Vertical 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 $x \in X$.

Then:


 * $E_x \in \Sigma_Y$

where $E_x$ is the $x$-vertical section of $E$.

Proof
Let:


 * $\mathcal F = \set {E \subseteq X \times Y : E_x \in \Sigma_Y}$

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

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


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

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

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


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

We will then have the demand.

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

From Vertical Section of Cartesian Product, we have:


 * $\ds \paren {S_1 \times S_2}_x = \begin{cases}S_2 & x \in S_1 \\ \O & x \not \in S_1\end{cases}$

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


 * $\paren {S_1 \times S_2}_x \in \Sigma_Y$

That is:


 * $S_1 \times S_2 \in \mathcal F$

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

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

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


 * $X \times Y \in \mathcal F$

We show that $\mathcal F$ is closed under countable union.

Let $\sequence {E_n}_{n \in \N}$ be a sequence in $\mathcal F$.

We have $\paren {E_n}_x \in \Sigma_Y$ for each $n \in \N$.

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


 * $\ds \bigcup_{n \mathop = 1}^\infty \paren {E_n}_x \in \Sigma_Y$

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


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

So we have:


 * $\ds \paren {\bigcup_{n \mathop = 1}^\infty E_n}_x \in \Sigma_Y$

That is:


 * $\ds \bigcup_{n \mathop = 1}^\infty E_n \in \mathcal F$

We finally show that $\mathcal F$ is closed under complementation.

Let $E \in \mathcal F$.

We then have $E_x \in \Sigma_Y$.

Since $\Sigma_Y$ is closed under complementation, we have $Y \setminus E_x \in \Sigma_Y$.

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


 * $Y \setminus E_x = \paren {\paren {X \times Y} \setminus E}_x$

so that:


 * $\paren {\paren {X \times Y} \setminus E}_x \in \Sigma_Y$

giving:


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

So $\mathcal F$ is a $\sigma$-algebra.

As discussed, we therefore obtain:


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

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

That is:


 * for any $E \in \Sigma_X \otimes \Sigma_Y$ we have $E_x \in \Sigma_X$

as was the demand.