Proof of cuts measurable Prove that the set $ D=\{(x,x): x \in [0,\frac{1}{2}]\} $ has all cuts measurable but not measurable against $\sigma $-product body $\mathcal{A} \otimes \mathcal{A} $.
I started with that: $ D_{x_0}= \{x_0\}$ for $x_0\le \frac{1}{2} $ and $\emptyset $ for $x_{0} > \frac{1}{2}. $
Hence, the set D has all the cuts that are measurable because $\emptyset \in \mathcal{A} $ and $\{x_{0}\} \in \mathcal{A} $. However, I have a problem with proving that $D$ is not measurable in regard of $\sigma $-product body $\mathcal{A} \otimes \mathcal{A} $. Does anyone have any idea how to prove this?
 A: Ok, The measurable space is $([0,1],\mathcal{A})$ where $\mathcal{A}$ is the $\sigma$-field generated by the countable subsets of $[0,1]$.
Consider $f: [0,1] \rightarrow [0,1] \times [0,1]$  defined by $f(x)=(x,x)$. It is easy to see that $f$ is a measurable function (it is enough to consider the pre-image of rectangles $A \times B$ where $A$ and $B$ are countable or have countable complement).
Now, if $D \in \mathcal{A} \otimes \mathcal{A} $ then we should have
$ \left [0,\frac{1}{2} \right ] = f^{-1}(D) \in \mathcal{A}$.
So $D \notin \mathcal{A} \otimes \mathcal{A} $.
Remark: Since $\mathcal{A}$ is the $\sigma$-field generated by the countable subsets of $[0,1]$, we have that $\mathcal {A} = \{ E \subset [0,1] : E \textrm{ or } E^c \textrm { is countable }\}$
In fact, it is easy to check that  $\{ E \subset [0,1] : E \textrm{ or } E^c \textrm { is countable }\}$ is a $\sigma$-field and it is clealy the smallest $\sigma$-field containing the countable subsets of $[0,1]$. So $\{ E \subset [0,1] : E \textrm{ or } E^c \textrm { is countable }\}$ is the  $\sigma$-field generated by the countable subsets of $[0,1]$.
