Vertical 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 $x \in X$.

Then:

$f_x$ is $\Sigma_Y$-measurable

where $f_x$ is the $x$-vertical 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:

$\paren {f_x}^{-1} \sqbrk D \in \Sigma_Y$ for each Borel set $D \subseteq \R$.

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

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

$\paren {f_x}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}_x$

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

$\paren {f^{-1} \sqbrk D}_x \in \Sigma_Y$

so:

$\paren {f_x}^{-1} \sqbrk D \in \Sigma_Y$

So:

$\paren {f_x}^{-1} \sqbrk D \in \Sigma_Y$ for each Borel set $D \subseteq \R$.

so:

$f_x$ is $\Sigma_Y$-measurable.

$\blacksquare$