Horizontal Section of Measurable Set is Measurable
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:
So, for any $E \in \Sigma_X \otimes \Sigma_Y$ we have:
- $\paren {f_y}^{-1} \sqbrk E \in \Sigma_X$
So:
- $E^y \in \Sigma_X$
$\blacksquare$
Sources
- 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $5.1$: Constructions: Lemma $5.1.2$