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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 $m_1$ 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 $c**$.
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 $b 2^{H\over n}\}$ 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 $nn$ 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 $b 2^{\omega_1}$, 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 $i{2\over 3}$. Let $B_i=[|x_{\beta}(\alpha_i)=g(\beta)|]$
and note that $\mu_{\kappa}(B_i)\geq {3/4}$, so by Kunen's Lemma:
$$\mu_{\kappa}\left(\left\{r\in 2^{\kappa}:
|\{i2/3$ and $\mu_{\kappa}(c)\geq 1/3$.
This is because it is impossible that
$$|\{i0\}$ 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 $i0\}$)
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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\end{document}
**