Measure of Vertical Section of Measurable Set gives Measurable Function

Theorem
Let $\struct {X, \Sigma_X, \mu_X}$ and $\struct {Y, \Sigma_Y, \mu_Y}$ be $\sigma$-finite measure spaces.

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


 * $\map {f_E} x = \map {\mu_Y} {E_x}$

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
First suppose that $\mu_Y$ 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 Vertical Section of Cartesian Product, we have:


 * $\ds \paren {S_1 \times S_2}_x = \begin{cases}S_2 & x \in S_1 \\ \O & x \in S_2\end{cases}$

so:


 * $\ds \map \mu {\paren {S_1 \times S_2}_x} = \begin{cases}\map \mu {S_2} & x \in S_1 \\ 0 & x \in S_2\end{cases}$

We therefore see that:


 * $\map \mu {\paren {S_1 \times S_2}_x} = \map \mu {S_2} \map {\chi_{S_1} } x$

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 $\mu_Y$ is a finite measure, we have that:


 * $\map {\mu_Y} Y$ and $\map {\mu_Y} {E_x}$ 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:


 * $\map {\mu_Y} Y - \map {\mu_Y} {E_\circ} = f_{\paren {X \times Y} \setminus E}$ 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 $x \in X$:

Define the sequence $\sequence {g_N}_{N \mathop \in \N}$ of functions $g_N : X \to \overline \R$ by:


 * $\ds \map {g_N} x = \sum_{n \mathop = 1}^N \map {f_{D_n} } x$

Then by the definition of infinite series, we have:


 * $\ds \map {f_D} x = \lim_{N \mathop \to \infty} \map {g_N} x$

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


 * $g_N$ is $\Sigma_X$-measurable for each $N$.

From Pointwise Limit of Measurable Functions is Measurable, we have:


 * $f_D$ 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 $\mu_Y$ is $\sigma$-finite.

From Measure Space Sigma-Finite iff Cover by Sets of Finite Measure, 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:


 * $\map {\mu_Y} {Y_n} < \infty$ 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:


 * $\map {\mu_Y} {F_n} \le \map {\mu_Y} {Y_n}$ for each $n$.

So:


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

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


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

From Intersection Measure is Measure:


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

We also have:


 * $\map {\mu_Y^{\paren n} } Y = \map {\mu_Y} {F_n} < \infty$

so:


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

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


 * $\map {f_E^{\paren n} } x = \map {\mu_Y^{\paren n} } {E_x}$

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 $\mu_Y$, we have:

That is:


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

for each $x \in X$.

Define the sequence $\sequence {h_N}_{N \mathop \in \N}$ of functions $h_N : X \to \overline \R$ by:


 * $\ds \map {h_N} x = \sum_{n \mathop = 1}^N \map {f_E^{\paren n} } x$

Then by the definition of infinite series, we have:


 * $\ds \map {f_E} x = \lim_{N \mathop \to \infty} \map {h_N} x$

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


 * $h_N$ is $\Sigma_X$-measurable for each $N$.

From Pointwise Limit of Measurable Functions is Measurable, we have:


 * $f_E$ is $\Sigma_X$-measurable.

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