1$, there does not exist a set $S \subseteq \R^2$ meeting every isometric copy of $A$ in exactly one point. \end{conj} \mys For $|A|=2,3$ the result is easily proved. \end{frame} \begin{frame} {\color{red}Gao}, {\color{red}Miller} showed that certain families of four-point sets satisfy the conjecture. \mys Then Gao's student {\color{red}Mingzhi Xuan} proved the following. \mys \begin{thm}[Xuan] Every four-point set in the plane satisfies the conjecture, \end{thm} \end{frame} \section{Something Old-Something New} \begin{frame} In 1991 Mauldin and J proved the following. \begin{thm}[$\zf$] Let $X$, $Y$ be uncountable Polish spaces. There is a $\bp^1_1$ set $P\subseteq X\times Y$ with all sections $P_x$ countable, such that $P$ is not the union of countably many $\bs^1_2$ (partial) graphs of functions. In fact same conclusion for $\sB(\bs^1_2)$. \end{thm} Assuming $\bp^1_1$-determinacy the least pointclass $\bg$ was identified such that every $\bp^1_1$ with countable sections can be written as a countable union of graphs in $\bg$. Namely, \[ \bg= \bigcup_n \Game \omega\cdot n-\bp^1_1. \] \end{frame} \begin{frame} Recently, the method of proof used for this non-uniformization result has been used to get a result concerning a question of {\color{red}Maharam} about $\sigma$-finite measures. \mys Let $\mu$, $\nu$ be Borel measures on uncountable Polish spaces $X$, $Y$, and let $\phi\colon X\to Y$ be Borel. \mys \begin{defn} A {\color{blue}disintegration} of $\mu$ with respect to $\nu$, $\phi$ is a family $\{ \mu_y\}_{y \in Y}$ on Borel measures on $X$ satisfying: (1) $\forall B \in \sB(X)\ (y \mapsto \mu_y(B))$ is Borel. (2) $\forall y \in Y\ (\mu_y(X-\phi^{-1}(y))=0$. (3) $\forall B\in \sB(X)\ (\mu(X)=\int \mu_y(B) \, d\nu)$ \end{defn} \end{frame} \begin{frame} We say the disintegration $\{ \mu_y\}_{y \in Y}$ is {\color{blue}$\sigma$-finite} if for every $y \in Y$, $\mu_y$ is a $\sigma$-finite measure. \mys We say the disintegration is {\color{blue}uniformly $\sigma$-finite} if there are Borel sets $B_n$ in $X$ satisfying: (1) $\forall n\ \forall y \in Y \ \mu_y(B_n)<\infty$. (2) $\forall y \in Y\ (\mu_y(X-\bigcup_n B_n)=0$. \mys \begin{ques}[Maharam] Is every $\sigma$ finite disintegration uniformly $\sigma$-finite? \end{ques} \end{frame} \begin{frame} \begin{thm}[Backs, J, Mauldin] Assuming $V=L$, there is a $\sigma$-finite disintegration which is not uniformly $\sigma$-finite. \end{thm} \mys The proof uses the following improvement to the earlier non-uniformization result. \mys \begin{thm} Assume $V=L$. Then there is a $\bp^1_1$ set $G \subseteq \ww \times \ww$ satisfying \begin{enumerate} \item $\forall x \in X\ |G_x| \leq \omega$ \item $G$ is not the union of countably many $\bp^1_1$ (or $\bs^1_2$) graphs. \item $\forall n \in \omega\ \forall B\in \sB(Y)\ \{ x \in X \colon |G_x|=n \} \in \sB(X)$ \end{enumerate} \end{thm} \end{frame} \end{document}