Tonelli's Theorem/Lemma 1

Lemma

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

Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$.

Let $E \in \Sigma_X \otimes \Sigma_Y$.

Let:

$f = \chi_E$

where $\chi_E$ is the characteristic function of $E$.

Then:

$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$

where:

$f^y$ is the $y$-horizontal section of $f$

$f_x$ is the $x$-vertical section of $f$.

Proof

From Integral of Characteristic Function: Corollary, we then have:

$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \map {\paren {\mu \times \nu} } E$

From Horizontal Section of Characteristic Function is Characteristic Function of Horizontal Section, we have:

$f^y = \chi_{E^y}$

From Vertical Section of Characteristic Function is Characteristic Function of Vertical Section, we have:

$f_x = \chi_{E_x}$

We therefore have:

\(\ds \int f^y \rd \mu\)

\(=\)

\(\ds \int \chi_{E^y} \rd \mu\)

\(\ds \)

\(=\)

\(\ds \map \mu {E^y}\)

Integral of Characteristic Function: Corollary

and:

\(\ds \int f_x \rd \nu\)

\(=\)

\(\ds \int \chi_{E_x} \rd \nu\)

\(\ds \)

\(=\)

\(\ds \map \nu {E_x}\)

Integral of Characteristic Function: Corollary

Then:

\(\ds \int_Y \paren {\int_X f^y \rd \mu} \rd \nu\)

\(=\)

\(\ds \int_Y \map \mu {E^y} \rd \nu\)

\(\ds \)

\(=\)

\(\ds \map {\paren {\mu \times \nu} } E\)

Definition of Product Measure

and:

\(\ds \int_X \paren {\int_Y f_x \rd \nu} \rd \mu\)

\(=\)

\(\ds \int_X \map \nu {E_x} \rd \mu\)

\(\ds \)

\(=\)

\(\ds \map {\paren {\mu \times \nu} } E\)

Definition of Product Measure

So, we have:

$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$

$\blacksquare$