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.

Proof
First we prove the case of:


 * $f = \chi_E$

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

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


 * $f_x = \chi_{E_x}$

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


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

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

so that:


 * $\map g x = \map {\mu_Y} {E_x}$ for each $x \in X$.

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


 * $g$ is $\Sigma_X$-measurable

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

Now consider the case of 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
 * $a_1, a_2, \ldots, a_n$ real numbers.

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


 * $f_x$ is a positive simple function

with:


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

where:


 * $\paren {E_1}_x, \paren {E_2}_x, \ldots, \paren {E_n}_x$ are pairwise disjoint $\Sigma_Y$-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:


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

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

giving:


 * $\ds \map g x = \sum_{k \mathop = 1}^n a_k \map {\mu_Y} {\paren {E_k}_x}$

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


 * $g$ is $\Sigma_X$-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 positive simple functions $\sequence {f_n}_{n \mathop \in \N}$ such that $f_n \to f$ pointwise.

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


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

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


 * $\ds f_x = \lim_{n \mathop \to \infty} \paren {f_n}_x$

From the Monotone Convergence Theorem, we then have:


 * $\ds \map g x = \int f_x \rd \mu_Y = \lim_{n \mathop \to \infty} \int \paren {f_n}_x \rd \mu_Y$

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


 * $\ds \map {g_n} x = \int \paren {f_n}_x \rd \mu_Y$

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 $f$.