Integral of Horizontal Section of Measurable Function gives Measurable Function

From ProofWiki
Jump to navigation Jump to search

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:

\(\ds \int f^y \rd \mu\) \(=\) \(\ds \int \chi_{E^y} \rd \mu\)
\(\ds \) \(=\) \(\ds \map {\mu} {E^y}\) Integral of Characteristic Function

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:

\(\ds \int f^y \rd \mu\) \(=\) \(\ds \int \paren {\sum_{k \mathop = 1}^n a_k \chi_{\paren {E_k}^y} } \rd \mu\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \int \paren {a_k \chi_{\paren {E_k}^y} } \rd \mu\) Integral of Positive Measurable Function is Additive
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n a_k \paren {\int \chi_{\paren {E_k}^y} \rd \mu}\) Integral of Positive Measurable Function is Positive Homogeneous
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n a_k \map {\mu} {\paren {E_k}^y}\) Integral of Characteristic Function: Corollary

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 $f$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Integral_of_Horizontal_Section_of_Measurable_Function_gives_Measurable_Function&oldid=551438"