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.  
 A: 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.)
A: 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.
