Measure of Horizontal Section of Measurable Set gives Measurable Function

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Theorem

Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be $\sigma$-finite measure spaces.

For each $E \in \Sigma_X \otimes \Sigma_Y$, define the function $f_E : Y \to \overline \R$ by:

$\map {f_E} x = \map {\mu} {E^y}$

for each $y \in Y$ where:

$\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$
$E^y$ is the $y$-horizontal section of $E$.


Then $f_E$ is $\Sigma_Y$-measurable for each $E \in \Sigma_X \otimes \Sigma_Y$.


Proof

From Horizontal Section of Measurable Set is Measurable, the function $f_E$ is certainly well-defined for each $E \in \Sigma_X \otimes \Sigma_Y$.


First suppose that $\mu$ is a finite measure.

Let:

$\mathcal F = \set {E \in \Sigma_X \otimes \Sigma_Y : f_E \text { is } \Sigma_Y\text{-measurable} }$

We aim to show that:

$\mathcal F = \Sigma_X \otimes \Sigma_Y$

at which point we will have the demand, since for all $E \in \Sigma_X \otimes \Sigma_Y$ we will have that $f_E$ is $\Sigma_Y$-measurable.

Since we clearly have:

$\mathcal F \subseteq \Sigma_X \otimes \Sigma_Y$

we only need to show:

$\Sigma_X \otimes \Sigma_Y \subseteq \mathcal F$


We first show that:

$S_1 \times S_2 \in \mathcal F$

for $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$.

From Measure of Horizontal Section of Cartesian Product, we have:

$\map {\mu} {\paren {S_1 \times S_2}^y} = \map {\mu} {S_1} \map {\chi_{S_1} } x$

Since $S_1$ is $\Sigma_X$-measurable, we have:

$\chi_{S_1}$ is $\Sigma_X$-measurable.

Then from Pointwise Scalar Multiple of Measurable Function is Measurable, we have:

$f_{S_1 \times S_2}$ is $\Sigma_X$-measurable.

So:

$S_1 \times S_2 \in \mathcal F$


With a view to apply Dynkin System with Generator Closed under Intersection is Sigma-Algebra, we first show that $\mathcal F$ is a Dynkin system.

We verify the three conditions of a Dynkin system.

Since:

$S_1 \times S_2 \in \mathcal F$

for each $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$, we have:

$X \times Y \in \mathcal F$

hence $(1)$ is shown.

Let $D \in \mathcal F$.

We aim to show that $\paren {X \times Y} \setminus D \in \mathcal F$.

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

$\paren {\paren {X \times Y} \setminus E}^y = X \setminus E^y$

Note that since $\mu$ is a finite measure, we have that:

$\map {\mu} X$ and $\map {\mu} {E_y}$ are finite.

So:

\(\ds \map {f_{\paren {X \times Y} \setminus D} } y\) \(=\) \(\ds \map {\mu} {\paren {\paren {X \times Y} \setminus E}^y}\)
\(\ds \) \(=\) \(\ds \map {\mu} {X \setminus E^y}\)
\(\ds \) \(=\) \(\ds \map {\mu} X - \map {\mu} {E^y}\) Measure of Set Difference with Subset

Since $E \in \mathcal F$, we have:

$f_E$ is $\Sigma_X$-measurable.

From Constant Function is Measurable and Pointwise Difference of Measurable Functions is Measurable, we have:

$\map {\mu} X - \map {\mu} {E^\circ} = f_{\paren {X \times Y} \setminus E}$ is $\Sigma_X$-measurable.

so:

$\paren {X \times Y} \setminus E \in \mathcal F$

and $(2)$ is verified.

Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\mathcal F$.

From Intersection of Horizontal Sections is Horizontal Section of Intersection, we have that $\sequence {\paren {D_n}^y}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets such that:

$f_{D_i}$ is $\Sigma_X$-measurable for each $i$.

Write:

$\ds D = \bigcup_{n \mathop = 1}^\infty D_n$

We want to show that:

$f_D$ is $\Sigma_X$-measurable

so that:

$D \in \mathcal F$

at which point we will have $(3)$.

We have, for each $y \in Y$:

\(\ds \map {f_D} y\) \(=\) \(\ds \map {\mu} {D^y}\)
\(\ds \) \(=\) \(\ds \map {\mu} {\paren {\bigcup_{n \mathop = 1}^\infty D_n}^y}\)
\(\ds \) \(=\) \(\ds \map {\mu} {\bigcup_{n \mathop = 1}^\infty \paren {D_n}^y}\) Union of Horizontal Sections is Horizontal Section of Union
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {\mu} {\paren {D_n}^y}\) since $\mu$ is countably additive
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {f_{D_n} } y\)

From Infinite Series of Measurable Functions is Measurable, we have that:

$f_D$ is $\Sigma_Y$-measurable.

So:

$\ds D = \bigcup_{n \mathop = 1}^\infty D_n \in \mathcal F$

Since $\sequence {D_n}_{n \mathop \in \N}$ was an arbitrary sequence of pairwise disjoint sets in $\mathcal F$, $(3)$ is verified.


Define:

$\mathcal G = \set {S_1 \times S_2 : S_1 \in \Sigma_X \text { and } S_2 \in \Sigma_Y}$

Since $\mathcal F$ is a Dynkin system, from the definition of a Dynkin system generated by a collection of subsets we have:

$\map \delta {\mathcal G} \subseteq \mathcal F$

We show that $\mathcal G$ is a $\pi$-system, at which point we may apply Dynkin System with Generator Closed under Intersection is Sigma-Algebra.

Let $A_1, A_2 \in \Sigma_X$ and $B_1, B_2 \in \Sigma_Y$.

Then from Cartesian Product of Intersections, we have:

$\paren {A_1 \times B_1} \cap \paren {A_2 \times B_2} = \paren {A_1 \cap A_2} \times \paren {B_1 \cap B_2}$

From Sigma-Algebra Closed under Countable Intersection, we have:

$A_1 \cap A_2 \in \Sigma_X$

and:

$B_1 \cap B_2 \in \Sigma_Y$

so:

$\paren {A_1 \times B_1} \cap \paren {A_2 \times B_2} \in \mathcal G$

So $\mathcal G$ is a $\pi$-system.

From Dynkin System with Generator Closed under Intersection is Sigma-Algebra, we have:

$\map \delta {\mathcal G} = \map \sigma {\mathcal G}$

so:

$\map \sigma {\mathcal G} \subseteq \mathcal F$

From the definition of product $\sigma$-algebra, we have:

$\map \sigma {\mathcal G} = \Sigma_X \otimes \Sigma_Y$

So:

$\Sigma_X \otimes \Sigma_Y \subseteq \mathcal F$

Hence:

$\mathcal F = \Sigma_X \otimes \Sigma_Y$

as required.


Now suppose that $\mu$ is $\sigma$-finite.

Then there exists a sequence of $\Sigma_X$-measurable sets $\sequence {X_n}_{n \mathop \in \N}$ with:

$\ds X = \bigcup_{n \mathop = 1}^\infty X_n$

with:

$\map {\mu} {X_n} < \infty$ for each $n$.

From Countable Union of Measurable Sets as Disjoint Union of Measurable Sets, there exists a sequence of pairwise disjoint $\Sigma_Y$-measurable sets $\sequence {F_n}_{n \mathop \in \N}$ with:

$\ds X = \bigcup_{n \mathop = 1}^\infty F_n$

with:

$F_n \subseteq X_n$ for each $n$.

From Measure is Monotone, we have that:

$\map {\mu} {F_n} \le \map {\mu} {X_n}$ for each $n$.

So:

$\map {\mu} {F_n}$ is finite for each $n$.

Now, for each $E \in \Sigma_X$ define:

$\map {\mu^{\paren n} } E = \map {\mu} {E \cap F_n}$

From Intersection Measure is Measure:

$\mu^{\paren n}$ is a measure for each $n$.

We also have:

$\map {\mu^{\paren n} } X = \map {\mu} {F_n} < \infty$

so:

$\mu^{\paren n}$ is a finite measure for each $n$.

For each $n$, define a function $f_E^{\paren n} : Y \to \overline \R$:

$\map {f_E^{\paren n} } y = \map {\mu^{\paren n} } {E^y}$

for each $y \in Y$.

From our previous work, we have that $f_E^{\paren n}$ is $\Sigma_X$-measurable.

For each $y \in Y$, we have:

\(\ds \map {f_E} y\) \(=\) \(\ds \map {\mu} {E^y}\)
\(\ds \) \(=\) \(\ds \map {\mu} {E^y \cap Y}\) Intersection with Subset is Subset
\(\ds \) \(=\) \(\ds \map {\mu} {E^y \cap \paren {\bigcup_{n \mathop = 1}^\infty F_n} }\)
\(\ds \) \(=\) \(\ds \map {\mu} {\bigcup_{n \mathop = 1}^\infty \paren {E^y \cap F_n} }\) Intersection Distributes over Union

Since:

$F_i \cap F_j = \O$ whenever $i \ne j$

we have:

$\paren {E^y \cap F_i} \cap \paren {E^y \cap F_j} = \O$ whenever $i \ne j$

from Intersection with Empty Set.

So, using countable additivity of $\mu$, we have:

\(\ds \map {\mu} {\bigcup_{n \mathop = 1}^\infty \paren {E^y \cap F_n} }\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {\mu} {E^y \cap F_n}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {\mu^{\paren n} } {E^y}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {f_E^{\paren n} } y\)

That is:

$\ds \map {f_E} y = \sum_{n \mathop = 1}^\infty \map {f_E^{\paren n} } y$

for each $y \in Y$.

From Infinite Series of Measurable Functions is Measurable, we have:

$f_E$ is $\Sigma_Y$-measurable.

So we get the result in the case of $\mu$ $\sigma$-finite, and we are done.

$\blacksquare$


Sources

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