Horizontal Section of Measurable Set is Measurable

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Theorem

Let $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$ be measurable spaces.

Let $E \in \Sigma_X \otimes \Sigma_Y$ where $\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$.

Let $y \in Y$.


Then:

$E^y \in \Sigma_X$

where $E^y$ is the $y$-horizontal section of $E$.


Proof 1

Let:

$\FF = \set {E \subseteq X \times Y : E^y \in \Sigma_X}$

We will show that $\FF$ contains each $S_1 \times S_2$ with $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$.

We will then show that $\FF$ is a $\sigma$-algebra, at which point we will have:

$\map \sigma {\set {S_1 \times S_2 : S_1 \in \Sigma_X, \, S_2 \in \Sigma_Y} } \subseteq \FF$

from Sigma-Algebra Contains Generated Sigma-Algebra of Subset.

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

$\Sigma_X \otimes \Sigma_Y \subseteq \FF$

We will then have the demand.


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

From Horizontal Section of Cartesian Product, we have:

$\ds \paren {S_1 \times S_2}^y = \begin{cases}S_1 & y \in S_2 \\ \O & y \not \in S_2\end{cases}$

From the definition of a $\sigma$-algebra, we have $\O \in \Sigma_X$, so in either case we have:

$\paren {S_1 \times S_2}^y \in \Sigma_X$

That is:

$S_1 \times S_2 \in \FF$


It remains to show that $\FF$ is a $\sigma$-algebra.

Since $S_1 \times S_2 \in \FF$ for $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$.

Since $X \in \Sigma_X$ and $Y \in \Sigma_Y$, we obtain:

$X \times Y \in \FF$

We show that $\FF$ is closed under countable union.

Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\FF$.

We have $\paren {E_n}^y \in \Sigma_X$ for each $n \in \N$.

So, since $\Sigma_X$ is a $\sigma$-algebra, we have:

$\ds \bigcup_{n \mathop = 1}^\infty \paren {E_n}^y \in \Sigma_X$

From Union of Horizontal Sections is Horizontal Section of Union, we have:

$\ds \bigcup_{n \mathop = 1}^\infty \paren {E_n}^y = \paren {\bigcup_{n \mathop = 1}^\infty E_n}^y$

So we have:

$\ds \paren {\bigcup_{n \mathop = 1}^\infty E_n}^y \in \Sigma_X$

That is:

$\ds \bigcup_{n \mathop = 1}^\infty E_n \in \FF$

We finally show that $\FF$ is closed under complementation.

Let $E \in \FF$.

We then have $E^y \in \Sigma_X$.

Since $\Sigma_X$ is closed under complementation, we have $X \setminus E^y \in \Sigma_X$.

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

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

so that:

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

giving:

$\paren {X \times Y} \setminus E \in \FF$

So $\FF$ is a $\sigma$-algebra.


As discussed, we therefore obtain:

$\Sigma_X \otimes \Sigma_Y \subseteq \FF$

In particular, for any $E \in \Sigma_X \otimes \Sigma_Y$, we have $E \in \FF$.

That is:

for any $E \in \Sigma_X \otimes \Sigma_Y$ we have $E^y \in \Sigma_X$

as was the demand.

$\blacksquare$


Proof 2

Define $f_y : X \to X \times Y$ by:

$\map {f_y} x = \tuple {x, y}$

for each $x \in X$.

Note that we have $\tuple {x, y} \in E$ if and only if $x \in E^y$ from the definition of the horizontal section.

In other words:

$\paren {f_y}^{-1} \sqbrk E = E^y$

We now show that $f_y$ is $\Sigma_X/\paren {\Sigma_X \otimes \Sigma_Y}$-measurable.

We use Mapping Measurable iff Measurable on Generator.

By the definition of the product $\sigma$-algebra, $\Sigma_X \otimes \Sigma_Y$ is generated by the rectangles $A \times B$ for $A \in \Sigma_X$ and $B \in \Sigma_Y$.

From Horizontal Section of Cartesian Product, we have:

$\ds \paren {A \times B}^y = \begin{cases}A & y \in B \\ \O & y \not \in B\end{cases}$

So for each $A \in \Sigma_X$ and $B \in \Sigma_Y$, we have:

$\ds \paren {f_y}^{-1} \sqbrk {A \times B} = \begin{cases}A & y \in B \\ \O & y \not \in B\end{cases}$

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

$\O \in \Sigma_X$

So we have:

$\ds \paren {f_y}^{-1} \sqbrk {A \times B} \in \Sigma_X$

for any $A \in \Sigma_X$ and $B \in \Sigma_Y$.

So for any $G \in \set {A \times B : A \in \Sigma_X, \, B \in \Sigma_Y}$ we have:

$\paren {f_y}^{-1} \sqbrk G \in \Sigma_X$

So by Mapping Measurable iff Measurable on Generator:

$f_y$ is $\Sigma_X/\paren {\Sigma_X \otimes \Sigma_Y}$-measurable.

So, for any $E \in \Sigma_X \otimes \Sigma_Y$ we have:

$\paren {f_y}^{-1} \sqbrk E \in \Sigma_X$

So:

$E^y \in \Sigma_X$

$\blacksquare$


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