Tonelli's Theorem
Theorem
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 $f: X \times Y \to \overline \R_{\ge 0}$ be a positive $\Sigma_X \otimes \Sigma_Y$-measurable function.
Then:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \int_X \map f {x, y} \map {\d \mu} x \map {\d \nu} y = \int_X \int_Y \map f {x, y} \map {\d \nu} y \map {\d \mu} x$
Corollary
Let $\sequence {a_{n, m} }_{\tuple {n, m} \in \N^2}$ be a doubly subscripted sequence of non-negative real numbers.
Then:
- $\ds \sum_{n \mathop = 1}^\infty \paren {\sum_{m \mathop = 1}^\infty a_{n, m} } = \sum_{m \mathop = 1}^\infty \paren {\sum_{n \mathop = 1}^\infty a_{n, m} }$
Proof
We rewrite the demand as:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
where $f_x$ is the $x$-vertical section of $f$, and $f^y$ is the $y$-horizontal section of $f$.
From Horizontal Section of Measurable Function is Measurable, we have:
- $f^y$ is $\Sigma_X$-measurable.
From Vertical Section of Measurable Function is Measurable, we have:
- $f_x$ is $\Sigma_Y$-measurable.
From Integral of Horizontal Section of Measurable Function gives Measurable Function, we then have:
- $\ds y \mapsto \int_X f^y \rd \mu$ is $\Sigma_Y$-measurable
and from Integral of Vertical Section of Measurable Function gives Measurable Function, we have:
- $\ds x \mapsto \int_Y f_x \rd \nu$ is $\Sigma_X$-measurable.
So, both:
- $\ds \int_Y \paren {\int_X f^y \rd \mu} \rd \nu$
and:
- $\ds \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
are well-defined.
We first prove the case of:
- $f = \chi_E$
where $E$ is a $\Sigma_X \otimes \Sigma_Y$-measurable set.
Lemma 1
Let $E \in \Sigma_X \otimes \Sigma_Y$.
Let:
- $f = \chi_E$
where $\chi_E$ is the characteristic function of $E$.
Then:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
$\Box$
Now consider the case of a positive simple $f$.
Lemma 2
Let $f : X \times Y \to \overline \R$ be a positive simple function.
Then:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
$\Box$
Now take a general positive $\Sigma_X \otimes \Sigma_Y$-measurable function $f$.
We first show:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu$
From Measurable Function is Pointwise Limit of Simple Functions:
- there exists a increasing sequence of positive simple functions $\sequence {f_n}_{n \mathop \in \N}$ such that $f_n \to f$ pointwise.
From Simple Function is Measurable, we have:
- for each $n \in \N$, the function $f_n$ is $\Sigma_X \otimes \Sigma_Y$-measurable.
From Horizontal Section of Measurable Function is Measurable, we have:
- for each $n \in \N$, the function $\paren {f_n}^y$ is $\Sigma_X$-measurable, for each $y \in Y$.
From Horizontal Section preserves Increasing Sequences of Functions, we have:
- the sequence $\sequence {\paren {f_n}^y}_{n \mathop \in \N}$ is increasing for each $y \in Y$.
Now, from Horizontal Section preserves Pointwise Limits of Sequences of Functions, we have:
- $\paren {f_n}^y \to f^y$ pointwise for each $y \in Y$.
For each $n \in \N$ define a function $g_n : X \to \overline \R$ by:
- $\ds \map {g_n} y = \int_X \paren {f_n}^y \rd \mu$
for each $y \in Y$.
Since $\sequence {\paren {f_n}^y}_{n \mathop \in \N}$ is an increasing sequence of $\Sigma_X$-measurable functions, we can apply the Monotone Convergence Theorem to obtain:
- $\ds \lim_{n \mathop \to \infty} \map {g_n} y = \int_X f^y \rd \nu$
From Integral of Positive Measurable Function is Monotone, we have:
- $\sequence {g_n}_{n \mathop \in \N}$ is increasing.
From Integral of Horizontal Section of Measurable Function gives Measurable Function, we have:
- $\sequence {g_n}_{n \mathop \in \N}$ is $\Sigma_Y$-measurable.
For each $n \in \N$, we have:
\(\ds \int_{X \times Y} f_n \map \rd {\mu \times \nu}\) | \(=\) | \(\ds \int_Y \paren {\int_X \paren {f_n}^y \rd \mu} \rd \nu\) | from our work with simple functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_Y g_n \rd \nu\) |
Since $\sequence {f_n}_{n \mathop \in \N}$ is an increasing sequence of $\Sigma_X \otimes \Sigma_Y$-measurable functions converging pointwise to $f$, we have:
- $\ds \lim_{n \mathop \to \infty} \int_{X \times Y} f_n \map \rd {\mu \times \nu} = \int_{X \times Y} f \map \rd {\mu \times \nu}$
from the Monotone Convergence Theorem.
Since $\sequence {g_n}_{n \mathop \in \N}$ is an increasing sequence of $\Sigma_X \otimes \Sigma_Y$-measurable functions with:
- $\ds g_n \to \int_X f^y \rd \mu$ pointwise.
So:
- $\ds \lim_{n \mathop \to \infty} \int_Y g_n \rd \nu = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu$
by the Monotone Convergence Theorem.
So we obtain:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu$
We now show:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
From Measurable Function is Pointwise Limit of Simple Functions:
- there exists a increasing sequence of positive simple functions $\sequence {f_n}_{n \mathop \in \N}$ such that $f_n \to f$ pointwise.
From Simple Function is Measurable, we have:
- for each $n \in \N$, the function $f_n$ is $\Sigma_X \otimes \Sigma_Y$-measurable.
From Vertical Section of Measurable Function is Measurable, we have:
- for each $n \in \N$, the function $\paren {f_n}_x$ is $\Sigma_Y$-measurable, for each $x \in X$.
From Vertical Section preserves Increasing Sequences of Functions, we have:
- the sequence $\sequence {\paren {f_n}_x}_{n \mathop \in \N}$ is increasing for each $x \in X$.
Now, from Vertical Section preserves Pointwise Limits of Sequences of Functions, we have:
- $\paren {f_n}_x \to f_x$ pointwise for each $x \in X$.
For each $n \in \N$ define a function $h_n : X \to \overline \R$ by:
- $\ds \map {h_n} x = \int_Y \paren {f_n}_x \rd \nu$
for each $x \in X$.
Since $\sequence {\paren {f_n}_x}_{n \mathop \in \N}$ is an increasing sequence of $\Sigma_Y$-measurable functions, we can apply the Monotone Convergence Theorem to obtain:
- $\ds \lim_{n \mathop \to \infty} \map {h_n} x = \int_Y f_x \rd \nu$
From Integral of Positive Measurable Function is Monotone, we have:
- $\sequence {h_n}_{n \mathop \in \N}$ is increasing.
From Integral of Vertical Section of Measurable Function gives Measurable Function, we have:
- $\sequence {h_n}_{n \mathop \in \N}$ is $\Sigma_X$-measurable.
For each $n \in \N$, we have:
\(\ds \int_{X \times Y} f_n \map \rd {\mu \times \nu}\) | \(=\) | \(\ds \int_X \paren {\int_Y \paren {f_n}_x \rd \nu} \rd \mu\) | from our work with simple functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_X h_n \rd \mu\) |
Since $\sequence {f_n}_{n \mathop \in \N}$ is an increasing sequence of $\Sigma_X \otimes \Sigma_Y$-measurable functions converging pointwise to $f$, we have:
- $\ds \lim_{n \mathop \to \infty} \int_{X \times Y} f_n \map \rd {\mu \times \nu} = \int_{X \times Y} f \map \rd {\mu \times \nu}$
from the Monotone Convergence Theorem.
Since $\sequence {h_n}_{n \mathop \in \N}$ is an increasing sequence of $\Sigma_X \otimes \Sigma_Y$-measurable functions with:
- $\ds h_n \to \int_X f^y \rd \mu$ pointwise.
So:
- $\ds \lim_{n \mathop \to \infty} \int_Y h_n \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
by the Monotone Convergence Theorem.
So we obtain:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
giving:
- $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \int_X \map f {x, y} \map {\d \mu} x \map {\d \nu} y = \int_X \int_Y \map f {x, y} \map {\d \nu} y \map {\d \mu} x$
as was the demand.
$\blacksquare$
Also see
- Fubini's Theorem, a similar result pertaining to $\mu$-integrable functions.
Source of Name
This entry was named for Leonida Tonelli.
Sources
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Tonelli's theorem
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $13.8$
- 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $5.2$: Fubini's Theorem: Proposition $5.2.1 \text{ (b)}$