Measure of Vertical Section of Measurable Set gives Measurable Function

== 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 $f_E : X \to \overline \R$ by: 
:$\map {f_E} x = \map {\nu} {E_x}$
for each $x \in X$ where: 
:$\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$
:$E_x$ is the $x$-vertical section of $E$.
Then $f_E$ is $\Sigma_X$-measurable for each $E \in \Sigma_X \otimes \Sigma_Y$. 
== Proof ==
From Vertical 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 $\nu$ is a finite measure.
Let: 
:$\mathcal F = \set {E \in \Sigma_X \otimes \Sigma_Y : f_E \text { is } \Sigma_X\text{-measurable} }$
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 $\Sigma_X$-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 Vertical Section of Cartesian Product, we have:
:$\map {\nu} {\paren {S_1 \times S_2}_x} = \map {\nu} {S_2} \map {\chi_{S_1} } x$
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 Vertical Section of Set is Vertical Section of Complement, we have: 
:$\paren {\paren {X \times Y} \setminus E}_x = Y \setminus E_x$
Note that since $\nu$ is a finite measure, we have that: 
:$\map {\nu} Y$ and $\map {\nu} {E_x}$ are finite.
So:
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: 
:$\map {\nu} Y - \map {\nu} {E_\circ} = f_{\paren {X \times Y} \setminus E}$ 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 Vertical Sections is Vertical Section of Intersection, we have that $\sequence {\paren {D_n}_x}_{n \mathop \in \N}$ 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 $x \in X$: 
From Infinite Series of Measurable Functions is Measurable, we have that: 
:$f_D$ is $\Sigma_X$-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 $\nu$ is $\sigma$-finite.
Then there exists a sequence of $\Sigma_Y$-measurable sets $\sequence {Y_n}_{n \mathop \in \N}$ with: 
:$\ds Y = \bigcup_{n \mathop = 1}^\infty Y_n$
with: 
:$\map {\nu} {Y_n} < \infty$ 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:
:$\ds Y = \bigcup_{n \mathop = 1}^\infty F_n$
with:
:$F_n \subseteq Y_n$ for each $n$. 
From Measure is Monotone, we have that: 
:$\map {\nu} {F_n} \le \map {\nu} {Y_n}$ for each $n$.
So:
:$\map {\nu} {F_n}$ is finite for each $n$.
Now, for each $E \in \Sigma$ define: 
:$\map {\nu^{\paren n} } E = \map {\nu} {E \cap F_n}$
From Intersection Measure is Measure:
:$\nu^{\paren n}$ is a measure for each $n$.
We also have: 
:$\map {\nu^{\paren n} } Y = \map {\nu} {F_n} < \infty$
so:
:$\nu^{\paren n}$ is a finite measure for each $n$. 
For each $n$, define a function $f_E^{\paren n} : X \to \overline \R$: 
:$\map {f_E^{\paren n} } x = \map {\nu^{\paren n} } {E_x}$
for each $x \in X$.
From our previous work, we have that $f_E^{\paren n}$ is $\Sigma_X$-measurable. 
For each $x \in X$, we have: 
Since:
:$F_i \cap F_j = \O$ whenever $i \ne j$
we have: 
:$\paren {E_x \cap F_i} \cap \paren {E_x \cap F_j} = \O$ whenever $i \ne j$
from Intersection with Empty Set.
So, using countable additivity of $\nu$, we have: 
That is: 
:$\ds \map {f_E} x = \sum_{n \mathop = 1}^\infty \map {f_E^{\paren n} } x$
for each $x \in X$.
From Infinite Series of Measurable Functions is Measurable, we have that: 
:$f_E$ is $\Sigma_X$-measurable.
So we get the result in the case of $\nu$ $\sigma$-finite, and we are done.
== Sources ==
* : $5.1$: Constructions: Proposition $5.1.3$
Category:Vertical Section of Sets