Horizontal Section of Measurable Function is Measurable

Theorem
Let $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$ be measurable spaces.

Let $f : X \times Y \to \overline \R$ be a $\Sigma_X \otimes \Sigma_Y$-measurable function where $\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$.

Let $y \in Y$.

Then:


 * $f^y$ is $\Sigma_X$-measurable

where $f^y$ is the $y$-horizontal section of $f$.

Proof
By the definition of a $\Sigma_X$-measurable function, we have that:


 * $f^{-1} \sqbrk D \in \Sigma_X \otimes \Sigma_Y$ for each Borel set $D \subseteq \R$.

We aim to show that:


 * $\paren {f^y}^{-1} \sqbrk D \in \Sigma_X$ for each Borel set $D \subseteq \R$.

Let $D \subseteq \R$ be a Borel set.

From Preimage of Horizontal Section of Function is Horizontal Section of Preimage, we have:


 * $\paren {f^y}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}^y$

From Horizontal Section of Measurable Set is Measurable, we have:


 * $\paren {f^{-1} \sqbrk D}^y \in \Sigma_X$

so:


 * $\paren {f^y}^{-1} \sqbrk D \in \Sigma_X$

So:


 * $\paren {f^y}^{-1} \sqbrk D \in \Sigma_X$ for each Borel set $D \subseteq \R$.

so:


 * $f^y$ is $\Sigma_X$-measurable.