% LaTex 2.09 \documentstyle[12pt]{article} \input amssym.def \input amssym %%%%%% if amssym.def and amssym.tex are not available %%%%%% delete the 2 lines above and uncomment 4 lines below %\def\Bbb#1{{\bf #1}} %\def\goth#1{{\bf #1}} %\def\upharpoonright{|} %\def\blacksquare{{\Box}} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \def\reals{{\Bbb R}} \def\poset{{\Bbb P}} \def\aa{{\cal A}} \def\rmand{{\mbox{ and }}} \def\rmiff{{\mbox{ iff }}} \def\res{\upharpoonright} \def\qed{\nopagebreak\par\noindent\nopagebreak$\blacksquare$\par} \def\proof{\par\noindent {\bf Proof:}\par} \begin{document} \bigskip \begin{center} {\Large Baire measures on uncountable product spaces \footnote{As I was just about finished writting this up (October 1998), I discovered that everything here is already known. See the last section of \par R.J.Gardner and W.F.Pfeffer, {\it Borel Measures}, in {\bf Handbook of Set Theoretic Topology}, North-Holland, 1984, 961-1044. }} \end{center} \bigskip \begin{center} by Arnold W. Miller\footnote{ I want to thank Joel Robbin for many helpful conversations, especially for some helpful comments which fixed my first ``proof'' of Theorem \ref{ch}. } \end{center} \begin{quote} \begin{center} Abstract \end{center} We show that assuming the continuum hypothesis there exists a nontrivial countably additive measure $\pi$ on the Baire subsets of the space ${\reals^{\omega_1}}$ which vanishes on each element of an open cover. Contrariwise, we show that assuming MA$_{\omega_1}$, that there is no such measure. \end{quote} Let ${\reals^{\omega_1}}$ be the $\omega_1$ product of uncountably many copies of the real line. The family of Baire sets in this context is the smallest $\sigma$-algebra which contains the basic open sets, i.e., sets of the form: $\prod_{\alpha<\omega_1}U_\alpha$ where each $U_\alpha$ is an open subset of $\reals$ and all but finitely many $U_\alpha$ are the whole real line. Other definitions of this family are that it is the smallest $\sigma$-algebra generated by the closed $G_\delta$-sets. It is also true that the closed $G_\delta$-sets are the preimages of zero under a continuous real-valued mapping. Basically the Baire sets are the Borel sets which depend on only countably many coordinates. For example, singletons are not Baire. A Baire measure is a countably additive measure into the unit interval $[0,1]$ which is defined on the Baire sets. Denis Saveliev $\cite{saveliev}$ has asked \bigskip Question: Does every Baire measure on ${\reals^{\omega_1}}$ which vanishes on a cover of open Baire sets have to be trivial? In particular what happens if we assume Martin's Axiom? \bigskip Here I will answer this question by proving: \begin{theorem} \label{ma} (Fremlin \cite{fremlin}) Assume MA$_{\omega_1}$. Then every Baire measure on ${\reals^{\omega_1}}$ which vanishes on a cover of open Baire sets is trivial. \end{theorem} \begin{theorem}\label{ch} (Moran \cite{moran}, Kemperman and Maharam \cite{kemp}) Assume CH. Then there exists a nontrivial Baire measure on ${\reals^{\omega_1}}$ which vanishes on a cover of open Baire sets. \end{theorem} \begin{center} {\bf Proof of Theorem \ref{ma}} \end{center} Assume MA$_{\omega_1}$ and let $\mu$ be a nontrivial Baire measure. Let $\poset$ be the following partially ordered set. We say that $p\subseteq {\reals^{\omega_1}}$ is a finite box iff $p$ is a product $p=\prod_{\alpha<\omega_1} p(\alpha)$ where for all but finitely many coordinates $p(\alpha)=\reals$ and for these finitely many coordinates $p(\alpha)$ is a compact interval in the reals. Define $$\poset=\{p\subseteq {\reals^{\omega_1}}: \mbox{ $p$ is a finite box and } \mu(p)>0\}$$ Order $\poset$ by inclusion. Note that it two condition are compatible iff their intersection has positive measure. Thus $\poset$ has the countable chain condition (ccc). (Given uncountably many elements of $\poset$, $\{p_\alpha:\alpha<\omega_1\}$, there must be some natural number $n$ such that uncountably many of the $p_\alpha$ have measure greater than $1\over n$, but there can be at most $n$ pairwise incompatible elements of $\poset$ of measure $1\over n$.) Suppose for contradiction, there were a cover of ${\reals^{\omega_1}}$ by Baire open sets of measure zero. We may assume this cover has cardinality $\omega_1$, say it is $\{U_\beta:\beta<\omega_1\}$, where the $U_\beta$ are particularly simple. Namely, each is an $\omega_1$ product of open intervals in $\reals$ with rational end points and all but finitely many are $(-\infty,\infty)$. For each $\alpha<\omega_1$ define $$D_\alpha=\{p\in\poset: p\cap U_\alpha=\emptyset\}.$$ Each $D_\alpha$ is dense in $\poset$. To see this note that the complement of the set $U_\alpha$ can be written as a countable union of finite boxes. To see this, let $F$ be the finitely many coordinates where $U_\alpha$ is a nontrivial open interval and for each $\beta\in F$ write the complement of this open interval as a union of compact intervals $J^\beta_n$ for $n\in\omega$. For each $\beta\in F$ and $n\in\omega$ let $$p_n^\beta=\{x\in{\reals^{\omega_1}}: x_\beta\in J^\beta_n\}.$$ Then the complement of $U_\alpha$ is the countable union of finite boxes $$\langle p_n^\beta:\beta\in F, n\in\omega\rangle.$$ Given an arbitrary $p\in\poset$ there exists some $\beta\in F,\; n\in\omega$ such that $q=p\cap p_n^\beta$ has positive measure and so $q\in D_\alpha$. This shows that $D_\alpha$ is dense. For each $\alpha$ and natural number $n$, let $$E_\alpha^n=\{p\in\poset: \mbox{ the diameter of $p(\alpha)$ is less than $1/n$ }\}.$$ To see that $E_\alpha^n$ is a dense subset of $\poset$, let $J_k$ for $k<\omega$ be a cover of $\reals$ by closed intervals of length less than $1/n$. Given an arbitrary $p\in\poset$ let $$p_k=\{x\in p:x_\alpha\in J_k\}.$$ Each $p_k$ is a closed Baire set and since $p$ has positive measure at least on $p_k$ must have positive measure. Hence, there exists $k$ such that $p_k\leq p$ and $p_k\in E_\alpha^n$. By MA$_{\omega_1}$ there exists a $\poset$-filter $G$ such that for all $\alpha<\omega$ and $n$, both $$G\cap E_\alpha^n\not=\emptyset \rmand G\cap D_\alpha\not=\emptyset.$$ First we claim that $\bigcap G$ is nonempty, in fact, a singleton $\{x\}$. To see that it is nonempty, note that we cannot directly use compactness since our space ${\reals^{\omega_1}}$ is not compact. However, for each fixed $\alpha$ the intersection $\bigcap \{p(\alpha): p\in G\}$ is a nonempty singleton since $G$ meets $E_\alpha^n$ for all $n$. Let $x\in{\reals^{\omega_1}}$ be defined by $$\{x(\beta)\}=\bigcap \{p(\alpha): p\in G\}$$ for every $\beta<\omega_1$. By its definition, $x\in p$ for every $p\in G$. Since $G$ meets each $D_\alpha$ it must be that $x\notin U_\alpha$ and this contradicts the fact that $U_\alpha$ were supposed to be a cover of ${\reals^{\omega_1}}$. \qed \begin{center} {\bf Proof of Theorem \ref{ch}} \end{center} We will begin by showing that there exists such a measure $\pi$ on the Baire subsets of $2^\omega \times \omega^{\omega_1}.$ Assuming CH let $2^\omega=\{x_\alpha:\; \alpha<\omega_1\}$. For any $\alpha<\omega_1$ let $\langle \sigma_n^\alpha\in 2^{n+1}:n\in\omega\rangle$ be defined by $$\sigma_n^\alpha(i)=\left\{ \begin{array}{cl} x_\alpha(i) & \mbox{ if }in$. Now we indicate what $\pi$ is on all of $2^\omega \times \omega^{\omega_1}$. Let $\aa$ be the algebra of Baire sets $B\subseteq 2^\omega \times \omega^{\omega_1}$ whose support in $\omega_1$ is finite. This means, that there exists a finite $F\subseteq\omega_1$ such that: \medskip for every $(x,y),(x,y^\prime)\in 2^\omega \times \omega^{\omega_1}$ if $y\res F=y^\prime\res F$, then \par $(x,y)\in B \rmiff (x,y^\prime)\in B.$ \medskip For each $s:F\to \omega$ define $$C_s=\cap\{[\sigma^\alpha_{s(\alpha)}]:\alpha\in F\}\subseteq 2^\omega.$$ A lot of the $C_s$ are empty, however they partition $2^\omega\setminus F$. (For each $x\in 2^\omega\setminus F$ define $s:F\to \omega$ by $s(\alpha)$ is the largest $n$ such that $x(n)=x_\alpha(n)$. Then $x\in C_s$.) Define \begin{eqnarray} B_s &=& \{x\in 2^\omega : \exists y\in\omega^{\omega_1} \;\; s\subseteq y, \;\;(x,y)\in B, \rmand x\in C_s \} \nonumber \\ B^* &=& \bigcup\{B_s:\;\;s:F\to\omega\} \nonumber \\ \pi(B) &=& \lambda(B^*) \nonumber \\ \nonumber \end{eqnarray} To verify that this is well-defined we should check that if $F\subseteq F^\prime\subseteq\omega_1$ for any $F^\prime$ finite that $$\sum_{s:F\to\omega}\lambda(B_s)= \sum_{t:F^\prime\to\omega}\lambda(B_t).$$ But for any fixed $s:F\to\omega$ $$C_s=\cup\{C_t: \;\; t:F^\prime\to\omega, \;\; s\subseteq t\} \cup(\{x_\alpha\in C_s : \alpha\in F^\prime\}.$$ Now since $F$ is the support of $B$ for each $s:F\to\omega$, $B_s$ will differ from $$\bigcup \{B_t:\;\;t:F^\prime\to\omega, s\subseteq t\}$$ by at most a finite set, hence $$\lambda(B_s)=\sum\{\lambda(B_t):\;\;t:F^\prime\to\omega,\;\; s\subseteq t\}.$$ Thus the set $B^*$ (which depends on $F$) is the same up to a finite set. To verify that $\pi$ extends to the $\sigma$-algebra generated by $\aa$, according to Halmos \cite{halmos}, we need to show that whenever $B_n$ are disjoint elements of $\aa$ such that their union $B=\cup\{B_n:n\in\omega\}$ is in $\aa$, we have $$\pi(B)=\sum_{n=0}^\infty\pi(B_n).$$ For each $B_n$ suppose that it is supported by the finite set $F_n$ and $B$ is supported by the finite set $F$. Without loss of generality, we may assume $F\subseteq F_n$ for all $n$. To verify our equation it is enough to see that for any $x$ except possibly the $x_\alpha$ for $\alpha\in \bigcup \{F_n:n\in\omega\}$ that $$x\in B^* \rmiff x\in\bigcup\{B_n^*:n\in\omega\}.$$ Let us verify this. Suppose $x\in B^*$. Then for some $s:F\to\omega$ and $y\supseteq s$ we have that $(x,y)\in B$ and $x\in C_s$. Since the support of $B$ is $F$ we can redefine $y$ so that for every $$\alpha\in \bigcup\{F_n:n\in\omega\}\setminus F$$ $y(\alpha)$ is the first $n$ such that $x(n)\not=x_\alpha(n)$ (i.e. so that $x\in [\sigma^\alpha_{y(\alpha}]$). Since $(x,y)\in B$ there exists $n$ such that $(x,y)\in B_n$. But since we are assuming $x\not=x_\alpha$ for any $\alpha\in F_n$, there exists $t:F_n\to\omega$ such that $t\supseteq s$ and if $t=y\res F_n$, then we have that $x\in C_t$ and hence $x\in B_n^*$. Now let us check the converse. Suppose $x\in B_n^*$ for some $n$. Then for some $t:F_n\to\omega$ and some $y\supseteq t$ we have $(x,y)\in B_n$ and $x\in C_t$. But then since $F\subseteq F_n$ we have that $(x,y)\in B$ and $x\in C_s$ where $s=t\res F$ and so $x\in B^*$. It follows that $\pi$ extends to the $\sigma$-algebra generated by $\aa$, i.e., the Baire subsets of $2^\omega \times \omega^{\omega_1}$. Finally, we note that for every element $(x,y)\in 2^\omega \times \omega^{\omega_1}$ there exists a clopen set $C$ with $\pi(C)=0$ and $(x,y)\in C$. Let $x=x_\alpha$ and suppose that $y(\alpha)=n$. Then let $s:\{\alpha\}\to \omega$ be defined by $s(\alpha)=n$ and let $t=x\res(n+1)$. Then $(x,y)\in C=[t]\times [s]$ and $\pi([t]\times [s])=0$. To obtain the same result for ${\reals^{\omega_1}}$, first note that we have such a measure on $\omega^{\omega_1}$, since $2^\omega\times \omega^{\omega_1}$ is a closed Baire set in $\omega^{\omega}\times \omega^{\omega_1}$ and we can identify $\omega^{\omega}\times \omega^{\omega_1}$ with $\omega^{\omega+\omega_1}= \omega^{\omega_1}$. Now to ``lift'' $\pi$ to $\hat{\pi}$ on the Baire subsets of ${\reals^{\omega_1}}$, note that for every Baire $B\subseteq {\reals^{\omega_1}}$ the set $B\cap\omega^{\omega_1}$ is Baire in $\omega^{\omega_1}$. So we can let $\hat{\pi}(B)=\pi(B\cap \omega^{\omega_1})$. The measure $\hat{\pi}$ vanishes on an open neighborhood of every point. To check this, suppose $x\in{\reals^{\omega_1}}$. Then if there exists some $\alpha$ such that $x(\alpha)\notin \omega$, then letting $\epsilon>0$ be a quarter of the distance from $x(\alpha)$ and the nearest element of $\omega$, we have that $$x\in U=\{y\in{\reals^{\omega_1}}: |y(\alpha)-x(\alpha)|<\epsilon\}$$ and $\hat{\pi}(U)=0$. On the otherhand if $x\in\omega^{\omega_1}$ then for some finite $F\subseteq \omega_1$ we have that $\pi([x\res F])=0$, and so if we let $$U=\{y\in{\reals^{\omega_1}}: |y(\alpha)-x(\alpha)|<1/4 \mbox{ for all }\alpha\in F\}$$ then $x\in U$ and $\hat{\pi}(U)=0$. \qed \begin{theorem} (Bockstein \cite{bock} \cite{ek}) If $C\subseteq \omega^{\omega_1}$ is a closed $G_\delta$ set, then $C$ is Baire. \end{theorem} \proof Suppose for contradiction that $C$ is not countably supported. This means that for each $\alpha<\omega_1$ there exists $f^i_\alpha:\omega_1\to \omega$ for $i=0,1$ such that $$f^0_\alpha \res\alpha=f^1_\alpha\res\alpha \rmand f^0_\alpha\in C \rmand f^1_\alpha\notin C.$$ Suppose $C=\bigcap\{U_n:n\in\omega\}$ where $U_n$ are open. For each $\alpha$ choose finite $F^\alpha_n$ for $n\in\omega$ so that $$[f^1_\alpha\res F^\alpha_0] \cap C=\emptyset$$ $$[f^0_\alpha\res F^\alpha_n] \subseteq U_n$$ By the push down lemma there exists $H$ and a stationary set of $\alpha$ such that $F_\alpha^0\cap\alpha=H$. Hence, it easy to find $\alpha<\beta$ such that $$F^n_\alpha \cap F^0_\beta \subseteq \alpha$$ for all $n$. Now if we define $g:\omega_1\to\omega$ so that $$g\res F^n_\alpha = f^0_\alpha \res F^n_\alpha$$ and $$g\res F^0_\beta = f^1_\beta \res F^0_\beta$$ then we get the contradiction that $g\notin C$ but $g\in\bigcap\{U_n:n\in\omega\}$. \qed \begin{theorem} (A.H.Stone \cite{stone}) The space $\omega^{\omega_1}$ is not normal. \end{theorem} \proof Let $$C_i=\{f\in \omega^{\omega_1}: \forall n\not= i\;\; |f^{-1}\{i\}|\leq 1\}$$ i.e., one-to-one except when equal to $i$. Then $C_0$ and $C_1$ are disjoint closed sets which cannot be separated by disjoint open sets. To see this suppose that $U_0$ and $U_1$ are disjoint open sets which separate $C_0$ and $C_1$. For each $\alpha<\omega_1$ choose $f_\alpha^i\in C_i$ for $i=0,1$ such that $$f_\alpha^0\res\alpha=f_\alpha^1\res\alpha.$$ For each $\alpha$ there exists a finite $F_\alpha$ such that $$[f_\alpha^0\res F_\alpha]\subseteq U_0 \rmand [f_\alpha^1\res F_\alpha]\subseteq U_1.$$ But by the $\Delta$-systems lemma it easy to find $\alpha<\beta$ such that $F_\alpha\cap F_\beta\subseteq \alpha$. But this means we can define $g:\omega_1\to\omega$ so that $$g\res F_\alpha=f_\alpha^0 \res F_\alpha \rmand g\res F_\beta=f_\beta^1 \res F_\beta$$ which is a contradiction since then $g\in U_0\cap U_1$. \qed \begin{theorem} The usual product measure $\mu$ on $2^{\omega_1}$ makes all Borel subsets measurable (not just the Baire sets). \end{theorem} \proof Let $U\subseteq 2^{\omega_1}$ be any open set. Suppose $$U=\bigcup_{\alpha<\omega_1}[s_\alpha].$$ Then we may choose $\alpha_0$ so that for every $\beta$ $$\mu(\bigcup_{\alpha<\alpha_0}[s_\alpha])= \mu([s_\beta]\cup\bigcup_{\alpha<\alpha_0}[s_\alpha]).$$ Let $\Gamma=\bigcup_{\alpha<\alpha_0}\mbox{domain}(s_\alpha)$. Because this is the product measure it must be that for every $\beta$ $$\mu([s_\beta\res\Gamma]\setminus \bigcup_{\alpha<\alpha_0}[s_\alpha])=0.$$ So let $$U_0= \bigcup_{\alpha<\alpha_0}[s_\alpha])$$ and let $$U_1= U_0\cup\bigcup_{\alpha_0\leq\beta<\omega_1}[s_\beta\res\Gamma]).$$ Then $U_0$ and $U_1$ are Baire open sets, $U_0\subseteq U\subseteq U_1$, and $\mu(U_1\setminus U_0)=0$. Hence, $U$ is measurable. \qed \begin{thebibliography}{99} \bibitem{bock} M.Bockstein, {\it Un th\'eor\`me de s\'eparabilit\'e pour les produits topologiques}, Fundamenta Mathematicae, 35(1948), 242-246. \bibitem{ek} R.Engelking and M.Karlowicz, {\it Some theorems of set theory and their topological consequences}, Fundamenta Mathematicae, 57(1965), \bibitem{fremlin} D.H.Fremlin, {\it Uncountable powers of $\reals$ can be almost Lindel\"{o}f}, Manuscripta mathematics, 22(1977), 77-85. \bibitem{gardner} R.J.Gardner and W.F.Pfeffer, {\it Borel Measures}, in {\bf Handbook of Set Theoretic Topology}, North-Holland, 1984, 961-1044. \bibitem{kemp} J.H.B.Kemperman, D.Maharam, {\it $\reals^{{\goth c}}$ is not almost Lindel\"{o}f}, Proceedings of the American Mathematical Society, 24(1970), 772-773. \bibitem{halmos} P.R.Halmos, {\bf Measure Theory}, Van Nostrand Reinhold Company, (1950), see Theorem A on page 54. \bibitem{moran} W.Moran, {\it The additivity of measures on completely regular spaces}, Journal of the London Mathematical Society, 43(1968), 633-639. \bibitem{saveliev} D.Saveliev, email to M.E.Rudin, July 1998. \bibitem{stone} A.H.Stone, {\it Paracompactness and product spaces}, Bulletin of the American Mathematical Society, 54(1948), 977-982. \end{thebibliography} \bigskip\noindent Address: University of Wisconsin-Madison, Department of Mathematics, Van Vleck Hall, 480 Lincoln Drive, Madison, Wisconsin 53706-1388, USA. \medskip\noindent e-mail: miller@math.wisc.edu \par \medskip\noindent web page: http://www.math.wisc.edu/$\sim$miller \end{document}