Tonelli's Theorem/Lemma 1
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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:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
$\blacksquare$