Existence of Product Measures

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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:

\(\ds \map {\rho_1} {\bigcup_{n \mathop = 1}^\infty E_n}\) \(=\) \(\ds \int_X \map \nu {\paren {\bigcup_{n \mathop = 1}^\infty E_n}_x} \rd \mu\)
\(\ds \) \(=\) \(\ds \int_X \map \nu {\bigcup_{n \mathop = 1}^\infty \paren {E_n}_x} \rd \mu\) Union of Vertical Sections is Vertical Section of Union
\(\ds \) \(=\) \(\ds \int_X \paren {\sum_{n \mathop = 1}^\infty \map \nu {\paren {E_n}_x} } \rd \mu\) since $\nu$ is countably additive
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\int_X \map \nu {\paren {E_n}_x} \rd \mu}\) Integral of Series of Positive Measurable Functions
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {\rho_1} {E_n}\)

and:

\(\ds \map {\rho_2} {\bigcup_{n \mathop = 1}^\infty E_n}\) \(=\) \(\ds \int_Y \map \mu {\paren {\bigcup_{n \mathop = 1}^\infty E_n}^y} \rd \nu\)
\(\ds \) \(=\) \(\ds \int_Y \map \mu {\bigcup_{n \mathop = 1}^\infty \paren {E_n}^y} \rd \nu\) Union of Horizontal Sections is Horizontal Section of Union
\(\ds \) \(=\) \(\ds \int_Y \paren {\sum_{n \mathop = 1}^\infty \map \mu {\paren {E_n}^y} } \rd \nu\) since $\mu$ is countably additive
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\int_Y \map \mu {\paren {E_n}^y} \rd \nu}\) Integral of Series of Positive Measurable Functions
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {\rho_2} {E_n}\)

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

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

We finally verify $(3)$.

We have:

\(\ds \map {\rho_1} \O\) \(=\) \(\ds \int_X \map \nu {\O_x} \rd \mu\)
\(\ds \) \(=\) \(\ds \int_X \map \nu \O \rd \mu\) Vertical Section of Empty Set
\(\ds \) \(=\) \(\ds \int_X 0 \rd \mu\) Empty Set is Null Set
\(\ds \) \(=\) \(\ds 0\) Measurable Function Zero A.E. iff Absolute Value has Zero Integral

and:

\(\ds \map {\rho_2} \O\) \(=\) \(\ds \int_Y \map \mu {\O^y} \rd \nu\)
\(\ds \) \(=\) \(\ds \int_Y \map \mu \O \rd \nu\) Horizontal Section of Empty Set
\(\ds \) \(=\) \(\ds \int_Y 0 \rd \nu\) Empty Set is Null Set
\(\ds \) \(=\) \(\ds 0\) Measurable Function Zero A.E. iff Absolute Value has Zero Integral

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:

\(\ds \map {\rho_1} {E_1 \times E_2}\) \(=\) \(\ds \int_X \map \nu {\paren {E_1 \times E_2}_x} \rd \mu\)
\(\ds \) \(=\) \(\ds \int_X \map \nu {E_2} \chi_{E_1} \rd \mu\) Measure of Vertical Section of Cartesian Product
\(\ds \) \(=\) \(\ds \map \nu {E_2} \int_X \chi_{E_1} \rd \mu\) Integral of Positive Measurable Function is Positive Homogeneous
\(\ds \) \(=\) \(\ds \map \mu {E_1} \map \nu {E_2}\) Integral of Characteristic Function: Corollary

and:

\(\ds \map {\rho_2} {E_1 \times E_2}\) \(=\) \(\ds \int_Y \map \mu {\paren {E_1 \times E_2}^y} \rd \nu\)
\(\ds \) \(=\) \(\ds \int_Y \map \mu {E_1} \chi_{E_2} \rd \nu\) Measure of Horizontal Section of Cartesian Product
\(\ds \) \(=\) \(\ds \map \mu {E_1} \int_Y \chi_{E_2} \rd \nu\) Integral of Positive Measurable Function is Positive Homogeneous
\(\ds \) \(=\) \(\ds \map \mu {E_1} \map \nu {E_2}\) Integral of Characteristic Function: Corollary

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, 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, 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:

\(\ds X \times Y\) \(=\) \(\ds \paren {\bigcup_{n \mathop = 1}^\infty E_n} \times \paren {\bigcup_{n \mathop = 1}^\infty F_n}\)
\(\ds \) \(=\) \(\ds \bigcup_{\tuple {n, m} \in \N \times \N} \paren {E_n \times F_m}\) Cartesian Product of Unions: General Result

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 $n, m$.

Hence:

$\rho$ is $\sigma$-finite

completing the proof.

$\blacksquare$


Sources

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