<n:r\in B_i\}|\geq {5\over 8}n\right\}
   \right)\geq 1/3$$
\end{lemma}

For any sentence $\theta$ in the forcing language let
$[|\theta|]$ be the boolean value of $\theta$, i.e.
$[|\theta|]=\Sigma \{b\in {\Bbb B_{\kappa}}:b\forces\theta\}$

\begin{lemma}
Suppose $f\in (2^{\omega_1})^{M[G]}$, then
$\exists n<\omega\;\exists g\in M[G_n]$
such that domain$(g)\in ([\omega_1]^{\omega_1})^M$ and
$\forall \alpha\in domain(g)\;\mu_{\kappa}([| f(\alpha)=g(\alpha)|])
\geq {3/4}$.
\end{lemma}
\proof
Working in $M$ for each $\alpha<\omega_1$ let
$C_{\alpha}\subseteq 2^{\kappa}$ be a clopen set such
that $\mu_{\kappa}([|f(\alpha)=1|]\bigtriangleup C_{\alpha})\leq 1/4$,
where $\bigtriangleup$ denotes symmetric difference.
Since $C_{\alpha}$ is clopen there exists $F_{\alpha}\in [\kappa]^{<\omega}$
and $T_{\alpha}\subseteq 2^{F_\alpha}$ such that
$C_{\alpha}=\bigcup_{t\in T_{\alpha}}[t]$, i.e. a finite union
of cylinders over finitely many coordinates.
Since $\kappa=\bigcup_{n\in\omega}\omega_n$ there exists $n\in\omega$ and
$Y\in[\omega_1]^{\omega_1}$ such that for every $\alpha\in Y$
we have $F_{\alpha}\subseteq\omega_n$.  Define $g:Y\to 2$ by
$g(\alpha)=1$ iff $C_\alpha\in G$.  Then $g\in M[G_n]$ since
$C_\alpha\in G$ iff $C_\alpha\in G_n$.  Also
$[| f(\alpha)\neq g(\alpha)]\leq ([|f(\alpha)=1|]\bigtriangleup C_\alpha)$
which has measure less than $1/4$.
This proves the Lemma. \qed

In $M[G]$ we have that $\kappa^+=c=\aleph_{\omega+1}$.  Now suppose
$\{x_{\alpha}\in 2^{\kappa^+}:\alpha<\omega_1\}$ is a
Sierpi\'{n}ski set in $2^c$ in $M[G]$.
Let $\{y_{\alpha}\in 2^{\omega_1}: \alpha<\kappa^+\}$ be defined
by $y_{\alpha}(\beta)=x_{\beta}(\alpha)$.  By the Lemma
and the fact that $$M[G_n]\models 2^{\omega_1}=\omega_n$$
there exists $Q\in [\kappa^+]^{\omega}\cap M$ (in fact one
of size $\kappa^+$) and $g:Y\to 2$ with $g\in M[G_n]$ and
$Y\in ([\omega_1]^{\omega_1})^M$ such that $\forall \alpha\in Q
\;\forall \beta\in Y$
$$\mu_{\kappa}([|y_{\alpha}(\beta)=g(\beta)|])\geq {3\over 4}$$
equivalently
$$\mu_{\kappa}([|x_{\beta}({\alpha})=g(\beta)|])\geq {3\over 4}$$
Now $\{x_{\beta}\res Q: \beta\in Y\}$ must be a Sierpi\'{n}ski set in $2^Q$.
Consequently letting $Q=\{\alpha_n: n\in\omega\}$, by the strong law of
large numbers for all but countably many $\beta\in Y$
$$\lim_{n\rightarrow\infty}{ |\{i<n:x_{\beta}(\alpha_i)=0\}| \over n }
={1\over 2}$$
Choose $\beta\in Y$ and $n\in\omega$ so that if
$$b=[|{7\over 16}<    { |\{i<n:x_{\beta}(\alpha_i)=0\}| \over n }
     <{9\over 16}|]$$
then $\mu_{\kappa}(b)>{2\over 3}$.  Let $B_i=[|x_{\beta}(\alpha_i)=g(\beta)|]$
and note that $\mu_{\kappa}(B_i)\geq {3/4}$, so by Kunen's Lemma:
$$\mu_{\kappa}\left(\left\{r\in 2^{\kappa}:
|\{i<n:r\in B_i\}|\geq {5\over 8}n\right\}
   \right)\geq 1/3$$
Let $$c=\left\{r\in 2^{\kappa}: |\{i<n:r\in B_i\}|\geq
{5\over 8}n\right\}=[|\{i<n:x_{\beta}(\alpha_i)=g(\beta)\}|\geq{5\over 8}n|]$$
I claim that $b\conj c=0$, which contradicts the fact that
$\mu_{\kappa}(b)>2/3$ and $\mu_{\kappa}(c)\geq 1/3$.
This is because it is impossible that
$$|\{i<n:x_{\beta}(\alpha_i)=0\}|\geq{5\over 8}n\;\mbox{ or }\;
|\{i<n:x_{\beta}(\alpha_i)=1\}|\geq{5\over 8}n$$
and
$${7\over 16}<    { |\{i<n:x_{\beta}(\alpha_i)=0\}| \over n }
     <{9\over 16}$$
     This proves the theorem. \qed


In a model due to Bartoszynski and Ihoda (1989)\cite{B} there exists
a subset of $2^\omega$ of cardinality $\omega_1$
which is a Sierpi\'{n}ski set, but also
every Sierpi\'{n}ski set in $2^\omega$ has cardinality $\omega_1$.
This does not mean that there is no Sierpi\'{n}ski set in
$2^{\frak c}$.
To construct their model start with a model
$M$ of ZFC$+$MA$+\neg$CH, then add $\omega_1$ random reals, i.e.
force with the measure algebra of $2^{\omega_1}$.
In this model there is a Sierpi\'{n}ski set in $2^{\omega_2}$.
To see this let $D_{\alpha}\in[\omega_1]^{\omega_1}$
for $\alpha<\omega_2$ be
almost disjoint sets (uncountable sets with pairwise intersection
countable) and define
$x_{\alpha}\in 2^{\omega_2}$ for $\alpha<\omega_1$ by
$x_{\alpha}(\beta)=G(D_{\beta}(\alpha))$ where $G:\omega_1\to 2$
is the generic map and $D_{\beta}(\alpha)$ is the $\alpha^{th}$ element
of $D_{\beta}$.  It is not hard to show that
$\{x_{\alpha}:\alpha<\omega_1\}\subseteq 2^{\omega_2}$
 is a Sierpi\'{n}ski set.   We do not know of model for
 Theorem~\ref{consierp} where the continuum is $\omega_2$.


\begin{theorem}
   It is relatively consistent with ZFC that the continuum hypothesis
   is false but in some $\omega_1$-saturated universe there is
   a Loeb-Sierpi\'{n}ski set.
\end{theorem}
\proof
Suppose $M$ is a countable standard model of ZFC+$\neg$CH.  And let
$W\in M$ be an $\omega_1$-saturated nonstandard universe, and $H\in W$
some hyperfinite set.
Let $\Bbb B$ be the boolean algebra obtained by taking the
$\sigma$-algebra generated by the internal subsets of $H$ and dividing
out by the Loeb measure zero sets.    Then $\Bbb B$ has the
countable chain condition.  Let $G$ be $\Bbb B$-generic over $M$.  Working
in $M[G]$ find an elementary extension of $W$ say $W^\prime$ such that
the type $\Sigma(x)=\{x\in A: [A]\in G\}$ is realized in $W^\prime$.
It is easy to check that for any $\langle A_n:n\in\omega\rangle\in M$
such that
each $A_n\subseteq H$ is internal and the Loeb measure of $A_n$
less than $1/n$, then there exists $n\in\omega$ with
$[H- A_n]\in G$. It follows that
$x\notin\bigcap_{n\in\omega}A_n$.  Iterate this
forcing $\omega_1$ times using finite support at limits and obtain
$W_{\alpha}\in M_{\alpha}$ and $x_{\alpha}\in H\in W_{\alpha}$,
then using the countable chain condition, in the final model
$M_{\omega_1}$ the universe $W_{\omega_1}=\bigcup_{\alpha<\omega_1}W_{\alpha}$
is $\omega_1$-saturated and $\{x_{\alpha}:\alpha<\omega_1\}$ is
a Loeb-Sierpi\'{n}ski set.
\qed

We say that a nonstandard universe $W$ is an $\omega$-power iff there exists
an infinite ordinal $\alpha$ (a typical example is $\alpha=\omega+\omega$)
and a nonprincipal ultrafilter $U$ on $\omega$ such
that $W$ is the ultrapower $V_{\alpha}^{\omega}/U$, where $V_{\alpha}$ is
the set of all sets of rank less than $\alpha$.   All $\omega$-powers are
$\omega_1$-saturated.


\begin{theorem}\label{omegapower}
It is relatively consistent with ZFC that continuum hypothesis is false and
we have an $\omega$-power in which there exists a Loeb-Sierpi\'{n}ski set.
\end{theorem}
\proof
We will use the following lemmas.
\begin{lemma}\label{L1}
   If $M$ is a countable standard model of ZFC and $U\in M$ a nonprincipal
   ultrafilter on $\omega$, then there exists a generic extension $N$ of
   $M$ which satisfies the countable chain condition and
   $$N\models\exists Z\in\infsets\;\forall X\in U\; Z\subseteq^* X $$
   where $A\subseteq^* B$ means inclusion modulo finite.
\end{lemma}
\proof
This is a well-known forcing
$\{(s,X): s\in \finsets \mbox { and }
X\in U\}$, see  Mathias (1977)\cite{Ma}.
\qed

Suppose $F\subseteq P(\omega)$ is a field of sets and $\mu:F\to [0,1]$
is a finitely additive measure with $\mu(\omega)=1$ and
$\mu(n)=0$ for each $n\in\omega$.   Let
${\Bbb P}=\{b\in F:\mu(b)>0\}$ ordered by inclusion.
\begin{lemma} \label{L2}
   $\Bbb P$ has the countable chain condition.  If $G$ is $\Bbb P$-
   generic over a model $M$ of ZFC , then for every
   $\langle a_n:n\in\omega\rangle\in M\cap F^\omega$
    with $\lim_{n\rightarrow \omega}
   \mu(a_n)=0$ there exists $n\in\omega$ with $(\omega- a_n)\in G$.
\end{lemma}
\proof
The countable chain condition holds because there cannot be $n$ sets
of measure greater than $1/n$ and pairwise intersection measure zero.
The second sentence is an easy density argument.
\qed

Given $Z\in\infsets$ and $h\in\bs$ with $h(n)\geq 2$ all
$n\in Z$, then define $X=\Pi_{n\in Z}h(n)$ and give $X$ the product measure
$\mu$ determined as follows.  Let $A_{i,n}=\{x\in X:x(n)=i\}$
for $i<h(n)$ then $\mu(A_{i,n})=1/h(n)$ and
for $n_1,n_2,\ldots,n_k$ distinct
$$\mu(A_{i_1,n_1}\cap\ldots\cap A_{i_k,n_k})=
{1\over h(n_1)}\cdot{1\over h(n_2)}\cdots{1\over h(n_k)}$$
Let ${\Bbb B}$ be the measure algebra determined by
$X$ and $\mu$, i.e. the Borel subsets of $X$ modulo
the $\mu$ measure zero sets.  Then
$\Bbb B$ is isomorphic to the usual random real forcing and a generic filter
$G$ determines and is determined by a ``random real'' $r\in X$.

\begin{lemma}\label{L3}
   Suppose $h$ and $Z$ are elements of $M$ a countable standard model of ZFC,
   $r\in X$ is a random real over $M$, $g:Z\to\finsets$
   with $g\in M$, $g(n)\subseteq h(n)$ all $n$, and
   $\lim_{n\rightarrow\omega\;n\in Z} {|g(n)|\over h(n)}=p$.
   Then the following limit exists and equals $p$:
   $$\lim_{n\rightarrow\omega}
   {  |\{m<n:m\in Z\mbox{ and } r(m)\in g(m)\}|
   \over
      |\{m<n:m\in Z\}|       }=p$$
\end{lemma}
\proof
This follows from the strong law of large numbers for variable
distributions (see Feller (1968)\cite{F} X.7 page 258).  For each $n\in \omega$ let
${\bf X_n}$ be a
random variable
which is $1$ with probability $p_n$ and $0$ with probability $1-p_n$
where $p_n={|g(z_n)|\over h(z_n)}$ and $\{z_n:n<\omega\}$
is an enumeration of $Z$.
The sequence is mutually independent and
satisfies Kolmogorov's
Criterion (variances are less than $1$) so that the
strong law of large numbers holds. This implies that with probability $1$
the sequence ${S_n-m_n}\over n$ tends to zero where
$S_n={\bf X_0+X_1+\ldots +X_{n-1}}$ and $m_n=p_0+\ldots+p_{n-1}$.
Since any random real $r$ must be in any measure one set coded in the ground
model we must have that for
$$a_n=|\{m<n: r(k_m)\in g(k_m)\}|$$
that
$\lim_{n \rightarrow\infty}{a_n-m_n\over n}=0$.
 To finish the proof
it is enough to see that
      $$\lim_{n \rightarrow\infty}{a_n\over n}=p$$
This is true because $\lim_{n \rightarrow\infty}{m_n\over n}=p$ since

\begin{eqnarray*}
{m_n\over n}-p & = & {p_0+p_1+\dots+p_{n-1}-np\over n}\\
               & = & {(p_0-p)+\ldots+(p_m-p)\over n}
                     +{(p_{m+1}-p)+\ldots+(p_{n-1}-p)\over n}
\end{eqnarray*}
and choosing $m$ large makes the second quotient small and $n$ large drives the
first down.
\qed


Now suppose $H$ is an infinite hyperinteger in an $\omega$-power
via a nonprincipal ultrafilter $U$ from $M$
a countable standard model of ZFC, and
$h\in\bs$ represents $H$ in this
ultrapower, i.e. $[h]_{U}=H$.   We combine the
last three lemmas:
\begin{lemma}\label{L4}
   There exists a countable chain condition generic extension
   $N$ of $M$ and an ultrafilter $U^*\in N$ extending $U$ and
   $f\in\bs$ such that $[f]_{U^*}\in H=[h]_{U^*}$ and
   $[f]_{U^*}$ is not in any Loeb measure zero set coded in $M$,
   i.e. $\langle g_n:n\in\omega\rangle\in M$ such that
   $[g_n]_U\subseteq H$  and such that Loeb measure of
   $[g_n]$ is less than $1/n$ for each $n$, there exists $n\in\omega$ with
   $[f]_{U^*}\notin[g_n]_{U^*}$.
\end{lemma}
\proof
Apply Lemma \ref{L1} to obtain $Z\in\infsets$ so that $\forall
X\in U$ we have $Z\subseteq^* X$.
Let $r\in X=\Pi_{n\in Z}h(n)$ be a random real over $M[Z]$.
Working in $M[Z][r]$ define the field of sets
$$F=\left\{X\subseteq\omega:\exists g\in M\;g:\omega\to\finsets
\mbox{ and } X\cap Z=\{m\in\omega:r(m)\in g(m)\}\right\}$$
Define $\mu(X)$ for any $X\in F$ by
$$\mu(X)=\lim_{n\rightarrow\omega}
{ |X\cap Z\cap n|
       \over
   |Z\cap n| }$$
By Lemma \ref{L3} this limit exists and in fact equals the Loeb measure
of the $[g]_{U}$ that put $X$ into $F$.  Use Lemma \ref{L2}
to obtain $G$ $\Bbb P$-generic over $M[Z][r]$
(where ${\Bbb P}=\{b\in F:\mu(b)>0\}$)
and let $U^*$ be
any ultrafilter extending $G$. Let $f\in\bs$ be any map
extending $r$.  Since $\mu(Z)=1$ we have $U^*\supset U$.
If $\langle g_n:n\in\omega\rangle\in M$ is the code for a
 Loeb measure zero subset of $H$, then letting
 $X_n=\{m\in\omega: f(m)\in g_n(m)\}$ ($f=r$ on $Z$)
 we have that the Loeb measure of $[g_n]_U$ is $\mu(X_n)$ which
 goes to zero as $n\rightarrow\infty$.
 So by Lemma~\ref{L2} there exists $n\in\omega$ such that
 $(\omega- X_n)\in G\subseteq U^*$.  This means
 that
 $$\{m:f(m)\notin g_n(m)\}\in U^*$$
 so $[f]_{U^*}\notin [g_n]_{U^*}$.
 \qed

Now we prove Theorem \ref{omegapower}.
Start with any $M_0$ countable standard model of ZFC+$\neg$CH,
nonprincipal ultrafilter $U\in M_0$ on $\omega$,
and $H=[h]_U$ an infinite hyperinteger
with $h\in\bs\cap M$.
Iterate Lemma \ref{L4} with finite support $\omega_1$ times
to obtain $\langle f_{\alpha}:\alpha<\omega_1\rangle$ and
$\langle U_{\alpha}:\alpha<\omega_1\rangle$ with
$[f_{\alpha}]_{U_{\alpha}}\in[h]_{U_{\alpha}}=H$ and
$[f_{\alpha+1}]_{U_{\alpha+1}}$ not in any Loeb measure zero set
coded in $M_{\alpha}$.   Since the entire iteration has the countable
chain condition, at the end $U=\bigcup_{\alpha<\omega_1}U_{\alpha}$ is
an ultrafilter in $M_{\omega_1}$ and
$\{f_{\alpha}:\alpha<\omega_1\}$ will be a Loeb-Sierpi\'{n}ski set.
\qed
I do not know if a Loeb-Sierpi\'{n}ski set can have cardinality
$\omega_2$.




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