Borel Section property - dynkin's theorem A well known property of Borel sets in $\mathbb{R}^2$ is that there cross sections are always Borel. That is $A^y:={x: (x,y)\in A}$ is a Borel set in $\mathbb{R}$ for all $y$ whenever $A$ is a Borel set in $\mathbb{R}^2$. This supposedly readily follows from Dynkin's theorem, however I do not see why we cannot just show that it is a $\sigma$-algebra directly? When proving that the set of such sets that satisfy the section property, the disjoint condition is never used?
Ex: let $F=\{A\in\mathcal{B}^2: A^y\in\mathcal{B}, y\in \mathbb{R}\}$, $\emptyset\in F$ and $F$ is closed under compliments trivially. When you let $A_1,A_2,A_3..\in F$ we know that $\bigcup_n A_n^y=(\bigcup_n A_n)^y$ and so $\bigcup A_n\in F$. So we never needed the disjointness of $A_n's$ and thus we don't need to work with any $\pi$-systems.
 A: Let me answer you (including your comment) step by step.
A. Your proof that $F$ is a $\sigma$-algebra is correct (and you don't need Dynkin's theorem here). Let us review your proof in detail.
Let $\mathcal{B}^2$ be the Borel $\sigma$-algebra in  $\mathbb{R}^2$ and $\mathcal{B}$ be the Borel $\sigma$-algebra in  $\mathbb{R}$. Let
$$F=\{A\in\mathcal{B}^2: A^y\in\mathcal{B}, \textrm{ for all } y\in \mathbb{R}\}$$
Then:

*

*It trivial that $\emptyset\in F$

*If $A \in F$, then $A\in\mathcal{B}^2$ and, for all  $y\in \mathbb{R}$, $A^y\in\mathcal{B}$. So, $A^c\in\mathcal{B}^2$ and, for all  $y\in \mathbb{R}$, $(A^c)^y\in\mathcal{B}$. So $A^c \in F$.

*Let $\{A_n\}_{n\in \mathbb{N}}$, such that for all $n$, $A_n \in F$. Then, for all $n$, $A_n\in\mathcal{B}^2$ and, for all  $y\in \mathbb{R}$, $(A_n)^y\in\mathcal{B}$. Then, $\bigcup_n A_n \in \mathcal{B}^2$ and, for all  $y\in \mathbb{R}$,
$$ \left (\bigcup_n A_n \right)^y = \bigcup_n (A_n)^y \in\mathcal{B}$$
So $\bigcup_n A_n \in F$.

So, $F$ is a $\sigma$-algebra.
B. How do we use item A to prove that: "If $A$ is a Borel set in $\mathbb{R}^2$, then, for all  $y\in \mathbb{R}$,  $A^y:={x: (x,y)\in A}$ is a Borel set in $\mathbb{R}$"?
Answer: we prove that $F=\mathcal{B}^2$.
It obvious that $F\subseteq\mathcal{B}^2$. On the other hand, it is easy to see that for all $O$ open set in $\mathbb{R}^2$, for all $y\in \mathbb{R}$, $O^y$ is open in $\mathbb{R}$, and so $O^y \in \mathcal{B}$. So, for all $O$ open set in $\mathbb{R}^2$, $O \in F$. Since $F$ is a $\sigma$-algebra, we have that  $\mathcal{B}^2 \subseteq F$. So we can conclude that $F=\mathcal{B}^2$.
C. What would happen if we try to mimic the proof above for the Lebesgue $\sigma$-algebra in $\mathbb{R}^2$?
Well, the step A will work.

Let $\mathcal{L}^2$ be the Lebesgue $\sigma$-algebra in  $\mathbb{R}^2$ and $\mathcal{L}$ be the Lebesgue $\sigma$-algebra in  $\mathbb{R}$. Let
$$F=\{A\in\mathcal{L}^2: A^y\in\mathcal{L}, \textrm{ for all } y\in \mathbb{R}\}$$
Then  $F$ is a $\sigma$-algebra. The proof is analogous to the proof in item A above.

However, the step B can not be mimic for the Lebesgue $\sigma$-algebra in $\mathbb{R}^2$. In fact,  for the Lebesgue $\sigma$-algebra in $\mathbb{R}^2$, $F$ is a $\sigma$-algebra strictly smaller than $\mathcal{L}^2$.
