Integral of Vertical Section of Measurable Function gives Measurable Function

Theorem
Let $\struct {X, \Sigma_X, \mu_X}$ and $\struct {Y, \Sigma_Y, \mu_Y}$ be $\sigma$-finite measure spaces.

Let $f : X \times Y \to \overline \R_{\ge 0}$ be a positive $\Sigma_X \otimes \Sigma_Y$-measurable function, where $\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$.

Define the function $g : X \to \overline \R$ by:


 * $\ds \map g x = \int f_x \rd \mu_Y$

where $f_x$ is the $x$-vertical section of $f$.

Then:


 * $g$ is $\Sigma_X$-measurable.