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\title{Sets on which measurable functions are determined by their range}
 
\author{
Maxim R.~Burke%
\thanks{Research supported by NSERC. This research was initiated while the
\ch % The footnote symbols seem to make this self explanatory
%first named 
author was visiting the Department of Mathematics at the University
of Toronto.}
\\
{\footnotesize Department of Mathematics and Computer Science,} \\
{\footnotesize University of Prince Edward Island,} \\
{\footnotesize Charlottetown, P.E.I., Canada C1A 4P3} \\
{\footnotesize burke\AT upei.ca}
\and
Krzysztof Ciesielski% 
\thanks{Work partially supported by the
  NATO Collaborative Research Grant CRG~950347. %\endgraf
  The 
\ch % same comment as above
%second named 
  author thanks Professor Dikran Dikranjan from
  the University of Udine, Italy, for his hospitality
  in June of 1995, while some of this research was carried out. \endgraf
%
  AMS classification
  numbers: Primary  03E35, 26A15; Secondary  04A30, 03E50, 26A30. \endgraf
  Key words and phrases: continuous images, set of range uniqueness (SRU). 
}\\
{\footnotesize Department of Mathematics,}\\
{\footnotesize West Virginia University,}\\
{\footnotesize Morgantown, WV 26506-6310, USA}\\
{\footnotesize KCies\AT wvnvms.wvnet.edu}
}
%\date{}
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\date{Version of 96/01/14}

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\begin{document}
 
\maketitle
  
\begin{abstract}
We study sets on which measurable real-valued functions
on a measurable space with negligibles are determined by their
range. 
\end{abstract}

\section{Introduction}

In \cite[theorem 8.5]{BD}, it is shown that, 
under the Continuum Hypothesis (CH),
in any separable Baire topological space $X$ there is a set $M$ 
such that for any two continuous real-valued functions $f$ and $g$ on $X$, if
$f$ and $g$ are not constant on any nonvoid open set then $f[M]\sq g[M]$ 
implies $f=g$. In particular, if $f[M]=g[M]$ then $f=g$. Sets having
this last property with respect to entire 
(analytic) functions in the complex plane
were studied in \cite{DPR} where they were called 
{\em sets of range uniqueness (SRU's)}.
We study the properties of such sets in measurable spaces with
negligibles. (See below for the definition.)
We prove a generalization of the aforementioned result 
\cite[theorem 8.5]{BD} to
such spaces 
(Theorem~\ref{th_main})
and answer Question~1 from \cite{BD} by showing that CH cannot
be omitted from the hypothesis of their theorem
(Example~\ref{5.5}).
We also study the descriptive nature of SRU's for the nowhere constant
continuous functions on Baire Tychonoff topological spaces.

When $X=\R$, the result of \cite[theorem~8.5]{BD} states that,
under CH, 
there is a set 
$M\sq\R$ such that for any two nowhere constant continuous functions
$f,g\colon\R\to\R$, if $f[M]\sq g[M]$ then $f=g$. 
It is shown in \cite[theorem~8.1]{BD} that there is (in ZFC) a set $M\sq\R$
such that for any continuous functions
$f,g\colon\R\to\R$, if $f$ has countable level sets and $g[M]\sq f[M]$ then $g$
is constant on the connected components of 
$\{x\in\R\colon f(x)\not=g(x)\}$. 
In the case where $g$ is the identity function, these properties of $M$ are
similar to various properties that have been considered in the literature. 
Dushnik and Miller \cite{DM} showed that, under CH, 
there is an uncountable set $M\sq\R$ such that for any monotone (nonincreasing
or nondecreasing) function $f\colon\R\to\R$, if $\{x\in\R\colon f(x)=x\}$ is
nowhere dense, then $f[M]\cap M$ is countable. 
Building on this
result from \cite{DM}, B\"uchi \cite{Bu} showed that, under CH, there is a set
$M\sq\R$ of cardinality $\cont$ such that for any $X\sq M$ and for any Borel
function $f\colon X\to M$, 
$f[\{x\in X\colon f(x)\not= x\}]$ has cardinality less than
$\cont$. He calls such sets {\em totally heterogeneous}. 
He also observes that
if instead of saying ``for any Borel function $f\colon X\to M$'' we say ``for
any Borel function belonging to collection A'' for 
for special classes A of Borel functions, the use of CH can be avoided. This is
true, e.g., for the 
class of all Borel functions $f\colon\R\to\R$ such that $f[X]\sq M$ and for
each $y\in\R$, $f^{-1}(\{y\})$ is either countable or of positive Lebesgue
measure (a class which contains monotone functions, for example).
In the last section of the paper, we shall discuss the relationships between
some of these properties and the SRU property for various classes of functions. 

\section{Preliminaries}\label{Prelim}
 
Our set-theoretic terminology is standard: see \cite{Ku1} or \cite{Je}.
In particular, $\real$, $\rational$ and $\integer$
will stand for the sets of real numbers, rational numbers and integers,
respectively. 
We will write $\bor$ for the Borel $\sigma$-algebra of $\R$.

A triple $\la X,\Sg,\NN\ra$ is a {\em measurable space with
negligibles}\/ if $\Sg$ is a $\s$-algebra of subsets of $X$ and $\NN$ is
a proper $\s$-ideal of subsets of $X$ generated by $\Sg\cap\NN$. 
(See \cite{F} for the basic properties of such spaces.) 
By analogy with the case where $\Sg$ and 
$\NN$ are respectively the $\s$-algebra of
measurable sets and the $\s$-ideal of null sets for a measure on $X$, we will
call the elements of $\Sg$ {\em measurable sets}\/ and refer to the elements
of $\NN$ as {\em negligible sets}. We will also call the members of
$\Sg\setminus\NN$ {\em positive sets}. (So, in particular, positive
sets are measurable.) If $\Sg$ and $\NN$ are clear from the context,
we will write $X$ in places where it would be more
appropriate to write $\la X,\Sg,\NN\ra$.
In particular, when $M\sq X$ is not negligible, we shall identify $M$
with the space $\la M,\Sg_M,\NN_M\ra$ where 
$\Sg_M=\{E\cap M\colon E\in\Sg\}$ and
$\NN_M=\{N\cap M\colon N\in\NN\}$. 
We say that $\la X,\Sg,\NN\ra$ is {\em
$\h_1$-saturated}\/ if every pairwise disjoint family of positive sets is
countable. An {\em atom}\/ of $\la X,\Sg,\NN\ra$ is a positive set which
does not have two disjoint positive subsets.
$\la X,\Sg,\NN\ra$ is {\em nonatomic}\/
if it has no atoms.
The {\em completion}\/ of $\la X,\Sg,\NN\ra$ is the space 
$\la X,\hat\Sg,\NN\ra$
where $\hat\Sg=\{E\,\triangle\,N\colon E\in\Sg,\,N\in\NN\}$.  


Now, let $\la X,\Sg,\NN\ra$ be a measurable space with
negligibles. 
A function $f\colon X\to \R$ is 
{\em $\Sg$-measurable}\/ (or simply {\em measurable}\/ if $\Sg$ is clear
from the context) if $f^{-1}(U)\in\Sg$ for
each open set $U\sq\R$. 
The family of all measurable functions from $X$ into $\real$
will be denoted by $\M_\Sg(X)$.
We will often write $\M(X)$ in place of $\M_\Sg(X)$
when $\Sg$ is clear from the context.
If $E\sq X$, then $f\restr E$ denotes the restriction of $f$ to $E$.
For $f,g\in\M(X)$ we write $f\equiv g$ to mean that 
$\{x\in X\colon f(x)\not= g(x)\}$ is negligible. The 
{\em level sets}\/ of $f\in\M(X)$ are the 
sets $f^{-1}(y)$ for $y\in\range(f)$. We say that $f$ is {\em nowhere
constant}\/ if it is not constant on any positive set, or
equivalently, if its level sets are negligible.  

\defi{DefSRU}{ Let $\la X,\Sg,\NN\ra$ be a measurable space with negligibles
and let $\F\sq\M(X)$ be a family of measurable functions. 
A set $M\sq X$ is {\em an SRU (a set of range uniqueness) for $\F$}\/ if
whenever $f,g\in\F$ are nowhere constant functions, $f[M]=g[M]$
implies $f\equiv g$.  A set $M$ is a 
{\em strong SRU for $\F$}\/ if for any positive set
$E$ and any nowhere constant $f,g\in\F$, if 
$f[M\cap E]\sq g[M]$ then $f\restr E\equiv g\restr E$.
We will frequently have $\F=\M(X)$. 
It will be convenient, when $M$ is an SRU (resp.\ a strong SRU) for $\M(X)$,
to call $M$ an SRU (resp. a strong SRU) for $\la X,\Sg,\NN\ra$ or, when the
structure is clear from the context, simply an SRU (resp.\  a strong SRU). 
}

Note that if $\la X,\Sg,\NN\ra$ is a measurable space 
with negligibles and $\F\sq\M(X)$ then every strong SRU for $\F$ is an SRU
for $\F$, and if $\F\sq\G\sq\M(X)$ then any 
SRU (resp.\ strong SRU) for $\G$ is also an SRU (resp.\ a strong SRU) for $\F$.

All topological spaces considered 
in this paper are assumed to be Tychonoff.
A topological space is {\em Baire}\/ if it 
satisfies the Baire Category Theorem,
i.e., if every meager subset of $X$ has 
empty interior. 
A {\em Cantor set}\/ 
is a nonvoid zero-dimensional compact
metrizable space with no isolated points, i.e., a homeomorphic copy of the
Cantor middle third set. 
If $X$ is a topological space, we write $\C(X)$ for
the family of all continuous real-valued functions on $X$.
An {\em $s_0$-set}\/ (in $X$)
is a set $S\sq X$ with the property that for every Cantor
set $P\sq X$, there is a Cantor set $Q\sq P$ such that $Q\cap S=\e$. (See
\cite{Mi2} for the basic facts about $s_0$-sets.) 
We will say that $S\sq X$ is a {\em strong $s_0$-set}\/ if 
$f[S]$ is an $s_0$-set in
$\R$ for every $f\in\C(X)$. 

To any Baire topological space $X$ we can associate a natural
measurable space with negligibles $\la X,\Sg,\NN\ra$, 
where $\NN$ is the $\sigma$-ideal of meager subsets of $X$ and $\Sg$ is 
the $\sigma$-algebra of Baire subsets of $X$, i.e., the $\sigma$-algebra
generated by 
the family $\{f^{-1}((a,\infty))\colon a\in\real\ \&\ f\in\C(X)\}$.
When we consider a topological space as a measurable
space with negligibles, it is this structure we have
in mind, unless otherwise stated.
Note that $\C(X)\sq\M(X)$. When $X$ is 
Baire, the term ``nowhere constant''
applied to a continuous function $f\in\C(X)$ considered as a member of $\M(X)$
coincides with the usual meaning of ``not constant on any nonvoid open set.'' 
On the few occasions where we consider spaces which are not Baire, 
we will clarify the meaning of ``nowhere constant.'' 
(See also Remark~\ref{NotBaire}.)

We record 
the following simple observation for future reference.

\prop{prop4_2}{ Let $X$ be a Baire topological space. 
If $M\sq X$ is an SRU (strong SRU) for $\M(X)$ then 
$M$ is an SRU (strong SRU) for $\C(X)$.
Moreover, for any $f,g\in\C(X)$ and any positive $E\in\Sg$,

    \begin{center}\hfill
$f\restr E\equiv g\restr E\ \mbox{ if and only if }\ 
 f\restr E= g\restr E$. \endproof\end{center}
}

Proposition~\ref{prop4_2} will be our main tool for producing SRU's for Baire
spaces. However, we shall also see (Example~\ref{sruContNotBorel}) that the
converse to the first part of Proposition~\ref{prop4_2} can fail.

Finally, note that if in some measurable space with negligibles 
$\la X,\Sg,\NN\ra$ there 
are no nowhere constant measurable functions $f\colon X\to\R$ 
then every subset of $X$ is (vacuously) a strong SRU. This happens,
for example, if $X$ has an atom. (See also Corollary~\ref{2.8}.)
Of course, this situation is of no interest and we shall be interested in
spaces in which it does not happen. 

\defi{1.2}{ 
Let $\la X,\Sg,\NN\ra$ be a measurable space with negligibles.
We shall say that 
$X$ is {\em flexible}\/ if there is a nowhere
constant measurable function $f\colon X\to\R$.
If $X$ is a 
Baire 
topological space then we say that
$X$ is {\em locally flexible}\/ if for every open set 
$U\sq X$ there is a continuous function $f\colon X\to\R$ which is identically
equal to zero outside $U$ and nowhere constant in $U$.
}

Note that if $\la X,\Sg,\NN\ra$ is the 
natural measurable space with negligibles
associated with some atomless countably additive
nontrivial $\sigma$-finite measure on $X$ then $X$ is flexible. 

The notion of local flexibility is better suited than the notion of
flexibility to the study of SRU's the class $\C(X)$. This will be more
apparent in Section~\ref{secCont}. 
Notice, however, that there are compact topological spaces that are 
flexible but not locally flexible. The space $X=[0,1]\times(\omega_1+1)$
with its natural topology is one example. Its flexibility is witnessed by
the projection onto the first coordinate. The failure of local flexibility is
witnessed by the open set $U=[0,1]\times\omega_1$. 
(See \cite[examples 1 and 2]{BS}.) 

The next proposition is established as part of the proof of 
\cite[theorem 1]{BS}. 
It shows, for example, that a space with no isolated points which is either
separable or metrizable is locally flexible.  


\prop{4.6}{Let $X$ be a space with no isolated
points such that either {\rm(i)} $X$ is separable or {\rm(ii)} $X$ is normal
and has a dense set which is a countable union of closed discrete sets.  Then
$X$ is locally flexible. \qed} 


\section{ Properties of SRU's for $\M(X)$.}\label{sec2}

In this section we establish some results concerning the nature of SRU's.


\prop{2.1}{ Let $\la X,\Sg,\NN\ra$ be a measurable space with
negligibles, and let $M\sq X$.
\begin{description} 

\item[(a)] Let $X$ be flexible.
If $M\sq X$ is an SRU (a strong SRU), 
then $M$ is not empty, and for every positive $E$,
$M\cap E$ is an SRU (a strong SRU) for $\M(E)$.
In particular, an SRU must meet every positive
subset 
of~$X$. 

\item[(b)] If $M$ is an SRU and 
$f,g\colon X\to\R$ are nowhere
constant measurable functions such that $f\not\equiv g$, then
$f[M]\,\triangle\,g[M]$ is uncountable. 

If $M$ is a strong SRU, $E$ is
a positive set, and $f\restr E\not\equiv g\restr E$, 
then $f[M\cap E]\setminus g[M]$ is uncountable. 

\item[(c)]  If $M$ is 
an SRU (for $\M(X)$), 
$N\in\NN\cap\Sg$, and $C\sq X$ is countable
then $(M\setminus N)\cup C$ is an SRU (for $\M(X)$). 

If $M$ is a strong SRU (for $\M(X)$), $N\in\NN$, and $C\sq X$ is countable
then $(M\setminus N)\cup C$ is a strong SRU (for $\M(X)$). 
\end{description} 
}

\P 
(a) 
Let $f\colon X\to\R$ be a nowhere constant function from the definition
of flexibility of $X$ and let
$g\colon X\to\R$ be given by $g(x)=f(x)+1$. Then $g$ is a
nowhere constant measurable function which does not agree with $f$
anywhere. So clearly $M$ 
cannot be empty. 

Let $E$ be a positive set. We
will show that $M\cap E$ is an SRU for $\M(X)$. 
The proof for the strong SRU case
is similar. Let $h_1,h_2\colon E\to \R$ be nowhere constant measurable
functions such that $h_1[M\cap E]=h_2[M\cap E]$. For $i=1,2$, extend
$h_i$ to $X$ by letting $h_i(x)=f(x)$ when $x\in X\setminus E$. Then
$h_1[M]=h_2[M]$ and hence $h_1\equiv h_2$. In particular, 
$h_1\restr E\equiv h_2\restr E$. 


(b) We prove the first statement; the second is proven similarly. 
By way of contradiction suppose that $C=f[M]\,\triangle\,g[M]$ is countable. 
By (a), $M$ is not negligible, and hence $f[M]$ and $g[M]$ are uncountable. 
So, we can choose $y\in f[M]\cap g[M]$. 
Define $\bar f,\bar g\colon X\to\R$ as follows.
$\bar f(x)=f(x)$ if $x\not\in f^{-1}(C)$, 
and $\bar f(x)=y$ if $x\in f^{-1}(C)$. 
Similarly,
$\bar g(x)=g(x)$ if $x\not\in g^{-1}(C)$, 
and $\bar g(x)=y$ if $x\in g^{-1}(C)$. 
We have $\bar f\equiv f\not\equiv g\equiv \bar g$. 
Thus, it is enough to show that 
$\bar f[M]=\bar g[M]$, since this contradicts that $M$ is SRU for $\M(X)$.

By symmetry, it is enough to prove that $\bar f[M]\sq\bar g[M]$. 
So, let $x\in M$.
If $x\in f^{-1}(C)$ then $\bar f(x)=y\in g[M]\setminus C=\bar g[M]$.
If $x\not\in f^{-1}(C)$ then 
$\bar f(x)=f(x)\in f[M]\setminus C=g[M]\setminus C=\bar g[M]$.
Thus, $\bar f[M]\sq\bar g[M]$. 

(c) 
First we will prove this in the case when $C=\emptyset$. 

Let $f,g\colon X\to\R$
be nowhere constant measurable functions. 
Suppose $M$ is an SRU, 
$N$ is a measurable negligible set, and
$f[M\setminus N]=g[M\setminus N]$. Let 
$\bar f,\bar g\colon X\to\R$ be the 
functions which agree on $X\setminus N$ 
with $f,g$ respectively, 
and are identically equal to $0$ on $N$. 
Then $\bar f,\bar g\in\M(X)$ and $\bar f[M]=\bar g[M]$. 
Since $M$ is an SRU, we have
$\bar f\equiv\bar g$ and hence $f\equiv g$. Thus $M\setminus N$ is an
SRU for $\M(X)$. 

Next suppose $M$ is a strong SRU for $\M(X)$, 
$N$ is negligible, $E$ is a positive set and
$f[(M\setminus N)\cap E]\sq g[M\setminus N]$. 
Choose $N'\in\Sg\cap\NN$ such that $N\sq N'$.
Then 
$f[M\cap(E\setminus N')]\sq 
f[(M\setminus N)\cap E]\sq g[M\setminus N]\sq g[M]$. Since
$E\setminus N'$ is positive
and $M$ is a strong SRU for $\M(X)$, we have $f\restr[E\setminus N']\equiv
g\restr[E\setminus N']$ and hence $f\restr E\equiv g\restr E$. Thus
$M\setminus N$ is a strong SRU for $\M(X)$.

Now, to prove the general case we can assume by what we proved above that 
$N=\emptyset$. But then, the desired result follows easily from (b).
\endproof

Proposition~\ref{2.1}(c)  suggests that possibly
$M\,\triangle\,N$ is an SRU (strong SRU)
for $X$ if $N\in\NN\cap\Sg$ and $M$ is an SRU (strong SRU)
for $X$. However, this is false already
for $X=\real$, since 
if $K\sq\R$ is any Cantor set, then
$M\cup K$ is not
an SRU for $\C(\real)$ for any strong SRU for $\M(\real)$.
(See Theorem~\ref{thm:MP}(5).)
However, we do not know the answer to the following question.

\pr{problemB}{ If $X$ is a Baire topological space, is
Proposition~\ref{2.1}(c) true with $\M(X)$ replaced by $\C(X)$?}

\noindent (It is consistent that there is a strong SRU for $\C(\real)$
which is not an SRU for $\M(\real)$, as we will show in
Example~\ref{sruContNotBorel}.)
 
The next proposition gives special circumstances in which no SRU can exist.
It will be useful later.

\prop{2.2}{ Let $\la X,\Sg,\NN\ra$ be a measurable space with negligibles. In
any of the following circumstances, there is no SRU for $\M(X)$.
\begin{description} 
\item[(a)] $\Sg$ is the collection of all subsets of $X$, the
cardinality of $X$ is at most $\cont$, and
$X$ is nonatomic.

\item[(b)] $X$ is flexible, $\cont$ is regular, $\NN\sq\Sg$, $\NN$ contains
all sets of cardinality less than $\cont$, and $X$ is covered by less than
$\cont$ negligible sets.

\item[(c)] There are more than $2^{\cont}$ pairwise nonequivalent (modulo
$\NN$) nowhere constant measurable functions $X\to\R$.
\end{description} 
}

\P If (a) holds, we argue as follows. 
Since $X$ is nonatomic, singletons are negligible. 
Thus any one-to-one function $f\colon X\to\R$ 
is a nowhere constant measurable function. 
Clearly there are such functions since $X$ has
cardinality at most $\cont$. Thus, $X$ is flexible.
Suppose $M\sq X$ were an SRU. By 
Proposition~\ref{2.1}(a), $M$ is not negligible. 
Fix any one-to-one function $f\colon X\to\R$.
Write $M$ as the union of two
disjoint sets $A$ and $B$ of equal cardinality and find a bijection
$h\colon X\to X$ such that 
$h[A]=B$ and $h[B]=A$.
Define $g\colon X\to\R$ by $g=f\circ h$. Then
$f[M]=g[M]$ but $f\not\equiv g$ and hence $M$ is not an SRU.

If (b) holds, suppose $M\sq X$ were an SRU. 
Since sets of cardinality less
than $\cont$ are negligible, $M$ has cardinality at least $\cont$. 
Since $X$ is
covered by less than $\cont$ negligible sets and $\cont$ is regular, 
there is a
negligible set $N\sq M$ of cardinality $\cont$. Let $h\colon N\to\R$ be any
surjection. Since $X$ is flexible and $N$ is measurable, there are many
distinct nowhere constant measurable extensions of $h$ to $X$. 
All of these extensions map $M$ onto $\R$, 
and hence $M$ is not an SRU.

If (c) holds, then for any $M\sq X$ there are, by the pigeonhole principle,
two nonequivalent nowhere constant measurable functions 
$f,g\colon X\to\R$ such
that $f[M]=g[M]$. Hence $M$ is not an SRU.
\endproof

To state the next result we need the following definition.

\defi{defCover}{ Let $\la X,\Sg,\NN\ra$ be a measurable space with
negligibles. A set $E\in\Sg$ is a {\em measurable cover of $M\sq X$}\/
if $M\sq E$ and every positive subset of $E$ meets $M$.}

%We will need the following result.

\lem{2.3}{ Let $M\sq X$ and let $f\colon M\to\R$ be a measurable function.
\begin{description} 
\item[(a)] The function $f$ extends to a
measurable function $\bar f\colon X\to\R$.

\item[(b)] Suppose $M$ has a measurable cover.
If $f$ is nowhere constant in $M$, and 
$X$ is flexible, then $f$ extends to a 
nowhere constant measurable function $\bar f\colon X\to\R$. 
\end{description} 

}

\P (a) Of course this is well-known. Here is a sketch of the proof. 
For each rational number $q$, let $E_q\in\Sg$ be such that 
$\{x\in M\colon f(x)<q\}=E_q\cap M$. 
For $x\in E=\bigcup\{E_q\colon q\in\Q\}$, we let
$\bar f(x)=\inf\{q\in\Q\colon x\in E_q\}$ and for 
$x\in X\setminus E$ we let
$\bar f(x)=0$. It is straightforward to check that $\bar f$ is as
desired. 

(b) Fix a nowhere constant measurable function $g\colon X\to\R$.
First notice that every positive subset of $E$
must have a nonnegligible intersection with $M$.
This is the case, since otherwise there exists a positive $F\sq E$
and $N\in\Sg\cap\NN$ with $F\cap M\sq N$
and then $F\setminus N$ is a positive
subset of $E$ which does not meet $M$.

Extend $f$ to $E$ using (a). 
By the above remark the extension of $f$ to $E$ is
necessarily nowhere constant. Now extend $f$ to $X$ by letting it
agree with $g$ outside of $E$.
\endproof

We leave the straightforward proof of the following consequence of
Lemma~\ref{2.3}(b) to the reader. 

\cor{2.4}{ Let $\la X,\Sg,\NN\ra$ be a flexible measurable space with
negligibles and suppose that every subset of $X$ has a measurable cover. If
$M\sq
X$ is an SRU (a strong SRU) for $\M(X)$, then $M$ is also an SRU (a strong
SRU) for $\M(M)$. \endproof}

The next proposition and its corollary show that in an
$\h_1$-saturated measurable spaces with negligibles, the assumption of
flexibility is not very 
restrictive when we are considering SRU's for $\M(X)$. 

Recall that
a {\it Souslin algebra}\/ is a ccc nonatomic complete
Boolean algebra in which the intersection of any 
countable collection of dense
open sets is a dense open set (or equivalently, 
every countable collection of
maximal antichains has a common refinement). 
See \cite[pp.~220,~274]{Je} for the basic
properties of these algebras. 

\prop{2.7}{ Let $\la X,\Sg,\NN\ra$ be a nonatomic, $\h_1$-saturated, 
measurable
space with negligibles.  
Then $X$ is flexible if and only if $\Sg_E/\NN_E$ is not a Souslin algebra 
for any positive $E$. In particular, if Souslin's Hypothesis is true then
$X$ is flexible.}

\P Suppose $\Sg_E/\NN_E$ is a Souslin algebra for some positive
$E\in\Sg$. Let $f\colon E\to\R$ be measurable. For each $q\in\Q$, let $\AA_q$
be the collection of positive subsets $A$ of $E$ such that
$A\sq f^{-1}(-\infty,q)$ or $A\sq f^{-1}(q,\infty)$. If we denote by
$\AA_q^\bullet$ the image of $\AA_q$ in $\Sg_E/\NN_E$, then $\AA_q^\bullet$ is
dense open in $\Sg_E/\NN_E$. Since
$\Sg_E/\NN_E$ is a Souslin algebra, there is an $A\in\bigcap_{q\in\Q}\AA_q$.
Clearly $f$ is constant on $A$, and thus $E$ (and hence $X$ as well) is not
flexible. 

Suppose, conversely, that none of the algebras $\Sg_E/\NN_E$ is a Souslin
algebra. Since $\Sg/\NN$ is ccc, to show that $X$ is flexible it suffices to
show that for each positive set $E$ there is a positive $E'\sq E$ which is
flexible. So fix a positive set $E$, and let $\AA_n\sq\Sg_E\setminus\NN_E$,
$n<\omega$, be partitions of $E$ which have no common refinement. Give each
$\AA_n$ the discrete topology and let $f\colon E\to\Pi_n\AA_n$  be given by
$f(x)=$ the unique sequence in $\Pi_n\AA_n$ whose $n$-th term contains $x$ for
each $n$. Clearly $f$ is measurable and its range is 
homeomorphic to a subspace of $\R$. Let $\AA=\{f^{-1}(y)\colon y\in\Pi_n\AA_n$,
$f^{-1}(y)$ positive$\}$. $\AA$ is an antichain which refines each
$\AA_n$, and hence $\AA$ is not maximal. Thus $E'=E\setminus\bigcup\AA$ is
positive and is flexible since $f\restr E'$ is nowhere constant. \endproof

The following structural result follows easily from the previous proposition.

\cor{2.8}{ If $\la X,\Sg,\NN\ra$ is an $\h_1$-saturated measurable space with
negligibles, then $X$ admits an essentially unique decomposition into three
pieces $X=A\cup S\cup F$, where $A$ is a countable union of
atoms, $\Sg_S/\NN_S$ is a Souslin algebra if $S$ is not empty, and $F$ is
either empty or flexible. \endproof}



\section{ Existence of SRU's for $\M(X)$}\label{sec_4}

The next lemma
states that sufficiently generic 
generalized Lusin sets are strong SRU's.
Since the conditions given here will be used several times in the rest of the
paper, we make the following definition.

\defi{kStrongSRU}{Let $\la X,\Sg,\NN\ra$ be a measurable space with
negligibles and let $\k$ be an uncountable cardinal. A set $L\sq X$ is called a
{\em $\kappa$-strong SRU}\/ (for $X$) 
if for each pair 
$f,g\in\M_{\Sg}(X)$ of nowhere constant 
functions there exists a subset $C_{f,g}$ of $L$
satisfying the following properties.
\begin{description}
\item[(i)] $\card(L\cap E\setminus f^{-1}(g[C_{f,g}]))\geq\k$ for each
positive set $E$, and $\card(L\cap N)<\k$ for each negligible set $N$.

\item[(ii)] $\card(C_{f,g})<\k$, and $f(x)\not=g(y)$ for every
distinct $x,y\in L\setminus C_{f,g}$.
\end{description}
}

Notice that if $\k$ is regular, then we can drop the
``$\setminus f^{-1}(g[C_{f,g}])$'' in clause (i) of
Definition~\ref{kStrongSRU}, as it is taken care of by the
other parts of clauses (i) and (ii). 
(The other parts of (i) and (ii) imply that
$L\cap f^{-1}(g[C_{f,g}])=\bigcup\{L\cap f^{-1}(y)\colon y\in g[C_{f,g}]\}$
is the union of less than $\k$ many sets of cardinality less 
than~$\k$.)


\lem{3.1}{ Let $\la X,\Sg,\NN\ra$ be a measurable space with
negligibles and $\k$ be an uncountable cardinal. If $L\sq X$ is a
$\kappa$-strong SRU, then for any nowhere constant 
functions 
$f,g\in\M_{\hat\Sg}(X)$ and any positive set $E$ such that 
$f\restr E\not\equiv g\restr E$,  
$f[L\cap E]\setminus g[L]$ has cardinality at least $\k$. 
In particular $L$ is a strong SRU for $\M_{\hat\Sg}(X)$.} 

\P 
For each pair $u,v\in\M_{\Sg}(X)$ of nowhere constant functions fix a set
$C_{u,v}$ as given by Definition~\ref{kStrongSRU}.  
Let $f,g,E$ be as in the hypothesis of the lemma.
Let $N$ be a negligible set on whose complement $f$ and $g$ agree
with $\Sg$-measurable functions $\bar f$ and $\bar g$ respectively. 
By shrinking $E$ we can assume that $E\cap N=\e$ and 
$E\sq\{x\in X\colon f(x)\neq g(x)\}$. 
But then $E$ is the countable union of the measurable sets
$\{x\in E\colon f(x)<q<g(x)\}$ and 
$\{x\in E\colon g(x)<q<f(x)\}$, where $q\in\Q$. So, at least one of
these sets is positive, and shrinking $E$ even further 
we can assume that $f[E]\cap g[E]=\e$. 

Now, since $L$ is a $\kappa$-strong SRU,
the set $K=(L\cap E)\setminus\left(\bar f^{-1}(\bar g[C_{\bar f,\bar g}])\cup
C_{\bar f,\bar g}\cup C_{\bar f,\bar f}\right)$ 
has cardinality at least
$\k$. Since $K\sq L\setminus C_{\bar f,\bar f}$, 
$\bar f$ is one-to-one on $K$ and hence 
$\bar f[K]$ has cardinality at least $\k$. We have also 
$\e=\bar f[K]\cap\bar g[L]=f[K]\cap\bar g[L]$,
since $K\subset E$ is disjoint from
$N\cup\bar f^{-1}(\bar g[C_{\bar f,\bar g}])\cup C_{\bar f,\bar g}$.
Hence, $f[K]\cap g[L]\sq g[L\cap N]$ has cardinality less than $\k$. 
So, $f[L\cap E]\setminus g[L]\supset f[K]\setminus(f[K]\cap g[L])$
has cardinality at least $\k$.
\endproof

\thm{th_main}{ Let $\la X,\Sg,\NN\ra$ be a measurable space with
negligibles. Suppose that $\Sg$ has cardinality $\cont$
and no positive set can be covered by less than $\cont$ negligible sets.
Then there is a $\cont$-strong SRU for $X$.
In particular, there is a strong SRU for $\la X,\hat\Sg,\NN \ra$.}

\P Let $\la f_\alpha\colon\alpha<\cont\ra$ be a list of all
nowhere constant functions from $\M_{\Sg}(X)$, let 
$\la E_\alpha\colon\alpha<\cont\ra$ be a
list of all positive sets in which each set is listed $\cont$ times
and let 
$\la N_\alpha\colon\alpha<\cont\ra$ be a list of $\NN\cap\Sg$.

Inductively choose points $x_\alpha\in X$ as follows. Let
$S_\alpha=\{x_\beta\colon\beta<\alpha\}$. Since the level sets of 
$f_\alpha$ are negligible, and $E_\alpha$ cannot be covered by less
than $\cont$ negligible sets, there is a point
$x_\alpha\in E_\alpha$ 
which does not belong to any sets of the form $N_\beta$
or $f_\beta^{-1}(f_\g(x_\d))$ 
with $\beta,\g\leq\alpha$ and $\d<\alpha$.

We shall show that 
$L=\{x_\alpha\colon\alpha<\cont\}$ is a 
$\cont$-strong SRU with witnessing sets
given by $C_{f,g}=S_\alpha$ for any $\alpha<\cont$ such that
$f,g\in\{f_\beta\colon\beta\leq\alpha\}$. 
First notice that $x_\alpha\neq x_\beta$ for every 
$\beta<\alpha<\cont$, since
$x_\alpha\not\in f_\alpha^{-1}(f_\alpha(x_\beta))$.

To see clause (ii) of Definition \ref{kStrongSRU}, 
suppose that $x$ and $y$ are distinct
elements of $L\setminus S_\alpha$. Thus
$x=x_\beta$, $y=x_\g$ for some distinct 
$\beta,\g>\alpha$, say $\beta<\g$.
Then $x_\g\not\in g^{-1}(f(x_\beta))$ and 
hence $f(x)=f(x_\beta)\not=g(x_\g)=g(y)$.

Clause (i) of Definition~\ref{kStrongSRU} now follows easily.
\endproof


\cor{cor:main}{ Consider the measurable space with negligibles 
$\la \R,\bor,\NN\ra$. If either
\begin{description}
\item[(i)] $\NN$ is the $\sigma$-ideal of null sets in $\real$ and the
           union of less than $\continuum$ many null sets does not
           cover $\real$; or
\item[(ii)] $\NN$ is the $\sigma$-ideal of meager sets in $\real$ and the
           union of less than $\continuum$ many meager sets does not
           cover $\real$,
\end{description}
then there exists a strong SRU for the completion of $\la \R,\bor,\NN \ra$.
\qed }

Note that the conclusion of Corollary~\ref{cor:main}
cannot be proved in ZFC for either of the ideals. 
This follows from Proposition~\ref{2.2}(b),
since in the Cohen model every subset of $\real$ of
cardinality $<\cont$ is null and $\R$ is the union of less
then $\cont$ many null sets, while in the random real
model every subset of $\real$ of
cardinality $<\cont$ is meager and $\R$ is the union of less
then $\cont$ many meager sets. (See e.g., \cite{Mi1}.)
But what about the existence of a strong SRU, or
an SRU, just for $\la\real,\bor,\NN \ra$,
with $\NN$ as in the corollary? 

To answer this question we will need the following theorem of
P.~Corazza~\cite{C}. 

\prop{prop_Corazza}{ {\rm \cite{C}} It is consistent with ZFC that
$\cont=\omega_2$, every subset of $\real$ of cardinality less than
$\cont$ has strong (so Lebesgue) measure zero,
and for every subset $M$ of $\real$ 
cardinality $\cont$ there is a
uniformly continuous function 
$f\colon\real\to[0,1]$ such that $f[M]=[0,1]$. \qed}

From this we can easily deduce the following.

\cor{cor_MainIndep}{ It is consistent with ZFC that there is no
SRU for $\la \real,\bor,\NN \ra$, where 
$\NN$ is the $\sigma$-ideal of null sets in
$\real$.
}

\P This happens in the model from Proposition~\ref{prop_Corazza}.
First recall, that in this model 
the real line is covered by
less than $\cont$ sets of measure zero,
i.e., $\real=\bigcup_{\xi<\omega_1}N_\xi$
for some null sets $N_\xi$. (See e.g. \cite{C} or \cite{Mi3}.)
Now, if $M$ were an SRU then, by Proposition~\ref{2.1}(a),
$M$ cannot be null. So, it has cardinality $\cont$. 
Hence, there is $\xi<\omega_1$ such that $M'=M\cap N_\xi\in\NN$
has cardinality $\cont$. 
Let $h\colon\R\to [0,1]$ be a uniformly continuous map
such that $h[M']=[0,1]$.
There are many pairwise nonequivalent extensions of $h\restr M'$ to nowhere
constant Borel functions from $\R$ into $[0,1]$. Since these extensions all
have the same image of $M$, $M$ is not an SRU, contradiction. \qed

We do not know whether 
there is a model of ZFC in which there is no 
SRU for $\la\real,\bor,\NN\ra$, where 
$\NN$ is the $\sigma$-ideal of meager sets.
(See Problem~\ref{prAAA}.) However, 
in models where every set of reals of cardinality $\cont$
maps uniformly continuously onto $[0,1]$,
there is no SRU for $\la\real,\bor,\NN\ra$
of cardinality $\continuum$. (See 
Theorem~\ref{thm:MP}(5) and the comments
before Lemma~\ref{S_0}.) On the other hand,
the following observations,
that arose in conversations with S.~Todorcevic,
show that there is an $\aleph_1$-strong SRU for $\la\real,\bor,\NN\ra$
in 
\ch 
the models from \cite{C} and \cite{Mi3}.

For any set $L$, let $L^{(2)}=\{\la a,b\ra\in L^2\colon a\not=b\}$.

\defi{DefTod}{ {\rm \cite[\S6]{To}}
Let $\NN$ be the ideal of subsets of $\R$ consisting either of the meager sets
or of the sets of Lebesgue measure zero.
A set $L\sq\R$ is called {\it
$2$-Lusin}\/ for $\NN$ if $L$ is uncountable, but for every $N\in\NN$,
the set
$N\cap L^{(2)}$ does not contain an uncountable disjoint set, where
two ordered pairs $\la a,b\ra$ and $\la c,d\ra$ 
are {\it disjoint} if $\{a,b\}\cap\{c,d\}=\e$.}

\prop{propTodor}{ Consider the measurable space with negligibles $\la
\R,\bor,\NN \ra$ where $\NN$ is either the ideal of meager sets or the ideal of
sets of Lebesgue measure zero. If $L$ is a $2$-Lusin set for $\NN$ which has 
uncountable intersection with every positive set, then 
$L$ is an $\aleph_1$-strong SRU for $\la\real,\bor,\NN \ra$.
}

\proof
Let $L$ be a $2$-Lusin set and let $f,g\colon\R\to\R$ be nowhere constant
Borel functions. We claim there is a countable set $C_{f,g}$ such that 
$f(x)\not=g(y)$ for any distinct $x,y\in L\setminus C$. 
(Then the rest of Definition~\ref{kStrongSRU} is easily checked.)
By way of contradiction assume that this is not the case.
Inductively choose distinct points 
$x_\alpha,y_\alpha\in L\setminus\bigcup_{\beta<\alpha}\{x_\beta,y_\beta\}$ such
that $f(x_\alpha)=g(y_\alpha)$. The set 
$F=\{\la x,y\ra\in\R^2\colon f(x)=g(y)\}$ is Borel and
contains all the points $\la x_\alpha,y_\alpha\ra$. Since $L$ is $2$-Lusin, it
follows that $F$ is not negligible. By Fubini's theorem (by which we mean
\cite[theorem~15.4]{Ox} if $\NN$ is the meager ideal), there is an $x\in\R$ 
such that $F_x=\{y\in\R\colon\la x,y \ra\in F\}$ is not negligible. But 
$g^{-1}(f(x))=F_x$, contradicting the fact that $g$ is nowhere constant.
\qed

Under CH there is a $2$-Lusin set having uncountable intersection with every
positive set (by a minor modification of the proof of
\cite[Proposition~6.0]{To}).  
It now follows from Proposition~\ref{propTodor}
that there is an $\aleph_1$-strong SRU's for $\la\real,\bor,\NN\ra$
in the iterated perfect set model for $\NN$ equal to either the meager or the
null ideal, and in the model of~\cite{C} for $\NN$ equal to the meager ideal.
The point is that in all of these situations, (i) the ground model coded
negligible Borel subsets of the plane are cofinal in the ideal of negligible
subsets of the plane in the extension and (ii) every positive Borel subset in
the extension contains a ground model coded positive Borel set, and hence a
ground model set which is $2$-Lusin and has uncountable intersection with every
positive set retains these properties in the extension.

\ch % Former example 4.9 deleted and appropriate changes made.

We now give a version of 
Theorem~\ref{th_main} for
several measurable spaces with negligibles simultaneously.

\thm{3.5}{ Let $\la X,\Sg,\NN_i\ra$, $i\in\N$, be $\h_1$-saturated
measurable spaces with negligibles. Suppose that $\Sg$ has
cardinality $\cont$ and, for each $i\in\N$, no set in $\Sg\setminus \NN_i$ can
be covered by less 
than $\cont$ members of $\NN_i$. Then there is a set $M\sq X$ which 
is a strong SRU for all the spaces
$\la X,\Sg,\NN_i\ra$ simultaneously.}

\P Let $\la\la i_\alpha,f_\alpha,g_\alpha,E_\alpha\ra\colon\alpha<\cont\ra$ be a
list of all quadruples $\la i,f,g,E\ra$ where $i\in\N$,
$f,g\colon X\to\R$ are nowhere constant measurable
functions with respect to $\la X,\Sg,\NN_i\ra$, 
$E\in\Sg\setminus\NN_i$, and $f[E]\cap g[E]=\e$.
%; let each such quadruple appear $\cont$ times in the list. 


Inductively choose points $x_\alpha\in X$ so that
$g_\alpha^{-1}(f_\alpha(x_\alpha))\in\bigcap_{i\in\N}\NN_i$, as follows.  
The set
$C_{g_\alpha}=\{y\in\R\colon g_\alpha^{-1}(y)\not\in\bigcap_{i\in\N}\NN_i\}$ is
countable since each $\la X,\Sg,\NN_i\ra$ is $\h_1$-saturated.  Hence
$f_\alpha^{-1}(C_{g_\alpha})\in\NN_{i_\alpha}$. The sets
$g_\beta^{-1}(f_\beta(x_\beta))$, for $\beta<\alpha$ are in
$\bigcap_{i\in\N}\NN_i$ by the 
induction hypothesis. The sets $f_\alpha^{-1}(g_\alpha(x_\beta))$, for
$\beta<\alpha$, are in $\NN_{i_\alpha}$ since
$f_\alpha$ is nowhere constant. 
Thus we may choose a point
$x_\alpha\in E_\alpha$ which avoids all these less that $\cont$ members of
$\NN_{i_\alpha}$. We have 
$g_\alpha^{-1}(f_\alpha(x_\alpha))\in\bigcap_{i\in\N}\NN_i$,
since $x_\alpha\not\in f_\alpha^{-1}(C_{g_\alpha})$.

Now let $i<\N$ and, 
with respect to $\la X,\Sg,\NN_i\ra$, let $f,g\colon
X\to\R$ be nowhere constant measurable functions and
$E\sq X$ a positive set such that $f\restr E\not\equiv g\restr E$.
By shrinking $E$ we may assume that $f[E]\cap g[E]=\e$.
Now, let $\alpha<\cont$ be such that $i_\alpha=i$,
$E_\alpha=E$, $f_\alpha=f$ and $g_\alpha=g$.
\ch
% As in Example~\ref{3.4} 
It is straightforward to verify that $f(x_\alpha)\not\in g[M]$.
\endproof

\rem{simulSRU} {
Consider the spaces $\la X,\Sg,\NN\ra$ where 
$X=\R$, $\Sg=\bor$, and $\NN$ is either the ideal of meager sets or the ideal
of sets of Lebesgue measure zero.

(a) It follows from Theorem~\ref{3.5} that 
if $\R$ can be covered neither by less than
$\cont$ meager sets, nor by less than $\cont$ sets of measure zero then
there is a set $M\sq X$ which is a strong SRU simultaneously
for both spaces under consideration. 
\ch
%(Actually, the argument in Example~\ref{3.4}
%shows that the covering assumption for the meager ideal suffices.)

(b) Another way to get a set which is a strong SRU for both
spaces simultaneously, is to force with an $\h_1$-stage finite
support iteration 
$\la P_\alpha,Q_\alpha\colon \alpha<\omega_1\ra$ where $Q_\alpha$ is a
$P_\alpha$-name for Cohen forcing if
$\alpha$ is even and for random forcing if $\alpha$ is odd. The generic set of
reals
$M=\{r_\alpha\colon \alpha<\omega_1\}$ is a strong SRU for both spaces
simultaneously. Since there are Lusin and 
Sierpi\'nski sets in this model, the
covering assumption of Theorem~\ref{3.5} fails for both ideals if $\cont>\h_1$.
The proof that $M$ is a strong SRU is left as an exercise for the reader.

(c) A set which is a strong SRU for both spaces simultaneously satisfies the
following stronger property. Let $f,g\colon\R\to\R$ be Borel functions. 
If $f[M]\,\triangle\,g[M]$ is
countable, then there is a Borel set $E$ 
such that $f[\R\setminus E]$ and $g[\R\setminus E]$ are countable and 
$f(x)=g(x)$ for all $x\in E$ except for $x$ belonging to a meager set of
measure zero.               
(Compare this with Theorem~\ref{AnyFct}.)

(d) No set 
$M\sq\R$ can be an SRU for the completion of both
spaces simultaneously. The reason is that there is a set $H\sq\R$ of measure
zero and whose complement is meager. One of $M\cap H$, $M\cap(\R\setminus H)$
would have the same cardinality as $M$, say the former. Then given any two
nonequivalent Lebesgue measurable functions 
$f,g\colon \R\to\R$, we can easily
modify $f\restr(M\cap H)$ and $g\restr(M\cap H)$ to arrange $f[M]=g[M]$.
}

The next example shows that an SRU need not be a strong SRU. (The assumptions
on the ideal $\NN$ are satisfied by the ideal of countable sets. They are also
consistently satisfied by the meager and null ideals.)

\ex{sruNotStrong}{Consider a measurable space with negligibles 
$\la X,\Sg,\NN\ra$ in which 
$X=\real$, 
$\Sg$ is the Borel $\sigma$-algebra of $\R$ and 
$\{-x\colon x\in N\}\in\NN$ for every $N\in\NN$.  
Assume that singletons are negligible and no positive set can be covered by
less than $\cont$ negligible sets. Then there is a set $M\sq\real$ which is an
SRU for $\la X,\hat\Sg,\NN \ra$ and such that $\{|x|\colon x\in M\}\sq M$. In
particular, $M$ is not a strong SRU for the piecewise linear functions.}

\P Clearly, such a set $M$ is not a strong SRU, for if we
let $f(x)=|x|$ and $g(x)=x$ for every $x\in\R$, we have that $f$ and $g$ are 
nowhere constant (since $\NN$ contains singletons),
$f[M]\sq g[M]$, and $f\not\equiv g$.

Note that since $\NN$ is a proper ideal, $\R$ is not negligible. Thus, from the
symmetry assumption on $\NN$, neither $(-\infty,0)$ nor $(0,\infty)$ is
negligible.
Let $h\colon\R\to\R$ be given by $h(x)=-x$ for all $x\in\R$.
To construct $M$, 
let $\F$ be the family of all triples $\la f,g,E\ra$ such that
$f$ and $g$ are nowhere constant Borel functions, $E$ is a positive 
Borel set, $f[E]\cap g[E]=\emptyset$ and either
\begin{description}
\item{(i)} $E\sq(0,\infty)$, or 
\item{(ii)} $E\sq(-\infty,0)$  %, $f\restr(0,\infty)\equiv g\restr(0,\infty)$ 
   and $f[E]\cap g[-E]=\emptyset$, where $-E=\{-x\colon x\in E\}$.
\end{description}
Let $\la\la f_\alpha,g_\alpha,E_\alpha\ra\colon\alpha<\cont\ra$ list all
elements of $\F$ with each triple appearing $\cont$ many times
and let $\la N_\alpha\colon \alpha<\cont\ra$ be a list of all negligible Borel
sets. 

By induction on $\alpha<\cont$ define a sequence
$\la x_\alpha\colon\alpha<\cont\ra$ of real numbers such that
\[
x_\alpha\in E_\alpha\setminus
\bigcup_{\beta<\alpha}
\left[N_\beta\cup  f_\alpha^{-1}(f_\alpha(x_\beta))\cup 
f_\alpha^{-1}(g_\alpha(\pm x_\beta))\cup 
\pm[f_\beta^{-1}(g_\beta(x_\beta))]\right].
\]
Such a choice can be made, since
each set
$N_\beta\cup  f_\alpha^{-1}(f_\alpha(x_\beta))\cup 
f_\alpha^{-1}(g_\alpha(\pm x_\beta))\cup 
\pm[f_\beta^{-1}(g_\beta(x_\beta))]$
is negligible and no positive set is the union of less than $\cont$ many
negligible sets. 

Let $M=\{x_\alpha\colon\alpha<\cont\}\cup\{|x_\alpha|\colon\alpha<\cont\}$.
Clearly $\{|x|\colon x\in M\}\sq M$. 

To see that $M$ is an SRU for $\la X,\hat\Sg,\NN \ra$
let $f,g\in\M_{\hat\Sg}(\real)$ be nonequivalent nowhere constant.
Let $N$ be a negligible Borel set on whose complement $f$ and $g$ agree
with $\Sg$-measurable functions $\bar f$ and $\bar g$ respectively. 
In particular, $\bar f$ and $\bar g$ are $\NN$-nowhere constant
and not equivalent.  
Note that 
\[
\mbox{there is a positive Borel set $E\sq X\setminus N$ such that either
$\la \bar f,\bar g,E\ra\in\F$ or $\la \bar g,\bar f,E\ra \in\F$.}
\]

It is clear if $\bar f\restr(0,\infty)\not\equiv\bar g\restr(0,\infty)$
since then $\la \bar f,\bar g,E\ra\in\F$ for any 
positive Borel set $E\sq(0,\infty)\setminus N$ such that
$\bar f[E]\cap\bar g[E]=\emptyset$. So, assume that 
$\bar f\restr(0,\infty)\equiv\bar g\restr(0,\infty)$.
Then $\bar f\restr(-\infty,0)\not\equiv\bar g\restr(-\infty,0)$.
Choose a positive Borel set $E\sq(-\infty,0)\setminus N$ such that
$\bar f[E]\cap\bar g[E]=\emptyset$.
By shrinking $E$, if necessary, we can also assume that
$(\bar f\circ h)\restr E=(\bar g\circ h)\restr E$, since 
$\bar f\restr(0,\infty)\equiv\bar g\restr(0,\infty)$.
Now 
%notice that 
either $\bar f\restr E\not\equiv(\bar f\circ h)\restr E$
or $\bar g\restr E\not\equiv(\bar g\circ h)\restr E$, since otherwise
we would have 
$\bar f\restr E\equiv(\bar f\circ h)\restr E=
(\bar g\circ h)\restr E\equiv \bar f\restr E$.
Assume the former case, the other being similar. 
Then we can shrink $E$ further
%, if necessary, 
to arrange
$\bar f[E]\cap (\bar f\circ h)[E]=\emptyset$.
But $(\bar f\circ h)[E]=(\bar g\circ h)[E]=\bar g[-E]$.
So, $f[E]\cap g[-E]=\emptyset$ and 
$\la \bar f,\bar g,E\ra \in\F$.

By symmetry we can assume that $\la \bar f,\bar g,E\ra\in\F$.
Now, let $\alpha<\cont$ be such that 
$\la \bar f,\bar g,E\ra=\la f_\alpha,g_\alpha,E_\alpha\ra$.
Then, 
$\bar f(x_\alpha)=f_\alpha(x_\alpha)\not\in g_\alpha[M]=\bar g[M]$,
because of the following. 
\begin{description}
\item{--} $f_\alpha(x_\alpha)\neq g_\alpha(\pm x_\beta)$ 
for $\beta<\alpha$ since
$x_\alpha\not\in f_\alpha^{-1}(g_\alpha(\pm x_\beta))$.

\item{--} $f_\alpha(x_\alpha)\neq g_\alpha(x_\alpha)$ 
since $x_\alpha\in E_\alpha$ and 
$f_\alpha[E_\alpha]\cap g_\alpha[E_\alpha] = \emptyset$.

\item{--} $f_\alpha(x_\alpha)\neq g_\alpha(|x_\alpha|)$. This follows
from the previous line if $x_\alpha>0$. If $x_\alpha<0$ 
then $E_\alpha\sq(-\infty,0)$ and
$f_\alpha[E_\alpha]\cap g_\alpha[-E_\alpha]=\emptyset$. So
and $f_\alpha(x_\alpha)\neq g_\alpha(-x_\alpha)=g_\alpha(|x_\alpha|)$,
since $x_\alpha\in E_\alpha$.

\item{--} $f(x_\alpha)\neq g(\pm x_\beta)$ for $\beta>\alpha$ since then
$x_\beta\not\in\pm[g_\alpha^{-1}(f_\alpha(x_\alpha))]$.
\end{description}

Since each $f_\alpha(x_\alpha)\neq f_\alpha(x_\beta)$ for 
every $\beta<\alpha<\cont$ and each element of $\F$ is listed
in our enumeration $\cont$ many times, we conclude that
$\bar f[M]\setminus \bar g[M]$ has cardinality continuum.
But $M\cap N$ has cardinality less than continuum 
since $N=N_\alpha$ for some $\alpha<\cont$. 
Hence, $f[M]\triangle\bar f[M]$ and $g[M]\triangle\bar g[M]$
have cardinality less than $\cont$ and so, 
$f[M]\setminus g[M]$ is nonempty. \qed

When $\la X,\Sg,\NN \ra$ is $\h_1$-saturated, we can give a version 
of Theorem~\ref{th_main} which shows that {\em any}\/ 
measurable function is essentially determined by its range on a strong SRU.
First we prove the following lemma. 

\lem{2.5}{ 
Let $\la X,\Sg,\NN\ra$ be a flexible $\h_1$-saturated measurable space
with negligibles and let $\k$ be a cardinal such
that the union of any collection of less than $\k$ many negligible sets is
negligible. 
If $M\sq X$ is a strong SRU then $M$ has the following properties.
\begin{description} 
\item[(a)] If $f,g\in\M(X)$
are such that $f[M]\setminus g[M]$ has
cardinality less than $\k$ then 
$E=X\setminus(f^{-1}(C_f)\cup\{x\in X\colon f(x)=g(x)\})$ is
negligible, where $C_f=\{y\in\R\colon f^{-1}(y) \mbox{ is positive}\}$.

\item[(b)] If $f,g\in\M(X)$
are such that $f[M]\,\triangle\,g[M]$ has
cardinality less than $\k$,
then there is a measurable set $U$ such 
that $f[X\setminus U]$ and $g[X\setminus U]$ are countable, and
$f\restr U\equiv g\restr U$.
\end{description} 
}  

\P (a) Let $C=f[M]\setminus g[M]$. Note that $C_f$, $C_g$ are
countable and $C$ has cardinality less than $\k$.  
By way of contradiction suppose that $E$ is positive.
Then $f\restr E$ is nowhere constant. 
Let $E_1\sq E$ be a positive set on which the ranges of $f$ and
$g$ are separated by some rational number $q$, say $f(x)<q<g(x)$
for all $x\in E_1$. 
Since the set $E_1\cap f^{-1}(y)$ is negligible for every
$y\in\real$ and $C\cup C_g$ has cardinality less than $\kappa$
we can find a positive subset $K$ of 
$E_1\setminus f^{-1}(C\cup C_g)$.
Let $A=g^{-1}(C_g)$. Define a measurable function $h\colon X\to\R$ so that
$h\restr(K\cup A)$ is any nowhere constant measurable function such
that $h[K]\cap f[K]=\e$, 
and $h$ agrees with $g$
on $X\setminus(K\cup A)$. 
Since $M$ is a strong SRU, there must be a point $x\in M\cap K$
such that $f(x)\not\in
h[M]$. From the definition of $K$, 
$f(x)\not\in C_g= g[A]$. 
So, $f(x)\not\in g[A]\cup g[K]\cup h[M\setminus(K\cup A)]=
g[A\cup K]\cup g[M\setminus(K\cup A)]\supseteq g[M]$. 
Thus $f(x)\in f[M]\setminus g[M]=C$, contradicting $x\in K$.

(b) 
Take $U=X\setminus(f^{-1}(C_f)\cap g^{-1}(C_g))$. By (a) applied as stated,
and also with $f$ and $g$ interchanged, 
$U\setminus \{x\in X\colon f(x)=g(x)\}=
\left(X\setminus(f^{-1}(C_f)\cup\{x\in X\colon f(x)=g(x)\})\right)\cup
\left(X\setminus(g^{-1}(C_g)\cup\{x\in X\colon f(x)=g(x)\})\right)$ 
is negligible.
\endproof

\thm{AnyFct}{
Let $\la X,\Sigma,{\cal N}\ra$ be an $\aleph_1$-saturated measurable
space with negligibles.  Suppose $\cont$ is regular, 
$\Sigma$ has cardinality $\cont$ and
the union of less than $\cont$ negligible sets is negligible.
Then there is a set $L\subseteq X$ such that for any two 
$\hat\Sigma$-measurable functions $f,g\colon X\to\real$, 
$f[L]\triangle g[L]$
has cardinality less than $\cont$ if and only if there is a measurable
set $E$ such that $f[\real\setminus E]$ and $g[\real\setminus E]$ 
are both countable and 
$f\restr E\equiv g\restr E$.
}

\P  First note that by Corollary~\ref{2.8}, we may assume that $X$ is
flexible,
since $f$ and $g$ have essentially countable range on the atomic and Souslin
parts of the space.
Let $L$ be a $\cont$-strong SRU (Theorem~\ref{th_main}). 
Then $L$ has the desired properties
by Lemma~\ref{2.5} and the assumption that 
the union of less than $\cont$ negligible sets is negligible.
\endproof

\rem{RemSaturation}{
The saturation assumption in Theorem~\ref{AnyFct} cannot be deleted. Consider
$X=\R^2$, 
$\Sg$ the Borel $\s$-algebra of $\R^2$, $\NN=$ the ideal of countable subsets
of $X$. 
By Theorem~\ref{th_main}, $X$ has a strong SRU $M$. By 
Proposition~\ref{2.1}(a),
$M\cap B\not=\e$ for every uncountable Borel set $B$. Let $\pi_1$, $\pi_2$ be
the projection 
maps $\R^2\to\R$ onto the first and second coordinates, respectively. Then
$\pi_1[M]=\pi_2[M]=\R$. However, $\pi_1(x)\not=\pi_2(x)$ except when $x$ is
on the main diagonal, and $\pi_1$ and $\pi_2$ are not both constant on any
set with more than one point.
}

The existence theorems we have given for SRU's in measurable spaces
with negligibles deal with structures in which there are only $\cont$
equivalence classes of measurable functions. As was pointed out in
Proposition~\ref{2.2}(c), there can be no SRU for a measurable space
with negligibles which has more than $2^\cont$ pairwise nonequivalent
measurable functions. This leaves the question of what happens when
there are $\k$ equivalence classes of measurable functions and
$\cont^+\leq\k\leq 2^\cont$. The next theorem 
and the remark which follows it
partially address this question. 

\thm{MoreThanC}{ Assume GCH. 
Let $X=2^{\omega_2}$, let
$\Sg$ be the $\s$-algebra of
Baire sets for the usual topology on $X$, and let $\NN$ 
be such that
$\la X,\Sg,\NN \ra$ is a 
measurable space with negligibles (i.e., $\NN$ is a
proper $\sigma$-ideal and $\Sg\cap\NN$ is cofinal in $\NN$). 
There is a countably closed $\h_2$-cc
forcing notion $\poset$ which preserves GCH and such that in the
extension $V^P$ there exists an $\h_1$-strong SRU for $X$.

In particular, it is consistent that there exists 
a strong SRU for a flexible measurable space with negligibles
in which there are $2^\cont$
equivalence classes of measurable functions.}

\proof 
The last statement follows from the rest of the theorem by taking
$\NN$ to be the $\sigma$-ideal of meager sets in $2^{\omega_2}$,
since any two projections onto subproducts of the form
$2^{[\alpha,\alpha+\omega)}\cong 2^\omega$, $\alpha<\omega_1$ a limit ordinal,
are nonequivalent nowhere constant members of $\M(X)$.
(We identify here $2^\omega$ with the Cantor middle third set $C\sq\R$.)
The space $2^\omega$ is flexible by Proposition~\ref{4.6}, since it is
separable.

\ch
We can assume that $X$ is flexible since 
otherwise the theorem is trivial.
 
We start with few remarks on the structure of 
Baire sets and Baire functions in $X$. 
First recall that for each Baire set $B\sq X$, there exists a countable set
$A\sq\omega_2$ on which $B$ ``lives'' in the sense that
\begin{equation}\label{conCod1}
p\in B\ \mbox{ iff }\ q\in B\ \ \ \ \ \ 
\mbox{ for every $p,q\in X$ with $p\restr A=q\restr A$}.
\end{equation}
(See \cite{Ku2} for example.) Similarly, since every Baire 
function $f\in\M(X)$ is fully described by the sets
$f^{-1}((a,\infty))$ ($a\in\Q$) we can find 
a countable set $A\sq\omega_2$
on which $f$ ``lives'' in the sense that
\begin{equation}\label{conCod2}
f(p)=f(q)\ \ \ \ \ \ 
\mbox{ for every $p,q\in X$ with $p\restr A=q\restr A$}.
\end{equation}
Note also that if $f$ and $A$ satisfy (\ref{conCod2}) 
then 
the function $f_A\colon 2^A\to\real$,
$f_A(p\restr A)=f(p)$ for $p\in X$, is well defined 
and it codes $f$. 

Now, for $A\sq\omega_2$ 
and $f\colon 2^A\to\R$ let $\hat f\colon X\to\R$
be defined by $\hat f(p)=f(p\restr A)$. 
Moreover, for $D\sq\omega_2$ let $\F(D)$ be the family 
of all $f\colon 2^A\to\R$ such that $A\in [D]^{\omega}$
and $\hat f$ is a $\Sg$-measurable 
$\NN$-nowhere constant.
When $s\colon C\times D\to\{0,1\}$
and $\gamma\in C$ we will write
$s_\gamma$ for the function
from $D$ into $\{0,1\}$ given by $s_\gamma(\delta)=s(\gamma,\delta)$
for every $\delta\in D$.
Define
\[
\poset=\{\la C,D,s,F\ra\colon 
C\in[\omega_1]^{\leq\omega}\ \&\ 
D\in[\omega_2]^{\leq\omega}\ \&\ 
s\colon C\times D\to\{0,1\}\ \&\
F\in[\F(D)]^{\leq\omega}\}
\]
and define a partial order on $\poset$ by 
$\la C,D,s,F\ra\leq\la C^\prime,D^\prime,s^\prime,F^\prime\ra$
provided $C^\prime\sq C$, $D^\prime\sq D$, $s^\prime\sq s$, 
$F^\prime\sq F$, and
for every $\gamma\in C\setminus C^\prime$ and 
$\alpha\in C\setminus\{\gamma\}$
\begin{equation}\label{con1}
(\forall A\in[D^\prime]^\omega)
(\forall f,g\in F^\prime)[\dom(f)=\dom(g)=2^A\ \implies\ 
f(s_\gamma\restr A)\not=g(s_\alpha\restr A)].
\end{equation}

It is easy to see that $\poset$ is countably closed. 
$\poset$ is $\h_2$-cc since, by standard $\triangle$-system arguments, for any
sequence of conditions
$\la\la C^\xi,D^\xi,s^\xi,F^\xi\ra\colon \xi<\omega_2\ra$
we can find $\xi<\zeta<\omega_2$ such that
$C^\xi=C^\zeta$ and $s^\xi\cup s^\zeta$ is a function.
(See~\cite[Ch.~VII,~sec.~6]{Ku1}.)
Then 
$\la C^\xi,D^\xi\cup D^\zeta,s^\xi\cup s^\zeta,F^\xi\cup F^\zeta\ra\in\poset$
extends both 
$\la C^\xi,D^\xi,s^\xi,F^\xi\ra$
and
$\la C^\zeta,D^\zeta,s^\zeta,F^\zeta\ra$
since condition (\ref{con1}) is viciously satisfied.
Thus, $\poset$ preserves cardinals and does not add any new 
countable sequences of ground model elements. In particular, $V^\poset$ 
contains neither any new real numbers nor any new code $f_A$ of 
any Baire functions $f$. 

Let $G\sq\poset$ be $V$-generic and let
\[
x=\bigcup\{s\colon\la C,D,s,F\ra\in G\}. 
\]
Clearly $x$ is a function from a subset of
$\omega_1\times\omega_2$ into 
$2$. To see that 
$\dom(x)=\omega_1\times\omega_2$
it is enough to notice that for every $\gamma<\omega_1$
and $\delta<\omega_2$ the sets $\{\la C,D,s,F\ra\in\poset\colon\gamma\in C\}$
and $\{\la C,D,s,F\ra\in\poset\colon \delta\in D\}$ are dense in $\poset$. For
the latter this is trivial. 
To see it for the former case, let $\la C,D,s,F\ra\in\poset$ and 
$\gamma\in\omega_1$. Pick 
\[
p\in X\setminus
\bigcup\{{\hat f}^{-1}(g(s_\alpha\restr A))\colon 
\alpha\in C\ \&\ f,g\in F\ \&\ 
\dom(g)=2^A\}.
\]
Then $\la C\cup\{\gamma\},D,t,F\ra\in\poset$ with
$t\restr{C\times D}=s$ and $t_\gamma=p\restr D$ extends 
$\la C,D,s,F\ra\in\poset$.

Let $L=\{x_\gamma\colon \gamma<\omega_1\}$ and notice that genericity 
easily imply that all $x_\gamma$'s are different. 
The proof is completed by verifying that for each pair $f,g\in\M_{\Sg}(X)$ of
nowhere constant functions, there is a countable set $C_{f,g}$ of
$L$ such that assumptions (i) and (ii) of
Definition~\ref{kStrongSRU} are satisfied with $\kappa=\h_1$.
Let $f,g\in\M_{\Sg}(X)$ be nowhere constant functions, let $E$ be a positive
Baire set, and let $N$ be a negligible Baire set. 
Let $A\sq\omega_2$ be countable and such that $\la f,A\ra$
and $\la g,A\ra$ satisfy (\ref{conCod2}) and choose a condition 
$\la C,D,s,F\ra\in G$ 
such that $f_A,g_A\in F$. (Such conditions are trivially dense.) 
Define $C_{f,g}=\{x_\gamma\colon \gamma\in C\}$. It is clear from
the definition of the order on $\poset$ that 
Definition~\ref{kStrongSRU}(ii) is satisfied.
Since $\kappa$ is regular, Definition~\ref{kStrongSRU}(i) is equivalent to:
$L\cap E$ is 
uncountable for each positive Baire set $E$ and that $L\cap N$ is countable 
for each negligible Baire set $N$. 
To verify this, let $E$ be a positive Baire set,
let $N$ be a negligible Baire set and let 
and let $\alpha<\omega_1$. 
Choose a countable set 
$A\sq\omega_2$ such that $\la B,A\ra$
and $\la N,A\ra$ satisfy (\ref{conCod1}). 
The desired properties follow easily from the density of the conditions
$\la C,D,s,F\ra$ such that $A\sq D$ and
$s_\gamma\in\{p\restr D\colon p\in E\}$ for some $\gamma\in C\setminus\alpha$, 
and the density of the conditions $\la C,D,s,F\ra$
such that $0\in C$ and that there are $f,g\in F$ 
with 
$\dom(f)=\dom(g)=2^A$ 
and $N={\hat f}^{-1}(g(s_0\restr A))$.
(Such an $\hat f$ exists since $X$ is flexible.)
\qed

\rem{AllSetsMble} {The measurable space with negligibles 
$\la\cont,P(\cont),\NN \ra$, 
where $\NN$ is the ideal of countable subsets of
$\cont$, has no SRU by Proposition~\ref{2.2}(a). 
Thus, in Theorem~\ref{th_main}, 
the cardinality restriction on $\Sigma$ cannot be relaxed to
$2^\cont$. If $\kappa$ is an atomlessly measurable cardinal and $\NN$ is the
null ideal of a witnessing measure, then $\la \kappa,P(\kappa),\NN \ra$ is an
$\aleph_1$-saturated measurable space with negligibles which illustrates the
same point.
}

\section{ SRU's for continuous functions}\label{secCont}

In this section we will examine properties of SRU's for $\C(X)$,
for $X$ a Baire 
topological space considered with its natural structure as a
measurable space with negligibles. 
(See Section~\ref{Prelim}).
Many of the results do not rely on $X$ being Baire, however. 
(See Remark~\ref{NotBaire}.)
Recall, that by Proposition~\ref{prop4_2} we can replace
the relation $\equiv$ in the definition of (strong) SRU with equality.
Proposition~\ref{prop4_2} and Theorem~\ref{th_main} ensure that, under suitable
hypotheses, strong SRU's exist in many Baire spaces.  We will need to assume,
for most of our results, that our spaces are locally flexible. 
Proposition~\ref{4.6} provides us with a healthy supply of such spaces.



We begin with several easy lemmas.

\lem{lemImage}{ Let $X$ be a Baire 
topological 
space considered with its natural measurable structure and let
$M\sq X$ be an SRU for $\C(X)$.
If $f\in\C(X)$ is nowhere constant, $K=f[M]$ and
$h\colon\real\to\real$ is a homeomorphism such that
$h[K]=K$ then $h(y)=y$ for every $y\in f[X]$.}

\proof 
Let $f$, $h$ and $K$ be as above
and by way of contradiction assume that
$h(y)\neq y$ for some $y\in f[X]$.
Then $g=h\circ f\neq f$ while $g\in\C(X)$ 
is nowhere constant and
$g[M]=h[f[M]]=h[K]=K=f[M]$, contradicting
the definition of an SRU for $\C(X)$. \qed


\lem{4.10}{ Let $a<b$ and $K\sq(a,b)$ be a Cantor set.
Then, every continuous function $g\colon K\to(a,b)$ has
an extension $G\colon\R\to\R$ such that $G[(a,b)]=(a,b)$,
$G(x)=x$ for all $x\in\R\setminus(a,b)$ and $G$ is 
countable-to-one on $\real\setminus K$.}

\P 
Let $c$ and $d$ be the maximum and the minimum of $K$, respectively.
Extend $g$ to a continuous function $g_1\colon[c,d]\to(a,b)$
by defining it linearly on any component of $[c,d]\setminus K$.
Let $U=\bigcup\{\inter(g_1^{-1}(y))\colon y\in[c,d]\}$.
If $f\colon [c,d]\to\R$ is the distance function from $[c,d]\setminus U$
then it is easy to find a constant $k>0$ such that 
$G=(g_1 + k f)\colon [c,d]\to(a,b)$. 
It is also not difficult to check that $G$ is continuous and 
countable-to-one on $[c,d]\setminus K$.
By extending it to the identity function on
$\R\setminus(a,b)$ and linearly on each 
of the intervals $(a,c)$ and $(d,b)$
we obtain the desired function.
\endproof 

\lem{LemComp}{ Let $X$ be a Baire topological space. Let $f\in\C(X)$ and
$g\in\C(\R)$ be nowhere constant.  
If either $X$ is locally connected or $g$ is countable-to-one, then $g\circ
f\in\C(X)$ is also nowhere constant.
}

\P 
If $g$ is countable-to-one, then
$(g\circ f)^{-1}(y)=\bigcup\{f^{-1}(z)\colon z\in g^{-1}(y)\}$
is meager, as it is a countable union of nowhere dense sets.

If $X$ locally connected and $(g\circ f)^{-1}(y)$ has nonempty interior, 
then there exists a nonvoid open connected
$U\sq (g\circ f)^{-1}(y)=f^{-1}(g^{-1}(y))$.
So, $f[U]\sq g^{-1}(y)$. 
But $f[U]\sq\R$ is connected, as an image of a connected set. So, either
$f[U]$ is a singleton, contradicting that $f$ is nowhere constant,
or $f[U]$ contains a nonvoid open interval, contradicting the fact
that $g^{-1}(y)$ is nowhere dense. \qed

\rem{RemComp}{
Notice also that Lemma~\ref{LemComp} 
may fail if we require only that $g$ is nowhere constant.
To see this, let $K\sq\R$ be a Cantor set and 
let $X=K\times K$. 
Moreover, let $c\colon K\to\{0\}$ be the constant map and let
$G$ be the extension of $c$ given by Lemma~\ref{4.10}. Then the conclusion
of Lemma~\ref{LemComp} fails for $g=G$ and $f=$ the projection of $X$ onto the
first coordinate.}


Recall from Section~\ref{Prelim} that $S\sq X$ is an {\em $s_0$-set}\/ if for
every Cantor set $P\sq X$, there is a Cantor set $Q\sq P$ such that $Q\cap
S=\e$. Also, $S$ is a {\em strong $s_0$-set}\/ if $f[S]$ is an $s_0$-set in
$\R$ for every $f\in\C(\R)$. 
Note that every set of cardinality less than $\cont$ is a strong $s_0$-set.
There are $s_0$-sets of cardinality $\cont$ in $\R$ \cite{Mi2}, but it is
consistent that there are no strong $s_0$-sets of cardinality $\cont$
\cite{Mi3}. 

\lem{S_0}{If $S\sq X$ is a strong $s_0$-set, then
$S$ is an $s_0$-set and $S$ is zero-dimensional.}


\P  The zero-dimensionality of $S$ follows easily from the fact that $X$ is
completely regular and the image of $S$ by a member of $\C(X)$ cannot contain
an interval. To see that $S$ is an $s_0$-set, let $K\sq X$ be a Cantor set, and
let $g\colon K\to\R$ be a homeomorphism onto its range. Since $K$ is
compact, $g$ extends to a function $G\in\C(\R)$ \cite[Exercise~3.2.J]{En}.
Since $G[S]$ is an $s_0$-set, there is a Cantor set $L\sq G[K]$ such that
$L\cap G[S]=\e$. Then $S\cap(G^{-1}(L)\cap K)=\e$. \qed


\thm{thm:MP}{ Let $X$ be a locally flexible
Baire topological 
space considered with its natural measurable structure.
If $M\sq X$ is an SRU for $\C(X)$ then $M$ has the
following properties.
\begin{description}
\item[(1)] $M$ is dense in $X$.
\item[(2)] Let $U\sq X$ be nonvoid open and let $f\in\C(X)$ be nowhere
           constant. 
           Then $f[M\cap U]$ is uncountable provided at least one of the
           following conditions hold.
           \begin{description}
                 \item[(a)] $U=f^{-1}(W)$ for some open $W\sq\R$;
                 \item[(b)] $M$ is a strong SRU.
           \end{description}
\item[(3)] $M\cap U$ is uncountable for every nonempty open set $U\sq X$.
\item[(4)] For every $f\in\C(X)$ there is a nowhere constant $\bar f\in\C(X)$
           such that $f[M]\sq\bar f[M]$.
\item[(5)] $M$ is a strong $s_0$-set. In particular, $M$ is
           a zero-dimensional $s_0$-set.
\item[(6)] If $X$ is a nonvoid analytic metric space
           then $M$ is not analytic.
\end{description}
}
 
\proof (1) By way of contradiction, assume that 
there is a nonvoid open set $U\sq X$ disjoint 
from $M$. Let $V=X\setminus\cl(U)$ and let $h_U$ and $h_V$
be functions from the definition of local flexibility
for $U$ and $V$ respectively.
Define $f=h_V+h_U$ and  $g=h_V-h_U$. Then 
$f$ and $g$ are continuous, nowhere constant.
Moreover, $f[M]=h_V[M]=g[M]$, while $f\neq g$, since
$h_U$ is nowhere constant on $U$. 

(2) 
By way of contradiction, assume that
$f[M\cap U]$ is countable for some 
nonvoid open $U$ and nowhere constant $f\in\C(X)$.
Let $Y=f[M\cap U]$.

First notice that
$Y$ has no isolated points. To see this, assume by way of contradiction 
that $Y$ has an isolated point $y$ and let 
$I\sq\R$ be open such that $I\cap Y=\{y\}$.
Then $V=U\cap f^{-1}(I)$ is nonempty and, by~(1),  
$f[V]\sq f[\cl(M\cap V)]\sq\cl(f[M\cap V])=\{y\}$
contradicting the fact that $f$ is nowhere constant. 

Next notice that 
\begin{description}
\item[($\star$)] there exists $a<b$ 
                 and a countable-to-one 
                 continuous 
                 function 
                 $h\colon\real\to\R$ such that
                 $h[Y]=Y$, $h(y_0)\neq y_0$ for some $y_0\in Y$,
                 and $h(x)=x$ for $x\in\R\setminus(a,b)$; 
                 moreover, $(a,b)\sq W$ if we are in the case~(a).
\end{description}

Note that this will finish the proof. Indeed,
put $g=h\circ f\in\C(X)$ and notice that $f\neq g$ on $U$, since
$U\cap f^{-1}(y_0)\neq\emptyset$.
Moreover, by Lemma~\ref{LemComp}, $g$ is nowhere constant
and $g[M\cap U]=h[f[M\cap U]]=h[Y]=Y=f[M\cap U]\sq f[M]$.
This gives a contradiction with $M$ being strong SRU, taking care of~(b).

If $U=f^{-1}(W)$ then
$g[M\setminus U]=h[f[M\setminus U]]=f[M\setminus U]
\sq f[M]$,
since $h(x)=x$ for all $x\in f[M\setminus U]\sq\R\setminus (a,b)$.
So, $g[M]=f[M]$
giving us a contradiction with (a) as well.

To prove ($\star$) consider two cases.

{\bf Case 1.} $\cl(Y)$ is somewhere dense. Let $(a,b)\sq\cl(Y)$
be a nonvoid open interval and assume that $(a,b)\sq W$ if $U=f^{-1}(W)$.
Then, there is
a nontrivial homeomorphism $h\colon\real\to\R$ such that
$h[Y]=Y$ and $h(x)=x$ for every $x\in\R\setminus(a,b)$.
Clearly $h(y)\neq y$ for some $y\in Y$.

{\bf Case 2.} $\cl(Y)$ is nowhere dense.
Let $a<b$ be such that $K=\cl(Y)\cap(a,b)$ 
is a Cantor set and that $(a,b)\sq W$ if $U=f^{-1}(W)$.
Let $g\colon K\to K\sq(a,b)$ be a homeomorphism such that
$g[K\cap Y]=K\cap Y$ and 
$g(y)\neq y$ for some $y\in K\cap Y$.
(See \cite[Exercise~4.3.H(3)]{En} for such a homeomorphism.)
Let $G$ be an extension of $g$ as in Lemma~\ref{4.10}.
Then $h=G$ satisfies~($\star$).

(3) By way of contradiction, assume that $M\cap U$ is countable for
some open nonempty $U\sq X$. For an open set $W\sq X$ let $f_W$
stand for the absolute value of the function from the 
definition of local flexibility for $W$. Put
$f=f_U-f_{\inter(X\setminus U)}$. Then $f$ is continuous and
nowhere constant, and $U\supset f^{-1}((0,\infty))\neq\emptyset$. 
An application of (2) case (a) gives a contradiction.  

(4) 
Let $f\in\C(X)$ and let $C_f=\{y\in\R\colon\inter(f^{-1}(y))\neq\emptyset\}$. 
By (1) for each $y\in C_f$ we can choose $x_y\in\inter(f^{-1}(y))\cap M$. 
Note that the set 
$H=[X\setminus\bigcup\{\inter(f^{-1}(y))\colon y\in C_f\}]
\cup\{x_y\colon y\in C_f\}$ is closed in $X$.
Let $g\in\C(X)$ be as in the definition of local flexibility of $X$ 
for the set $X\setminus H$ and let $\bar f=f+g\in\C(X)$.
Then $\bar f$ is nowhere constant and
$f[M]=f[M\cap H]=\bar f[M\cap H]\sq\bar f[M]$.

(5) 
The claims made in the second statement follow from Lemma~\ref{S_0}.
For the proof of the first statement, let us
first verify that if $f\in\C(\R)$, we cannot have a Cantor set contained in
$f[M]$. By way of contradiction, assume that there is a Cantor set
$K\sq f[M]$.
By (4) we can assume that $f$ is nowhere constant.
Let $g$ be a continuous two-to-one function from $K$ onto $[0,1]$
and let $G$ be an extension of $g$ as in Lemma~\ref{4.10}.
Then, $G\in\C(\R)$ is countable-to-one and, by Lemma~\ref{LemComp},
$F=G\circ f$ is nowhere constant. 
But then, $F[M]=G[f[M]]\supset G[K]\supset[0,1]$,
which contradicts Lemma~\ref{lemImage},
since there are many nontrivial homeomorphisms $h$ of $\R$ 
with $h(x)=x$ for all $x\in\R\setminus(0,1)$.

Now suppose that $f[M]\cap L\not=\e$ for every Cantor set
$L\sq K$. Since $K$ is homeomorphic to its square, and the level sets of the
projection of the square onto one of the coordinates are all Cantor sets, 
there is a continuous map $g\colon K\to K$ all of whose level sets are Cantor
sets. Extend this function $g$ to a member $G$ of $\C(\R)$ and note that
$(G\circ f)[M]\supseteq K$, contradicting the result established in the
previous paragraph.

(6) This follows immediately from (3) and (5).
\endproof

\pr{prAB}{ Are the assumptions (a) and (b) essential in (2) 
of Theorem~\ref{thm:MP}?}

We will now consider some specific properties of SRU's
for $\C(X)$ where $X$ is a separable metric space. We begin with
following lemma. 

\lem{lemSRUnotMeager1}{ Let $\la X,d \ra$ be a separable metric space, and let
$M\sq X$ a strong $s_0$-set. 
If $F$ is a closed subset of $X$ and $\ep>0$ then there exists
a continuous function $g\colon X\to[0,\ep]$ such that 
$g^{-1}(\{0\})=F$ and $g[M]$ is countable.}

\proof 
Let $h\colon X\to[0,\ep]$ be given by the formula
$h(x)=\min\{\ep,d(x,F)\}$ for $x\in X$ and for
$n<\omega$ let $I_n=[\ep/2^{n+1},\ep/2^n]$.
Since $M$ is a strong $s_0$-set, 
for every $n<\omega$ there exists a Cantor set
$C_n\sq I_n$ such that $C_n\cap h[M]=\e$. 
Let $f_n\colon I_n\to I_n$ be a non decreasing Cantor function
such that $f_n[C_n]=I_n$ 
and let $f\colon [0,\ep]\to[0,\ep]$ be an extension of all these functions. 
It is easy to see that $h=f\circ g\colon X\to[0,\ep]$ has the desired
properties. \qed

\lem{lemSRUnotMeager2}{ 
Let $\la X,d \ra$ be a separable Baire 
metric space without isolated points. 
Let $U\sq X$ be open and let $M\sq U$ be a meager strong $s_0$-set.
Then there is a function
$g\in\C(X)$ such that $g(x)=0$ for all $x\in X\setminus U$, $g(x)\geq 0$ for
all $x\in X$, $g\restr U$ is nowhere constant, and $g[M]$ is countable.}

\P
Let $K_n$ be closed nowhere dense
sets in $X$ such that $M\sq\bigcup_{n<\omega} K_n$. 
Since $X$ is Baire, there is a set
$\{d_n\colon n<\omega\}\sq U\setminus \bigcup_{n<\omega} K_n$ which is dense in
$U$. We will construct,
by induction on $n<\omega$,
a sequence $g_n\colon X\to[0,\ep_n)$ of continuous functions
such that for every $n<\omega$
\begin{description}
\item{(i)} $g_n[M]$ is countable

\item{(ii)} $g_n^{-1}(\{0\})=(X\setminus U)\cup\bigcup_{i<n}(K_i\cup\{d_i\})$

\item{(iii)} $\ep_n\in(0,2^{-n})$

\item{(iv)}  
$\left(\sum_{i<n}g_i\right)(d_j)\not\in
\left(\left(\sum_{i<n}g_i\right)(d_n),
\left(\sum_{i<n}g_i\right)(d_n)+\ep_n\right]$ for $n>0$ and $j<n$. 
\end{description}

The construction is easily carried out. It is simply a matter of
choosing $\ep_n$ satisfying (iii) and (iv), and then 
defining $g_n$ to be the function
$g$ from Lemma~\ref{lemSRUnotMeager1} applied with 
$F=(X\setminus U)\cup \bigcup_{i<n}(K_i\cup\{d_i\})$ and $\ep=\ep_n$. 

Let $g=\sum_{i<\omega}g_i$. Then, by condition (iii), $g$ is continuous.
Also, condition (ii) implies that for every $n<\omega$
\[
g[M\cap K_n]=\left(\sum_{i\leq n}g_i\right)[M\cap K_n]
\]
which is countable by (i). 
So, $g[M]$ is countable. 
Moreover, by (ii) and (iv), 
\[
g(d_j)=\left(\sum_{i<n}g_i\right)(d_j)\not\in
\left(\left(\sum_{i<n}g_i\right)(d_n),
\left(\sum_{i<n}g_i\right)(d_n)+\ep_n\right]
\ni \left(\sum_{i\leq n}g_i\right)(d_n)=g(d_n)
\]
for every $j<n<\omega$. So $g$ is nowhere constant in $U$.
\qed

\thm{SRUnotMeager}{Let $\la X,d \ra$ be a separable 
Baire metric space without isolated points. If $M\sq X$ is an SRU for $\C(X)$
and $U\sq X$ is a nonvoid open set, then $M\cap U$ is not meager.}

\P Let $M$ be an SRU for $\C(X)$. By Theorem~\ref{thm:MP}(5), $M$ is a strong
$s_0$-set. 
If $M\cap U$ is meager, then Lemma~\ref{lemSRUnotMeager2} gives a
nonnegative function $g_1\in\C(X)$ which is identically equal to zero outside
$U$, nowhere constant on $U$, and has a countable image of $M\cap U$. Let
$g_2\in\C(X)$ be a nonpositive function which is identically equal to zero on
$U$ and nowhere constant in the exterior of $U$. (For example, apply 
Lemma~\ref{lemSRUnotMeager2} with $U$ replaced by the exterior of $U$ and $M$
replaced by the empty set.) Let $g=g_1+g_2$. Then $g\in\C(X)$ is nowhere
constant. Let $W=g^{-1}((0,\infty))$. We have $g\restr W=g_1\restr W$ and hence
$W\sq U$. Also, $g[M\cap W]\sq g_1[M\cap U]$ is countable, contradicting
Theorem~\ref{thm:MP}(2)(a). \qed

\cor{5.4}{ A Sierpi\'{n}ski subset of $\R^n$ is not an SRU for
$\C(\R^n)$.}

\P Sierpi\'{n}ski sets are meager. \endproof

\rem{rem3}{ Theorems~\ref{SRUnotMeager} and~\ref{thm:MP}(5) show that
an SRU for $\C(\R)$ cannot be meager, but also cannot be too big. 
However, little can be said about the measure of an SRU for $\C(\R)$. 
Under CH, there is a strong SRU for $\C(\R)$ which is a Lusin set and hence has
strong measure zero and there is another one of full outer measure. (For the
full outer measure example, apply Theorem \ref{3.5} to $\la \R,\bor,\NN_i
\ra$, $i=1,2$, where $\NN_1$ and $\NN_2$ are the meager and null ideals.)}

\rem{rem333}{ It is not difficult to see that one can prove 
Theorems~\ref{SRUnotMeager} 
for any Baire space without isolated points in which
any dense $G_\delta$ set contains a countable dense subset. 
The proof requires only minor changes. This could be used
for example to show that any SRU for
$[0,1]^{\continuum}$ must be nowhere meager, a fact which also follows from the
next proposition.}

\prop{NonMeagerSRUprod}{ Let 
$X$ be a ccc topological space such that $X^\omega$ is 
\ch Baire.
%flexible.
If every SRU for 
\ch $\C(X^\omega)$ 
is nowhere meager then 
so is every SRU for 
\ch $\C(X^\kappa)$ 
for every infinite cardinal $\kappa$. }

\proof By way of contradiction assume that for some infinite 
cardinal number $\kappa$ there is $M\sq X^\kappa$ which is 
an SRU for 
\ch $\C(X^\kappa)$ and such that 
$U\cap M$ is meager for some nonvoid open set $U\sq X^\kappa$. 
Decreasing $U$, if necessary, we can assume that 
$U$ is a basic open set supported by
a finite set $T$ of coordinates.
Since $X^\kappa$ is ccc we can find an $F_\sigma$ meager set $F$
supported by a countable infinite set $S$ of coordinates
and such that $U\cap M\sq F$. We can also assume that 
$T\sq S$. 
Let $\pi$ be the projection map
of $X^\kappa$ onto $X^S$. Thus, $\pi[M\cap U]\sq\pi[F]$
is meager in $\pi[U]$. 
So, by our assumption, $\pi[M]$ is not an SRU for $X^S$,
i.e., there are two different nowhere constant
functions $f,g\in\C(X^S)$ with $f[\pi[M]]=g[\pi[M]]$. 
But $f\circ\pi,g\circ\pi\in\C(X^\kappa)$ are 
different and nowhere constant too, contradicting 
the assumption that $M$ was an SRU for $X^\kappa$. \qed

\cor{corNonMeagerSRU}{ For any cardinal $\kappa$
every SRU for $[0,1]^\kappa$ is nowhere meager 
in $[0,1]^\kappa$. \qed}

\rem{NotBaire}{As we pointed out in Section~\ref{Prelim}, if $X$ is a Baire
topological space, the usual notion of ``nowhere constant'' for a function
$f\in\C(X)$ 
(i.e., ``not constant on any nonvoid open set'') 
coincides with the meaning ``not constant on any
nonmeager Baire set'' we gave in Section~\ref{Prelim}.  
Most of the results in
Section~\ref{secCont} go through even for spaces that are not Baire, if the
usual notion 
of ``nowhere constant'' is used in the definitions of SRU and of locally
flexible. Lemma~\ref{LemComp} for the case where $g$ is
countable-to-one holds by a different argument: 
the level sets of $g$ are scattered and it is easy to
verify by induction on the Cantor-Bendixson height of a scattered set that its
preimage under a nowhere constant function is nowhere dense. 
The proof of Lemma~\ref{lemImage} goes through with no changes, as does the
proof of Theorem~\ref{thm:MP}. As for Theorem~\ref{SRUnotMeager}, note that
if $X$ is any nonvoid separable metric space with no isolated points and for
some open set $U\sq X$, $M\cap U$
is meager, then by Theorem~\ref{SRUnotMeager} applied to the completion $Y$
of $X$, there are two distinct nowhere constant functions $f,g\in\C(Y)$ such
that $f[M]=g[M]$. But then the restrictions of $f$ and $g$ to $X$ witness that
$M$ is not an SRU. In particular, if $X$ itself has a nonvoid meager open set
(i.e., is not Baire),
then there is no SRU for $\C(X)$. 
}

The next example shows that the assumption of CH cannot be removed from
\ch % the theorem from 
\cite[theorem~8.5]{BD}.

\ex{5.5}{ Suppose that 
\begin{description}
\item[(i)]  there is a nonmeager subset of $\R$
            of cardinality $\h_1$; and, 
\item[(ii)] any two nowhere meager subsets of $\R$ of cardinality
            $\h_1$ are order isomorphic. 
\end{description}
(See \cite{Sh} for a proof that this is consistent.) 
If $X\sq\R$ is a nowhere meager subset of $\R$ of 
cardinality $\h_1$ then there is no SRU for $\C(X)$.}

\P Suppose $M\sq X$ were an SRU for $\C(X)$. By 
Theorem~\ref{SRUnotMeager}
$M$ is nowhere meager. But then, using (ii),
it is easy to construct many distinct order isomorphisms of $M$ with itself.
Each of these will extend to a homeomorphism of $\R$, and hence $M$ is not an
SRU. \endproof

The models we are aware of for assumptions (i) and (ii) of 
Example~\ref{5.5} are obtained by 
$\k$-stage finite support iterations of ccc forcing notions, where $\k$ is
regular. As a result, Cohen reals are added cofinally in the construction and
the covering assumption of 
Theorem~\ref{th_main}
is satisfied for $X=\R$, $\Sg=$
the Borel $\s$-algebra, and $\NN=$ the meager ideal. Thus, there is a strong
SRU for $\C(\R)$ in these models.

The following remains an intriguing problem.

\pr{prAAA}{ Is the existence of an SRU for $\C(\R)$ provable in ZFC?}

We now give the example promised in 
Section~\ref{Prelim} to show that the
converse to the first statement of Proposition~\ref{prop4_2} is false.
First we prove the following lemma.
    
\lem{IrrHom} {Let $X$ be the space of irrational numbers (with
the usual topology). There is a countable dense set $A\sq X$ and a
homeomorphism $h\colon X\setminus A\to X\setminus A$ 
such that for any $m\in\Z\setminus \{0\}$ and any nowhere constant function
$g\in \C(X)$, 
$g\circ h^m\colon X\setminus A\to X\setminus A$ does not extend to a
continuous function on any nonempty open subset of $X$.} 

\P 
Fix any countable dense set $A\sq\R$. 
We define open subintervals 
$I_s=(a_s,b_s)$ of $\R$ for $s\in\Z^{<\omega}$ as follows.
Let $I_{\e}=\R$ and given $I_s$, define open subintervals 
$I_{s^\frown n}$, $n\in\Z$, of $I_s$ so that 
$I_s\setminus \bigcup_{n\in\Z}I_{s^\frown n}\sq A$, 
and $b_{s^\frown n}=a_{s^\frown (n+1)}$. 
(For $s\in\Z^{<\omega}$  and $n\in\Z$ the symbol $s^\frown n$
denotes the extension of $s$ by $n$, i.e.,
$s^\frown n=s\cup\{\la k,n\ra\}$, where $k=\dom(s)$.)
Also ensure that if we let
$U_n=\bigcup\{I_s\colon s\in\Z^n\}$, then 
$\bigcap_{n\in\omega}U_n=\R\setminus A$. 
Identify $\Z^\omega$ with $\R\setminus A$ via the map which sends
$u\in\Z^\omega$ to the unique member of 
$\bigcap\{I_{u\srestr n}\colon n\in\omega\}$. Let $h\colon\R\setminus
A\to\R\setminus A$ be 
the homeomorphism which corresponds via this identification to the
homeomorphism of $\Z^\omega$ such that for $u\in\Z^\omega$ 
and $k\in\Z$ we have 
$h(u)(k)=u(k)$ if $u(k)$ is odd, and $h(u)(k)=u(k)-2$ if $u(k)$ is even. Let 
$B\sq\R\setminus A$ 
be a countable dense set such that $h[B]=B$. We may assume
$X=\R\setminus B$ since this set is homeomorphic to the set of irrational
numbers. Note that $A\sq X$ and that $h$ induces a homeomorphism of 
$X\setminus A$. 
We will show that $X$, $h$ and $A$ satisfy the conclusion of the lemma.

Let $g\in \C(X)$ and $m\in\Z\setminus\{0\}$ 
be such that $g\circ h^m$ extends to a continuous function on
a nonvoid open subset $I\sq X$. 
By shrinking $I$, we may assume that 
$I=I_{s_0}\cap X$ for some ${s_0}\in\Z^{<\omega}$.
Let ${t_0}\in\Z^{<\omega}$ be such that $h^m[I_{s_0}]=I_{t_0}$.
We will show that $g$ is constant on $I_{t_0}$. 
First, we claim that for every $t\in\Z^{<\omega}$ such that $t_0\sq t$,
\[
\mbox{$g$ is constant on the set 
$S_t=\{a_{t^\frown(2m\ell)}\colon \ell\in\Z\}$.}
\]

So, let $\ell\in\Z$, $x=a_{t^\frown(2m\ell)}$ and $y=a_{t^\frown(2m(\ell+1))}$.
We will show that $g(x)=g(y)$. 
Let $s\in\Z^{<\omega}$ be such that $h^m[I_s]=I_t$.
Note that $s_0\sq s$, so $g\circ h^m$ extends to a continuous function on
$I_s\sq I_{s_0}$. 
For $z\in I_{s^\frown (2m(\ell+1))}\cap(X\setminus A)$, as $z\to
a_{s^\frown(2m(\ell+1))}$, $h^m(z)\to a_{t^\frown(2m\ell)}$ and hence
$g(h^m(z))\to g(a_{t^\frown(2m\ell)})=g(x)$.
For $z\in I_{s^\frown (2m(\ell+1)-1)}\cap(X\setminus A)$, as $z\to
a_{s^\frown(2m(\ell+1))}=b_{s^\frown(2m(\ell+1)-1)}$, $h^m(z)\to
b_{t^\frown(2m(\ell+1)-1)}=a_{t^\frown(2m(\ell+1))}$ and hence
$g(h^m(z))\to g(a_{t^\frown(2m(\ell+1))})=g(y)$.
Thus, we must have $g(x)=g(y)$, as desired.

Note that for each $t\supseteq t_0$ and each $n\in\Z$, the constant values of
$g$ on $S_{t^\frown n}$ and $S_{t^\frown(n+1)}$ must be the same since these
two sets share a cluster point. Let $c$ be the constant value taken by $g$ on 
$T_1$ where for each $m\in(\dom(t),\omega)$ we let $T_m=\bigcup\{S_t\colon 
t\supseteq t_0,\,t\in\Z^m\}$. 
By similar considerations to the case $m=1$, it follows by
induction on $m$, using the fact that $T_{m+1}$
has cluster points in $T_m$, that $g$ has the constant value $c$ on each $T_m$.
Since $\bigcup\{T_m\colon m\in(\dom(t),\omega)\}$ is dense in $I_{t_0}$,
$g$ is constant on $I_{t_0}$. \endproof

\ex{sruContNotBorel} {If $\R$ cannot be covered by less than $\cont$ meager
sets, then in any nonvoid perfect Polish space~$X$, there is a set $M\sq X$ and
a Borel isomorphism $h\colon X\to X$ such 
that $h[M]=M$, $\{x\in\R\colon h(x)=x\}$ is meager, and
if $f,g\in \C(X)$ are nowhere constant and $E\sq X$ is a nonvoid open set
such that $f\restr E\not= g\restr E$, then $f[M\cap E]\setminus g[M]$ has
cardinality $\cont$. In particular, there is a strong SRU for $\C(X)$ which is
not an SRU for the Borel isomorphisms.}

\P Since every nonvoid perfect Polish space contains a residual
copy of the irrationals, we may assume that $X$ is the space of irrational
numbers.  Let 
$\la \la f_\alpha,g_\alpha,E_\alpha\ra\colon\alpha<\cont \ra$ be a list of
all triples $\la f,g,E\ra$ where $f,g\in \C(X)$ are nowhere constant maps, $E$
is a nonempty open set, and $f[E]\cap g[E]=\e$; let each such triple appear
$\cont$ times in the list.  Let $A$ and $h$ be given by 
Lemma~\ref{IrrHom}.
Inductively choose points $x_\alpha\in E_\alpha\setminus A$ such that
$x_\alpha$ does not belong to any set of the form
$f_\alpha^{-1}(g_\alpha(h^m(x_\beta)))$ ($m\in\Z$, 
$\beta<\alpha$) or $f_\alpha^{-1}(f_\beta(x_\beta))$ ($\beta<\alpha$) or 
$h^{-m}([g_\beta^{-1}(f_\beta(x_\beta)]\setminus A)$ ($m\in\Z$, $\beta<\cont$)
or $\{x\in  X\setminus A\colon f_\alpha(x)=g_\alpha(h^m(x))\}$ 
($m\in\Z\setminus \{0\}$). Note that sets of the latter form are nowhere dense
by Lemma~\ref{IrrHom}, and the remaining sets are nowhere dense because
$f_\alpha$ and
$g_\beta$ ($\beta<\alpha$) are nowhere constant. 
Let $M=\{h^m(x_\alpha)\colon m\in\Z,\,\alpha<\cont\}$. 

To see that this works, let $f,g\in \C(X)$ be nowhere constant and let
$E\sq X$ be a nonvoid open set such that 
$f\restr E\not= g\restr E$. By shrinking $E$ we may assume that $f[E]\cap
g[E]=\e$.  For each $\alpha$ such that 
$\la f_\alpha,g_\alpha,E_\alpha\ra=\la f,g,E\ra$,
we chose  
$x_\alpha\in E$. We have that $f_\alpha(x_\alpha)\not\in g_\alpha[M]$, i.e.,
for each $\beta<\cont$ and each $m\in\Z$,
$f_\alpha(x_\alpha)\not=g_\alpha(h^m(x_\beta))$.  
(This is clear from the choice of $x_\alpha$. Consider separately the cases
$\beta<\alpha$, $\beta=\alpha$, $\beta>\alpha$.)  The rest of the properties of
the example now follow easily. \endproof


\section{Variations on the theme}\label{sec:var}
 
As mentioned in the introduction, properties similar to the SRU property have
been considered by various authors. We examine some of them in this section.

Let us begin with the results from \cite{DM} and \cite{Bu} mentioned in the
introduction. 
Dushnik and Miller \cite{DM} showed that, under CH,  
there is an uncountable set $M\sq\R$ such that for any monotone (nonincreasing
or nondecreasing) function $f\colon\R\to\R$, if $\{x\in\R\colon f(x)=x\}$ is
nowhere dense, then $f[M]\cap M$ is countable. 
In a model of set theory where
$\cont=\h_2$, $\R$ is covered by $\h_1$ meager sets and sets of cardinality
$\h_1$ are meager (e.g., in the random real model), there is no such set. To
see this, note that in any such model, for every uncountable set $M\sq\R$,
there is a set $M'\sq M$ of the same cardinality as $M$ such that $M'$ is
nowhere dense in $\R$. Then there is a monotone function $f\colon\R\to\R$ such
that $\{x\in\R\colon f(x)=x\}$ is equal to the closure of $M'$, and hence
$f[M]\cap M$ has the same cardinality as $M$ since it contains $M'$.

Consider now the result from \cite{Bu} that, under CH, there is a 
totally heterogeneous set. (See the introduction for the definition.) 
A classical diagonalization argument
shows that there is (in ZFC) a set $M\sq\R$ of cardinality $\cont$ such that
for any Borel function $f\colon M\to M$, $f[\{x\in M\colon f(x)\not=x\}]$ is
countable. 
It is not possible, however, to produce a totally heterogeneous set
in ZFC. To see this, consider a model in which every set $M\sq\R$ of
cardinality $\cont$ can be mapped continuously onto $[0,1]$ (e.g., the iterated
perfect set model \cite{Mi2}). It is easily seen that a totally heterogeneous
set of reals cannot contain a Cantor set, and hence its complement must have
cardinality $\cont$. In the model under consideration, if $M\sq[0,1]$ is any
set such that both $M$ and $[0,1]\setminus M$ have cardinality $\cont$, then
let $f\colon M\to[0,1]$ be a continuous surjection. Choose $M'\sq M$ such that
$f[M']=M$. Then $f[\{x\in M'\colon f(x)\not=x\}]$ contains $M\setminus M'$ and
hence has cardinality $\cont$. In particular, $M$ is not totally heterogeneous.

For the remainder of this section we will examine 
two generalizations of the notion of
an SRU for $\C(\R)$ for pairs of families $\F,\G\sq\C(\R)$.
We are interested particularly in the following subfamilies of $\C(\R)$
which we will abbreviate as follows:
\begin{description}
\item{$\bullet$} $\C=\C(\R)$; 
\item{$\bullet$} $\const$, the class of all constant functions $f\in\C(\R)$;
\item{$\bullet$} $\cm$, the class of all nowhere constant functions
                 $f\in\C(\R)$;
\item{$\bullet$} $\cc$, the class of all countable-to-one functions
                 $f\in\C(\R)$.
\end{description}

\defi{defTrun}{ A function $g\in\C$
is said to be a {\em truncation} of $f\in\C$ if
$g$ is constant on every connected component of
$\{x\in\real\colon f(x)\neq g(x)\}$.}

The following proposition is \cite[theorem~8.1]{BD}.

\prop{prop:magic1}{
There exists a set $M \subseteq \real$
such that for every $f\in\cc$ and $g\in\C$ if $g[M]\subseteq f[M]$
then $g$ is a truncation of $f$. \qed }
 
Notice that a strong SRU $M$ for $\la\R,\bor,\NN\ra$, $\NN=$ the meager ideal,
has a similar property. If $f\in\C$ 
and $g\in \C$ and $g[M]\sq f[M]$, then by Lemma~\ref{2.5} (with
$\kappa=\aleph_1$), the open intervals in $\{x\in\R\colon f(x)\not=g(x)\}$ on
which
$g$ is constant are dense in $\{x\in\R\colon f(x)\not=g(x)\}$. We do not need to
assume that $f\in\cc$, however we cannot
conclude that $g$ is a truncation of $f$, even if $f\in\cc$. (For example, we
could have $f(x)=x$ for all $x$, and $g(x)=x$ for $x\not\in[0,1]$,
$g\restr[0,1]=$ the Cantor ternary function.) Proposition \ref{prop:magic1}
also has the advantage of being a ZFC theorem.


\defi{GenSRU}{ 
For the families $\F,\G\sq\C$ we say that 
$M\sq\R$ is an {\em $\la\F,\G\ra$-truncation SRU}\/ if 
for every $f\in\F$ and $g\in\G$, if 
$g[M]\sq f[M]$ then $g$ is a truncation of $f$. 
A set $M\sq\R$ is an {\em $\la\F,\G\ra$-SRU}\/ if 
for every $f\in\F$ and $g\in\G$, if 
$g[M]\sq f[M]$ then $g=f$. }

We have the following general fact.
 
\prop{prop:basic}{
For every $\F,\G\sq\C$
\begin{description}
\item[(1)] If $M\sq\R$ is an $\la\F,\G\ra$-SRU then it is
           $\la\F,\G\ra$-truncation SRU;
\item[(2)] If $\G\sq\cm$ then every $\la\F,\G\ra$-truncation SRU
           is an $\la\F,\G\ra$-SRU.
\end{description}
}
 
\proof This is obvious since every function
is its own truncation and this is the only
truncation that could be nowhere constant. \qed
 
Proposition~\ref{prop:basic} shows in particular that
for $\G\sq\cm$, 
$M\sq\R$ is an $\la\F,\G\ra$-SRU if and only if it is an 
$\la\F,\G\ra$-truncation SRU.
Thus, for $\G\sq\cm$ we will examine only 
$\la\F,\G\ra$-SRU's.
 
In the next theorem we seek, for various choices of $\G$,
the largest family $\F$ for which
there exists an $\la\F,\G\ra$-SRU 
($\la\F,\G\ra$-truncation SRU).
The examples in (A$^\prime$)-(C$^\prime$) indicate that the families found in 
(A)-(C) are to some extent the best possible.

 
\thm{thm:main}{
\mbox{ }
\begin{description}
\item[(A)] $M=\R$ is a $\la\const,\C\ra$-SRU.
\item[(A$^\prime$)] There is no $\la\{f\},\G\ra$-SRU
                    if $f\in\C\setminus\const$ and $\const\sq\G$.
\item[(B)] There exists a $\la\cc,\C\ra$-truncation SRU.
\item[(B$^\prime$)] There is no $\la\cm,\C\ra$-truncation SRU.
                    More precisely, there is no 
                    $\la\{f\},\C\ra$-truncation SRU
                    for every $f\in\cm$ with the property that
                    $f^{-1}(y)$ is perfect for every $y\in\real$.
\item[(C)] If $\R$ is not the union less than $\cont$ many meager
           sets then there exists a $\la\C,\G\ra$-truncation SRU 
           for any $\G\sq\C$ of cardinality less than $\cont$.
\item[(C$^\prime$)] There exist $f,g,h\in\C$ such that there 
                    is no $\la\{h\},\{f,g\}\ra$-SRU.
\item[(D)] Any strong SRU for $\C(\R)$ is a $\la\C,\cm\ra$-SRU.
\item[(E)] There exists an $\la\{f\},\cm\ra$-SRU for every $f\in\C$.
\end{description}
}
 
{\sc Proof of (A)}. Obvious.


\bigskip
 
{\sc Proof of (A$^\prime$)}.
Let $f\in\C\setminus\const$ and $M\sq\real$.
If $M=\emptyset$ take an arbitrary $g\in\const\sq\G$.
If $M\neq\emptyset$ and $x_0\in M$ let $g\in\G$
be a constant function equal to $f(x_0)$.
Then $g\neq f$ but $g[M]\sq f[M]$.
 
\bigskip
 
{\sc Proof of (B)}. This is Proposition~\ref{prop:magic1}.
 
\bigskip
 
{\sc Proof of (B$^\prime$)}.
Let $f\in\cm$ be such that $f^{-1}(y)$ is perfect for every
$y\in\real$.
For example, if $F=(f_0,f_1)\colon[0,1]\to[0,1]^2$ is a classical
Peano curve (see e.g., \cite[Example~4.3.8]{CLO}) then we can define $f$ by
$f(n+r)=n+f_0(r)$ for every integer $n$ and $r\in[0,1)$.
 
Let $M\sq\real$. To see that it is not an 
$\la\{f\},\C\ra$-truncation SRU we
will find $g\in\C$ with $g[M]\sq f[M]$
which is not a truncation of $f$.
We have two cases to consider.
 
\medskip
 
\noindent Case 1. $f[M]$ is not dense in $\real$.
 
Take $c<d$ such that $(c,d)\cap f[M]=\emptyset$.
Since $f[\real]=\real$ there exist $a<b$ such that
$(a,b)\sq f^{-1}((c,d))$. So, $(a,b)\cap M=\emptyset$.
Choose $a_0<b_0$ such that $a<a_0<b_0<b$ and define
$g$ on $[a_0,b_0]$ to be nonconstant and such that
$g(x)\neq f(x)$ for every $x\in[a_0,b_0]$.
Put $g(x)=f(x)$ for every $x\in\real\setminus(a,b)$
and extend it to a continuous function on $\real$.
Then $g$ is not a truncation of $f$,
while $g[M]=f[M]$.
 
\medskip
 
\noindent Case 2. $f[M]$ is dense in $\real$.
 
If $f[M]=\real$ it is enough to take as $g$ an arbitrary
continuous function which is not a truncation of $f$.
So, without loss of generality we can assume that
$f[M]\neq\real$.
 
Choose $y_0\in\real\setminus f[M]$ and let
$P=f^{-1}(y_0)$. So, $P$ is perfect, nowhere dense
and $P\cap M=\emptyset$.
Choose a countable dense subset $D$ of $f[M]$
and notice the following fact.
\begin{description}
\item{($\star$)} For every $d_0,d_1\in D$, $d_0<d_1$,
                 there exists a continuous function
                 $g$ from $\real$ onto $[d_0,d_1]$
                 such that $g[\real\setminus P]\sq D\sq f[M]$
                 and $g$ is not constant at on any open interval
                 intersecting $P$.
\end{description}
To see it, let $h\colon[0,1]\to[0,1]$ be a classical
Cantor function, i.e., $h$ is nondecreasing,
constant on any component of $[0,1]\setminus C$ and
such that $h[C]=[0,1]$,
where $C$ is a classical Cantor ternary set. (See~\cite[p.~50]{Ro}.)
Extend $h$ to $\real$ by putting  $h(x)=0$ for
$x<0$ and $h(x)=1$ for $x>1$
and notice that $h[\real\setminus C]\sq\rational$.
Let $h_0\colon\real\to\real$ be a homeomorphism such that $h_0[P]=C$
and $h_1\colon\real\to\real$ be an order isomorphism such that
$h_1[\rational]=D$, $h_1(0)=d_0$ and $h_1(1)=d_1$.
Then $g=h_1\circ h\circ h_0$ satisfies ($\star$).
 
To finish the proof notice that for any function
$g$ satisfying ($\star$) we have
$g[M]\sq g[\real\setminus P]\sq D\sq f[M]$.
Now, if $g\colon\real\to[d_0,d_1]$
and $g^\prime\colon\real\to[d_0^\prime,d_1^\prime]$
are as in ($\star$) and such that
$[d_0,d_1]\cap[d_0^\prime,d_1^\prime]=\emptyset$
then for every $x\in P$ we have
$g(x)\neq g^\prime(x)$.
In particular, either $g$ or $g^\prime$ is not a truncation of $f$.
 
\bigskip
 
{\sc Proof of (C)}.
\ch Choose $\G\sq\C$ of cardinality less than $\cont$.
%$\G\in[\C]^{<\continuum}$.
Since any constant function is a truncation of any
other function we can assume without loss of
generality that $\G\cap\const=\emptyset$.
 
For $g\in\C$ let $\const(g)$ denotes the set of these points
at which $g$ is locally constant, i.e.,
\[
\const(g)=\{x\in\real\colon (\exists a,b\in\real)
[a<x<b\ \&\ g\mbox{ is constant on } (a,b)]\}.
\]
Then for every $g\in\G$ the set $P_g=\real\setminus\const(g)$
is nonempty and perfect.
In particular, it is not a union of less
then continuum many its nowhere dense subsets. 
 
Let
$\{\la f_\alpha,g_\alpha\ra\colon \alpha<\continuum\}=
\{\la f,g\ra\in\C\times\G\colon g\mbox{ is not a truncation of } f\}$.
We will construct, by induction on $\alpha<\continuum$,
a set $M=\{m_\alpha\colon\alpha<\continuum\}$
such that $g_\alpha(m_\alpha)\not\in f_\alpha[M]$ for every
$\alpha<\continuum$.
This will finish the proof.
 
To have $g_\alpha(m_\alpha)\not\in f_\alpha[M]$
we will choose $m_\alpha$ such that the following
inductive conditions are satisfied.
 
$g_\alpha(m_\alpha)\not\in\{f_\alpha(m_\alpha)\}$, i.e., such that
\begin{description}
\item{(I$_\alpha$)} $m_\alpha\in U_\alpha$, where
        $U_\alpha=\{x\in\real\colon f_\alpha(x)\neq g_\alpha(x)\}$.
\end{description}
 
$g_\alpha(m_\alpha)\not\in\{f_\alpha(m_\gamma)\colon\gamma<\alpha\}$,
i.e., such that
\begin{description}
\item{(II$_\alpha$)}
$m_\alpha\not\in\bigcup_{\gamma<\alpha} g^{-1}_\alpha(f_\alpha(m_\gamma))$.
\end{description}
 
$g_\alpha(m_\alpha)\not\in\{f_\alpha(m_\gamma)\colon\gamma>\alpha\}$,
i.e., such that
$f_\alpha(m_\gamma)\neq g_\alpha(m_\alpha)$
for every $\alpha<\gamma$.
By interchanging $\alpha$ and $\gamma$ in the last condition
we obtain
$f_\gamma(m_\alpha)\neq g_\gamma(m_\gamma)$
for every $\gamma<\alpha$.
So, it is enough to choose
\begin{description}
\item{(III$_\alpha$)}
$m_\alpha\not\in\bigcup_{\gamma<\alpha} f^{-1}_\gamma(g_\gamma(m_\gamma))$.
\end{description}
 
To make such a choice possible, we will also require that
\begin{description}
\item{($\star_\alpha$)}
$f^{-1}_\alpha(g_\alpha(m_\alpha))$ is nowhere dense in $P_g$ for
every  $g\in\G$.
\end{description}
We will achieve this by making sure that
$g_\alpha(m_\alpha)\not\in S^g_\alpha$ for every $g\in\G$,
where
\[
S^g_\alpha=\{y\in\real\colon f_\alpha^{-1}(y)\cap P_g
\mbox{ is not nowhere dense in } P_g\}.
\]
Notice that each $S^g_\alpha$ is at most countable.
So, we will guarantee ($\star_\alpha$)
by choosing
\begin{description}
\item{(IV$_\alpha$)}
$m_\alpha\not\in\bigcup_{g\in\G} g^{-1}_\alpha(S^g_\alpha)$.
\end{description}
 
Clearly it is enough to show that the choice of such $m_\alpha$
is possible. So, assume that for some
$\alpha<\continuum$ the construction is
done till step $\alpha$. We will choose $m_\alpha$.
 
Let
\[
V_\alpha=\bigcup_{\gamma<\alpha}
[g^{-1}_\alpha(f_\alpha(m_\gamma))\cup f^{-1}_\gamma(g_\gamma(m_\gamma))]
\cup \bigcup_{g\in\G} \bigcup_{y\in S^g_\alpha} g_\alpha^{-1}(y)
\]
and put $T_\alpha=U_\alpha\setminus V_\alpha$.
It is enough to show that
$T_\alpha\cap P_{g_\alpha}\neq\emptyset$.
But $g_\alpha$ is not a truncation of $f_\alpha$.
So, $|g_\alpha[U_\alpha]|=\continuum$
and the set $U_\alpha\cap P_{g_\alpha}$ is nonempty
and open in $P_{g_\alpha}$.
Since $V_\alpha$ is a union of less than
continuum many sets, it is enough
to argue that each of these sets is
nowhere dense in $P_{g_\alpha}$.
 
But sets $f^{-1}_\gamma(g_\gamma(m_\gamma))$ are nowhere dense
in $P_{g_\alpha}$ by ($\star_\gamma$), i.e., the inductive
assumption (IV$_\gamma$).
To finish the proof it is enough to notice that
$g_\alpha^{-1}(y)$ is nowhere dense
in $P_{g_\alpha}$ for every $y\in\real$,
which follows immediately from the definition of
$P_{g_\alpha}$.
 
\bigskip
 
{\sc Proof of (C$^\prime$)}.
Let $h(x)=\min\{0,x^2-1\}$,
$g(x)= 0$, and 
$f(x)=\max\{h(x),x-2\}$ for every $x\in\R$. Clearly $f\neq h\neq g$.
It is enough to show that for every $M\sq\real$ either $f[M]\sq h[M]$ 
or $g[M]\sq h[M]$. 
But if $M\setminus(-1,1)\neq\emptyset$ then 
$g[M]=\{0\}\sq h[M]$. 
Otherwise $M\sq(-1,1)$ and $f[M]=h[M]$.
 
\bigskip
 
{\sc Proof of (D)}.
Apply Lemma~\ref{2.5}
with $\kappa=\aleph_1$, $g\in\C$ and $f\in\cm$.
 
\bigskip
 
{\sc Proof of (E)}. Let $f\in\C$.
If there are $a<b$ such that $f$ is constant on $[a,b]$
it is enough to take $M=[a,b]$.
 
So, assume that $f\in\cm$ and let
$\{g_\alpha\colon\alpha<\continuum\}=\cc$.
As in the proof of (C) it is enough to find
$M=\{m_\alpha\colon\alpha<\continuum\}$
such that
\begin{description}
\item{(I$_\alpha$)} $m_\alpha\in U_\alpha=
        \{x\in\real\colon f(x)\neq g_\alpha(x)\}$;
\item{(II$_\alpha$)}
$m_\alpha\not\in\bigcup_{\gamma<\alpha} g^{-1}_\alpha(f(m_\gamma))$;
\item{(III$_\alpha$)}
$m_\alpha\not\in\bigcup_{\gamma<\alpha} f^{-1}(g_\gamma(m_\gamma))$.
\end{description}
But
$|f[U_\alpha]|=\continuum$,
since $U_\alpha\neq\emptyset$ and $f\in\cm$.
Also
$\left|\bigcup_{\gamma<\alpha} g^{-1}_\alpha(f(m_\gamma))\right|
<\continuum$ since
$|g^{-1}_\alpha(f(m_\gamma))|\leq\aleph_0$.
Thus,
\[
\left|f\left[\bigcup_{\gamma<\alpha}
g^{-1}_\alpha(f(m_\gamma))\right]\right|
<\continuum.
\]
Moreover,
\[
f\left[\bigcup_{\gamma<\alpha} f^{-1}(g_\gamma(m_\gamma))\right]=
\bigcup_{\gamma<\alpha} f[f^{-1}(g_\gamma(m_\gamma))]\sq
\{g_\gamma(m_\gamma)\colon\gamma<\alpha\}
\]
has cardinality $<\continuum$. So, the set
\[
U_\alpha\setminus\left[
\bigcup_{\gamma<\alpha} [g_\alpha^{-1}(f(m_\gamma))\cup
                               f^{-1}(g_\gamma(m_\gamma))]
\right]
\]
is nonempty. This finishes the proof of Theorem~\ref{thm:main}. \qed
 
\rem{rem1}{ There is no set $M\sq\R$ such that 
for every $f,g\in\C(\R)$ (not necessarily nowhere constant), 
if $f[M]=g[M]$ then $f=g$.}

\proof Such a set $M$ would be an SRU for $\C(\R)$, so, by theorem
\ref{thm:MP}, $M$ is dense and is disjoint from a Cantor set $K$. But then
we can build distinct Cantor-like functions $f$ and $g$ with
$f[\R\setminus K]=g[\R\setminus K]$ (a countable set). \qed



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%\end{changemargin}
\end{document}
