Integral of Horizontal Section of Measurable Function gives Measurable Function

Theorem
Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ 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:


 * $\ds \map g y = \int f^y \rd \mu$

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

Then:


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

Proof
First we prove the case of:


 * $f = \chi_E$

where $E$ is a $\Sigma_X \otimes \Sigma_Y$-measurable set.

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


 * $f^y = \chi_{E^y}$

From Horizontal Section of Measurable Function is Measurable, we also have:


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

Since $f \ge 0$, we may take $\mu$-integrals, giving:

so that:


 * $\map g y = \map {\mu} {E^y}$ for each $y \in Y$.

From Measure of Horizontal Section of Measurable Set gives Measurable Function, we then have:


 * $g$ is $\Sigma_Y$-measurable

in the case that $f$ is a characteristic function.

Now consider the case of positive simple $f$.

Write the standard representation of $f$ as:


 * $\ds f = \sum_{k \mathop = 1}^n a_k \chi_{E_k}$

with:


 * $E_1, E_2, \ldots, E_n$ pairwise disjoint $\Sigma_X \otimes \Sigma_Y$-measurable sets
 * $a_1, a_2, \ldots, a_n$ non-negative real numbers.

Then, we have, from Horizontal Section of Simple Function is Simple Function:


 * $f^y$ is a positive simple function

with:


 * $\ds f^y = \sum_{k \mathop = 1}^n a_k \chi_{\paren {E_k}^y}$

where:


 * $\paren {E_1}^y, \paren {E_2}^y, \ldots, \paren {E_n}^y$ are pairwise disjoint $\Sigma_X$-measurable sets
 * $a_1, a_2, \ldots, a_n$ non-negative real numbers.

From Simple Function is Measurable, we have:


 * $f$ is $\Sigma_X \otimes \Sigma_Y$-measurable.

From Vertical Section of Measurable Function is Measurable, we also have:


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

Since $f \ge 0$, we may take $\mu$-integrals, giving:

giving:


 * $\ds \map g y = \sum_{k \mathop = 1}^n a_k \map {\mu} {\paren {E_k}^y}$

From Pointwise Sum of Measurable Functions is Measurable: General Result, we have:


 * $g$ is $\Sigma_Y$-measurable

in the case that $f$ is a simple function.

Now take a general positive $\Sigma_X \otimes \Sigma_Y$-measurable function $f$.

From Measurable Function is Pointwise Limit of Simple Functions:


 * there exists a increasing sequence of simple functions $\sequence {f_n}_{n \mathop \in \N}$ such that $f_n \to f$.

From Horizontal Section preserves Increasing Sequences of Functions, we have:


 * the sequence $\sequence {\paren {f_n}^y}_{n \mathop \in \N}$ is increasing.

From Horizontal Section preserves Pointwise Limits of Sequences of Functions, we have:


 * $\ds f^y = \lim_{n \mathop \to \infty} \paren {f_n}^y$

From the monotone convergence theorem, we then have:


 * $\ds \map g y = \int f^y \rd \mu = \lim_{n \mathop \to \infty} \int \paren {f_n}^y \rd \mu$

For each $n \in \N$, define the function $g_n : Y \to \overline \R$ by:


 * $\ds \map {g_n} y = \int \paren {f_n}^y \rd \mu$

Since each $f_n$ is a positive simple function, we have that:


 * $\ds \sequence {g_n}_{n \mathop \in \N}$ is a sequence of $\Sigma_X$-measurable functions.

So$g$ is the limit of a sequence of $\Sigma_X$-measurable functions.

Then, from Pointwise Limit of Measurable Functions is Measurable, we have:


 * $g$ is $\Sigma_X$-measurable

for each positive $\Sigma_X \otimes \Sigma_Y$-measurable $f$.