Show that there exists a set $E \in \mathbb{R} \times \mathbb{R}$ such that the cross sections are borel sets, but $E \notin B\otimes B$

Question

From Axler's MIRA

Let $B$ denote the $\sigma$-algebra of Borel subsets of $\mathbb{R}$. Show that there exists a set $E \subset \mathbb{R} \times \mathbb{R}$ such that $[E]_a \in B$ and $[E]^a \in B$ for every $a \in \mathbb{R}$, but $E \notin B \otimes B$.

Here $[E]_a$ is the cross section with a fixed second coordinate and $[E]^a$ is the cross section with a fixed first coordinate. Currently my approach to this problem is to use a non-measurable set such as the vitali set $V$ so that $V \times V \notin B \oplus B$ and then having the projection onto an axis make it trivially a Borel subset, however everytime I try to write this up it ends up being wrong so I am wondering if I am even going in the right direction with this problem now.

edit: Here is what I wrote up so far btw

Answer 1

Following Sheldon Axler's hints, we construct $E$ without invoking the Continuum Hypothesis. Define an equivalence relation $\sim$ on $[0,1]$ by $x\sim y$ iff $x-y\in\mathbb{Q}$. Denote the quotient set $[0,1]/\sim$ by $\{A_{i}\mid i\in I\}$. By the Axiom of Choice, there exists a map $\theta:I\rightarrow\cup_{i\in I}A_{i}=[0,1]$ such that $\theta(i)\in A_{i}$ for each $i$. Denote $V=\theta(I)$. We go to show that $V$ is not Lebesgue measurable by contradiction.

Suppose the contrary that $V$ is Lebesgue measurable. Let $\mathbb{Q}_{1}=\mathbb{Q}\cap[-1,1]$. We assert that $[0,1]\subseteq\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)\subseteq[-1,2]$ and the union is disjoint. Denote $x_{i}=\theta(i)$. Let $x\in[0,1]$, then $x\sim x_{i}$ for some $i$. Clearly $r_{0}:=x-x_{i}\in\mathbb{Q}_{1}$. Therefore, $x=x_{i}+r_{0}\in V+r_{0}\subseteq\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)$. Let $x\in\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)$, then $x=x_{i}+r$ for some $i\in I$ and $r\in\mathbb{Q}_{1}$. Note that $x_{i}\in[0,1]$ and $r\in[-1,1]$, so $x\in[-1,2]$. Finally, we go to show that the union $\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)$ is disjoint. Suppose that there exist $r_{1}\neq r_{2}$ such that $(V+r_{1})\cap(V+r_{2})\ne\emptyset$. Choose $\xi\in(V+r_{1})\cap(V+r_{2})$. $\xi\in(V+r_{1})$ implies that $\xi=x_{i}+r_{1}$ for some $i$. $\xi\in(V+r_{2})$ implies that $\xi=x_{j}+r_{2}$ for some $j$. Hence, $x_{i}+r_{1}=x_{j}+r_{2}\Rightarrow x_{i}-x_{j}=r_{2}-r_{1}\in\mathbb{Q}\Rightarrow x_{i}\sim x_{j}\Rightarrow x_{i}=x_{j}$. It follows that $r_{1}=r_{2}$, which is a contradiction.

Let $l=m(V)$. Since Lebesgue measure is translation invariant, we have that \begin{eqnarray*} & & m\left(\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)\right)\\ & = & \sum_{r\in\mathbb{Q}_{1}}m(V+r)\\ & = & \sum_{r\in\mathbb{Q}_{1}}m(V)\\ & = & \begin{cases} 0, & \mbox{ if }l=0\\ \infty, & \mbox{ if }l>0 \end{cases}, \end{eqnarray*} which is impossible because $1=m\left([0,1]\right)\leq m\left(\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)\right)\leq m([-1,2])=3$.

Finally, we define $E=\{(x,x)\mid x\in V\}$. For each $t\in\mathbb{R}$, \begin{eqnarray*} E_{t} & = & \{x\mid(x,t)\in E\}\\ & = & \begin{cases} \{t\}, & \mbox{ if }t\in V\\ \emptyset, & \mbox{ if }t\notin V \end{cases}. \end{eqnarray*} Therefore $E_{t}$ is Borel. Similarly, we can prove that $E^{t}=\{y\mid(t,y)\in V\}$ is Borel. Next, we go to show that $E\notin\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$. Let $\pi_{X}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the canonical projection onto the first coordinate, i.e., $\pi_{X}(x,y)=x$. Let $\pi_{Y}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the canonical projection onto the second coordinate. We recall the universal property of joint measurability: For any map $f:\mathbb{R}\rightarrow\mathbb{R}^{2}$, $f$ is $\mathcal{B}/\mathcal{B}\otimes\mathcal{B}$-measurable iff $\pi_{X}\circ f$ and $\pi_{Y}\circ f$ are measurable. Define $f:\mathbb{R}\rightarrow\mathbb{R}^{2}$ by $f(t)=(t,t)$. Clearly $\pi_{X}\circ f=\pi_{Y}\circ f=\mbox{id}_{\mathbb{R}}$, where $\mbox{id}_{\mathbb{R}}:\mathbb{R}\rightarrow\mathbb{R}$, $\mbox{id}_{\mathbb{R}}(t)=t$, which is clearly a Borel function. Therefore, $t\mapsto(t,t)$ is $\mathcal{B}/\mathcal{B}\otimes\mathcal{B}$-measurable. Finally, $V=f^{-1}(E)$. Since $V\notin\mathcal{B}$, we must have $E\notin\mathcal{B}\otimes\mathcal{B}$.

Answer 2

We prove by assuming the Continuum Hypothesis. Let $\omega_1=\{\alpha\mid\alpha<\omega_1\}$ be the first uncountable ordinal. Assume that the Continuum Hypothesis holds, i.e., $\omega_1 = c := 2^\omega$. Choose a bijection $\theta:c\rightarrow\mathbb{R}.$ Denote $x_{\alpha}=\theta(\alpha)$, $\alpha<c$. Define $E=\{(x_{\alpha},x_{\beta})\mid\alpha<\beta<c\}$. We go to show that all $x$-sections and $y$-sections of $E$ are Borel. Let $t\in\mathbb{R}$ be fixed, then there exists a unique $\alpha_{0}<c$ such that $t=x_{\alpha_{0}}$.

Observe that \begin{eqnarray*} E_{t} & = & \{x\in\mathbb{R}\mid(x,t)\in E\}\\ & = & \{x_{\alpha}\mid\alpha<\alpha_{0}\} \end{eqnarray*} which is countable, so $E_{t}$ is a Borel set.

Also, \begin{eqnarray*} E^{t} & = & \{y\in\mathbb{R}\mid(t,y)\in E\}\\ & = & \{x_{\beta}\mid\alpha_{0}<\beta<c\} \end{eqnarray*} which is co-countable (i.e., its complement is countable. For, the complement $(E^{t})^{c}=\{x_{\beta}\mid\beta\leq\alpha_{0}\}$ is countable), so $E^{t}$ is also a Borel set.

Since the Lebesgue measure $\lambda$ on $\mathbb{R}$ is $\sigma$-finite, Fubini-Tonelli Theorem is applicable. We prove by contradiction that $E\notin\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$. Suppose the contrary that $E\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}),$ then $$ \int\left(\int1_{E}(x,y)dx\right)dy=\int\left(\int1_{E}(x,y)dy\right)dx. $$ For each $y$, $1_{E}(x,y)=1_{E_{y}}(x)$, so $\int1_{E}(x,y)dx=\lambda(E_{y})=0$ because $E_{y}$ is a countable set. Hence, $\int\left(\int1_{E}(x,y)dx\right)dy=\int0dy=0$. On the other hand, for each $x$, $1_{E}(x,y)=1_{E^{x}}(y)$, so $\int1_{E}(x,y)dy=\lambda(E^{x})=\infty$ because $E^{x}$ is co-countable. It follows that $\int\left(\int1_{E}(x,y)dy\right)dx=\int\infty dx=\infty$. We arrive a contradiction.