Existence of Product Measures

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

Then there exists a measure $\rho$ on the product space $\paren {X \times Y, \Sigma_1 \otimes \Sigma_2}$ such that:


 * $\forall E_1 \in \Sigma_X, E_2 \in \Sigma_Y: \map \rho {E_1 \times E_2} = \map \mu {E_1} \map \nu {E_2}$

and all such measures are $\sigma$-finite.

Further, two such measures are given by:


 * $\ds \map {\rho_1} E = \int_X \map \nu {E_x} \rd \mu$

and:


 * $\ds \map {\rho_2} E = \int_Y \map \mu {E^y} \rd \nu$

for each $E \in \Sigma_1 \otimes \Sigma_2$.

Proof
For each $E \in \Sigma_1 \otimes \Sigma_2$ define the function $f_E : X \to \overline \R$ by:


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

for each $x \in X$.

Define the function $g_E : Y \to \overline \R$ by:


 * $\map {g_E} y = \map \mu {E^y}$

for each $y \in Y$.

From Measure of Vertical Section of Measurable Set gives Measurable Function, we have that:


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

From Measure of Horizontal Section of Measurable Set gives Measurable Function, we have that:


 * $g_E$ is $\Sigma_Y$-measurable.

Note that $f_E \ge 0$ and $g_E \ge 0$ since $\mu \ge 0$ and $\nu \ge 0$.

We can therefore define the function $\rho_1 : \Sigma_X \otimes \Sigma_Y \to \overline \R$ by:


 * $\ds \map {\rho_1} E = \int_X \map \nu {E_x} \rd \mu$

and the function $\rho_2 : \Sigma_X \otimes \Sigma_Y \to \overline \R$ by:


 * $\ds \map {\rho_2} E = \int_Y \map \mu {E^y} \rd \nu$

for each $E \in \Sigma_X \otimes \Sigma_Y$.

We show that $\rho_1$ and $\rho_2$ are measures.

We verify each of the conditions $(1)$, $(2)$ and $(3)$ for $\rho_1$ and $\rho_2$.

From the definition of the integral of a positive measurable function, we have that:


 * $\map {\rho_1} E \ge 0$

and:


 * $\map {\rho_2} E \ge 0$

for each $E \in \Sigma_X \otimes \Sigma_Y$.

This verifies $(1)$ for $\rho_1$ and $\rho_2$.

Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma_X \otimes \Sigma_Y$-measurable sets.

From Intersection of Vertical Sections is Vertical Section of Intersection and Vertical Section of Measurable Set is Measurable, we have that:


 * $\sequence {\paren {E_n}_x}_{n \mathop \in \N}$ is a sequence of pairwise disjoint $\Sigma_Y$-measurable sets.

From Intersection of Horizontal Sections is Horizontal Section of Intersection and Horizontal Section of Measurable Set is Measurable, we have that:


 * $\sequence {\paren {E_n}^y}_{n \mathop \in \N}$ is a sequence of pairwise disjoint $\Sigma_X$-measurable sets.

We have:

and:

So $\rho_1$ and $\rho_2$ are countably additive.

This verifies $(2)$ for $\rho_1$ and $\rho_2$.

We finally verify $(3)$.

We have:

and:

verifying $(3)$ for $\rho_1$ and $\rho_2$.

So $\rho_1$ and $\rho_2$ are both measures.

We now show that:


 * $\forall E_1 \in \Sigma_X, E_2 \in \Sigma_Y: \map {\rho_1} {E_1 \times E_2} = \map \mu {E_1} \map \nu {E_2}$

and:


 * $\forall E_1 \in \Sigma_X, E_2 \in \Sigma_Y: \map {\rho_2} {E_1 \times E_2} = \map \mu {E_1} \map \nu {E_2}$

Let $E_1 \in \Sigma_X$ and $E_2 \in \Sigma_Y$.

We have:

and:

giving the demand for both $\rho_1$ and $\rho_2$.

Finally, we show that any measure $\rho$ satsfiying:


 * $\forall E_1 \in \Sigma_X, E_2 \in \Sigma_Y: \map \rho {E_1 \times E_2} = \map \mu {E_1} \map \nu {E_2}$

is $\sigma$-finite.

Since $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ are $\sigma$-finite measure spaces, we have:


 * $\mu$ and $\nu$ are $\sigma$-finite measures on $X$ and $Y$ respectively.

Since $\mu$ is $\sigma$-finite, by Measure Space Sigma-Finite iff Cover by Sets of Finite Measure, there exists a sequence of $\Sigma_X$-measurable sets $\sequence {E_n}_{n \mathop \in \N}$ such that:


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

with:


 * $\map \mu {E_n} < \infty$ for each $n$.

Since $\nu$ is $\sigma$-finite, by Measure Space Sigma-Finite iff Cover by Sets of Finite Measure, there exists a sequence of $\Sigma_Y$-measurable sets $\sequence {F_n}_{n \mathop \in \N}$ such that:


 * $\ds Y = \bigcup_{n \mathop = 1}^\infty F_n$

with:


 * $\map \mu {F_n} < \infty$ for each $n$.

We then have:

Since $E_n \in \Sigma_X$ and $F_m \in \Sigma_Y$, we have:


 * $E_n \times F_m \in \Sigma_X \otimes \Sigma_Y$

for each $n, m$.

For each $n, m$, we also have:


 * $\map {\rho} {E_n \times F_m} = \map {\mu_X} {E_n} \map {\mu_Y} {F_m} < \infty$

From Cartesian Product of Countable Sets is Countable, we have that:


 * $\N \times \N$ is countable.

So:


 * $\ds \bigcup_{\tuple {n, m} \in \N \times \N} \paren {E_n \times F_m}$ is the countable union of $\Sigma_X \otimes \Sigma_Y$-measurable sets with $\map {\rho} {E_n \times F_m} < \infty$ for each $n, m$.

So, from Measure Space Sigma-Finite iff Cover by Sets of Finite Measure, we have:


 * $\rho$ is $\sigma$-finite

completing the proof.