Horizontal Section of Measurable Set is Measurable/Proof 2

Proof
Define $f_y : X \to X \times Y$ by:


 * $\map {f_y} x = \tuple {x, y}$

for each $x \in X$.

Note that we have $\tuple {x, y} \in E$ $x \in E^y$ from the definition of the horizontal section.

In other words:


 * $\paren {f_y}^{-1} \sqbrk E = E^y$

We now show that $f_y$ is $\Sigma_X/\paren {\Sigma_X \otimes \Sigma_Y}$-measurable.

We use Mapping Measurable iff Measurable on Generator.

By the definition of the product $\sigma$-algebra, $\Sigma_X \otimes \Sigma_Y$ is generated by the rectangles $A \times B$ for $A \in \Sigma_X$ and $B \in \Sigma_Y$.

From Horizontal Section of Cartesian Product, we have:


 * $\ds \paren {A \times B}^y = \begin{cases}A & y \in B \\ \O & y \not \in B\end{cases}$

So for each $A \in \Sigma_X$ and $B \in \Sigma_Y$, we have:


 * $\ds \paren {f_y}^{-1} \sqbrk {A \times B} = \begin{cases}A & y \in B \\ \O & y \not \in B\end{cases}$

From the definition of a $\sigma$-algebra, we have:


 * $\O \in \Sigma_X$

So we have:


 * $\ds \paren {f_y}^{-1} \sqbrk {A \times B} \in \Sigma_X$

for any $A \in \Sigma_X$ and $B \in \Sigma_Y$.

So for any $G \in \set {A \times B : A \in \Sigma_X, \, B \in \Sigma_Y}$ we have:


 * $\paren {f_y}^{-1} \sqbrk G \in \Sigma_X$

So by Mapping Measurable iff Measurable on Generator:


 * $f_y$ is $\Sigma_X/\paren {\Sigma_X \otimes \Sigma_Y}$-measurable.

So, for any $E \in \Sigma_X \otimes \Sigma_Y$ we have:


 * $\paren {f_y}^{-1} \sqbrk E \in \Sigma_X$

So:


 * $E^y \in \Sigma_X$