Tonelli's Theorem/Lemma 2

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 $f : X \times Y \to \overline \R$ be a simple function.

Then:


 * $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \int_X \map f {x, y} \map {\d \mu} x \map {\d \nu} y = \int_X \int_Y \map f {x, y} \map {\d \nu} y \map {\d \mu} x$

where:
 * $f^y$ is the $y$-horizontal section of $f$
 * $f_x$ is the $x$-vertical section of $f$.

Proof
Write the standard representation of $f$ as:


 * $\ds f = \sum_{k \mathop = 1}^n a_k \chi_{E_k}$

with:


 * $E_1, E_2, \ldots, E_n$ pairwise disjoint $\Sigma_X \otimes \Sigma_Y$-measurable sets
 * $a_1, a_2, \ldots, a_n$ non-negative real numbers.

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


 * $f^y$ is a positive simple function

with:


 * $\ds f^y = \sum_{k \mathop = 1}^n a_k \chi_{\paren {E_k}^y}$

where:


 * $\paren {E_1}^y, \paren {E_2}^y, \ldots, \paren {E_n}^y$ are pairwise disjoint $\Sigma_X$-measurable sets
 * $a_1, a_2, \ldots, a_n$ are real numbers.

From the definition of the $\mu$-integral of a positive simple function, we have:


 * $\ds \map {I_\mu} {f^y} = \sum_{k \mathop = 1}^n a_k \map \mu {\paren {E_k}^y}$

From Integral of Positive Measurable Function Extends Integral of Positive Simple Function, we then have:


 * $\ds \int f^y \rd \mu = \sum_{k \mathop = 1}^n a_k \map \mu {\paren {E_k}^y}$

Similarly, from Vertical Section of Simple Function is Simple Function, we have:


 * $f_x$ is a positive simple function

with:


 * $\ds f_x = \sum_{k \mathop = 1}^n a_k \chi_{\paren {E_k}_x}$

where:


 * $\paren {E_1}_x, \paren {E_2}_x, \ldots, \paren {E_n}_x$ are pairwise disjoint $\Sigma_X$-measurable sets
 * $a_1, a_2, \ldots, a_n$ are non-negative real numbers.

From the definition of the $\nu$-integral of a positive simple function, we have:


 * $\ds \map {I_\nu} {f_x} = \sum_{k \mathop = 1}^n a_k \map \nu {\paren {E_k}_x}$

From Integral of Positive Measurable Function Extends Integral of Positive Simple Function, we then have:


 * $\ds \int f_x \rd \nu = \sum_{k \mathop = 1}^n a_k \map \nu {\paren {E_k}_x}$

We then have:

With a view to show:


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

write:

To show that:


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

write:

So:


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