Does Measurablity of Cuts imply Measurability in Product Space?

Thinking about some problem I arrived at the following question:

Let $\Omega \neq \emptyset$ a set and $\Sigma$ a $\sigma$-algebra on $\Omega$. Consider the product space $\Omega \times \Omega$ with the product $\sigma$-algebra $\Sigma \otimes \Sigma$. Now it is a well known fact that if $M \in \Sigma \otimes \Sigma$ then the cuts satisfy $$M_x = \lbrace y \in \Omega \mid (x, y) \in M\rbrace \in \Sigma, \quad M_y = \lbrace x \in \Omega \mid (x, y) \in M\rbrace \in \Sigma,$$ i.e. are measurable.

Now I got interested in the converse: Assuming we have $M_x = \lbrace y \in \Omega \mid (x, y) \in M\rbrace \in \Sigma$ for all $x \in \Omega$ and $M_y = \lbrace x \in \Omega \mid (x, y) \in M\rbrace \in \Sigma$ for all $y \in \Omega$, can we deduce that $M \in \Sigma \otimes \Sigma$?

I couldn't see a quick argument and I feel like it doesn't suffice. On the other hand I couldn't think of a counter example to this question.

Solution 1:

No this is, in general, not true.

Example 1: Consider $\Omega = (0,1)$ endowed with the $\sigma$-algebra

$$\Sigma = \{A \subseteq (0,1); \text{$A$ is countable or $A^c$ is countable}\}$$

which contains, in particular, all singletons. Set $A:= (0,1/2)$ and

$$M := A \times A = \{(x,x); x \in A\}.$$

Since the slices $M_x$ and $M_y$ are either empty or singletons, we clearly have $M_x \in \Sigma$ and $M_y \in \Sigma$ for all $x,y \in \Omega$. However, $M$ is not measurable. Indeed: The mapping

$$(\Omega,\Sigma) \ni x \mapsto T(x):= (x,x) \in (\Omega \times \Omega, \Sigma \otimes \Sigma)$$

is measurable, and therefore the set $M = A \times A$ cannot be measurable as $A \notin \Sigma$. (If $M$ was measurable, then $A = T^{-1}(M)$ would be measurable.)

Remark: More generally, this counterexample works for any uncountable set $\Omega$ (choose $A \subseteq \Omega$ such that $A$ and $A^c$ are uncountable.)

Example 2: Consider $\Omega = \mathbb{R}$ endowed with Borel $\sigma$-algebra $\Sigma$ and take a set $A \subseteq \mathbb{R}$ which is not Borel measurable. Define

$$M := \{(x,x); x \in A\}.$$

Clearly, $M_x = \{x\} \in \Sigma$ and $M_y = \{y\} \in \Sigma$ for any $x,y \in \Omega$. However, $M \notin \Sigma \otimes \Sigma$. (Just note that $M$ can be obtained by rotating the non-measurable set $A \times \{0\}$ by 45 degrees.)

Solution 2:

I remember seeing an example that the converse fails. However it uses the continuum hypothesis.

Under the continuum hypothesis we can define a well ordering $\preceq$ on $[0,1]$ such that the initial segment $I_x:=\{y\in[0,1] \mid y\prec x\}$ of every $x\in[0,1]$ is countable. Then define $E$ by $$ E := \{(x,y)\in[0,1]^2 \mid y\preceq x\}. $$ Observe that for each $x\in[0,1]$, $$ E_x =\{y\in[0,1] \mid y\preceq x\} = I_x\cup\{x\} $$ and for each $y\in[0,1]$, $$ (E^y)^c = \{x\in[0,1] \mid y \preceq x\}^c = \{x\in[0,1] \mid x \prec y\} = I_y. $$ The definition of $\preceq$ forces both of these sets to be countable. Therefore all cuts of $E$ are Lebesgue measurable.

However $$ \int_{[0,1]}\int_{[0,1]}\chi_E(x,y)\,dx\,dy = 1 $$ and $$ \int_{[0,1]}\int_{[0,1]}\chi_E(x,y)\,dy\,dx = 0. $$ Thus $E$ is not Lebesgue measurable because an assumption to the contrary would imply, by Tonelli's theorem, equality in the two double integrals above.