Measure of Vertical Section of Cartesian Product

Theorem

Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be measure spaces.

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

Let $x \in X$.

Then:

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

where:

$\paren {S_1 \times S_2}_x$ is the $x$-vertical section of $S_1 \times S_2$

$\chi_{S_1}$ is the characteristic function for $S_1$.

Proof

From Vertical Section of Cartesian Product, we have:

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

So:

$\map {\nu} {\paren {S_1 \times S_2}_x} = \begin{cases}\map {\nu} {S_2} & x \in S_1 \\ 0 & x \not \in S_1\end{cases}$

That is:

$\map {\nu} {\paren {S_1 \times S_2}_x} = \map {\nu} {S_2} \times \begin{cases}1 & x \in S_1 \\ 0 & x \not \in S_1\end{cases}$

giving:

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

from the definition of a characteristic function.

$\blacksquare$