Fubini'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 $f: X \times Y \to \overline \R$ be a $\mu \times \nu$-integrable function.

Define a function $I_f : X \to \R$ by:


 * $\ds \map {I_f} x = \begin{cases}\ds \int_Y f_x \rd \nu & \text {if } f_x \text { is } \nu\text{-integrable} \\ 0 & \text{otherwise}\end{cases}$

for each $x \in X$.

Define a function $J_f : Y \to \R$ by:


 * $\ds \map {J_f} y = \begin{cases}\ds \int_X f^y \rd \mu & \text {if } f^y \text { is } \mu\text{-integrable} \\ 0 & \text{otherwise}\end{cases}$

for each $y \in Y$.

Then:


 * $I_f$ is $\mu$-integrable and $J_f$ is $\nu$-integrable

and:


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

Proof
Since $f$ is $\mu \times \nu$-integrable, we have:


 * $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_{X \times Y} f^+ \map \rd {\mu \times \nu} - \int_{X \times Y} f^- \map \rd {\mu \times \nu}$

Lemma
We first aim to show that:


 * $\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu} - \int_{X \times Y} f^- \map \rd {\mu \times \nu} = \int_X I_f \rd \mu$

From Tonelli's Theorem, we have:


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

From Positive Part of Vertical Section of Function is Vertical Section of Positive Part, we have:


 * $\paren {f^+}_x = \paren {f_x}^+$

and so:


 * $\ds \int_X \paren {\int_Y \paren {f^+}_x \rd \nu} \rd \mu = \int_X \paren {\int_Y \paren {f_x}^+ \rd \nu} \rd \mu$

From Tonelli's Theorem, we also obtain:


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

From Negative Part of Vertical Section of Function is Vertical Section of Negative Part, we have:


 * $\paren {f^-}_x = \paren {f_x}^-$

and so:


 * $\ds \int_X \paren {\int_Y \paren {f^-}_x \rd \nu} \rd \mu = \int_X \paren {\int_Y \paren {f_x}^- \rd \nu} \rd \mu$

We therefore have:


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

From Almost All Vertical Sections of Integrable Function are Integrable, we have:


 * $f_x$ is $\nu$-integrable for $\mu$-almost all $x \in X$.

That is:


 * there exists a $\mu$-null set $N_1 \subseteq X$ such that whenever $f_x$ is not $\nu$-integrable, $x \in N_1$.

To rewrite this in terms of $I_f$, we pass the $\mu$-integral to an $\mu$-integral over $X \setminus N_1$.

Since $\sigma$-algebras are closed under relative complement, we have:


 * $X \setminus N_1$ is $\Sigma_X$-measurable.

From Characteristic Function Measurable iff Set Measurable, we have:


 * $1_{X \setminus N_1}$ is $\Sigma_X$-measurable.

so:


 * $\ds \paren {\int_Y \paren {f_x}^+ \rd \nu} \times 1_{X \setminus N_1}$ is $\Sigma_X$-measurable

and:


 * $\ds \paren {\int_Y \paren {f_x}^- \rd \nu} \times 1_{X \setminus N_1}$ is $\Sigma_X$-measurable.

From Characteristic Function of Null Set is A.E. Equal to Zero: Corollary, we have:


 * $1_{X \setminus N_1} = 1$ $\mu$-almost everywhere.

So, from Pointwise Multiplication preserves A.E. Equality, we have:


 * $\ds \paren {\int_Y \paren {f_x}^+ \rd \nu} \times 1_{X \setminus N_1} = \int_Y \paren {f_x}^+ \rd \nu$ $\mu$-almost everywhere

and:


 * $\ds \paren {\int_Y \paren {f_x}^- \rd \nu} \times 1_{X \setminus N_1} = \int_Y \paren {f_x}^- \rd \nu$ $\mu$-almost everywhere.

From A.E. Equal Positive Measurable Functions have Equal Integrals, we have:


 * $\ds \int_X \paren {\int_Y \paren {f_x}^+ \rd \nu} \rd \mu = \int_X 1_{X \setminus N_1} \paren {\int_Y \paren {f_x}^+ \rd \nu} \rd \mu$

and:


 * $\ds \int_X \paren {\int_Y \paren {f_x}^- \rd \nu} \rd \mu = \int_X 1_{X \setminus N_1} \paren {\int_Y \paren {f_x}^- \rd \nu} \rd \mu$

From the definition of the $\mu$-integral of an integrable function over a $\Sigma_X$-measurable set, we have:


 * $\ds \int_X 1_{X \setminus N_1} \paren {\int_Y \paren {f_x}^+ \rd \nu} \rd \mu = \int_{X \setminus N_1} \paren {\int_Y \paren {f_x}^+ \rd \nu} \rd \mu$

and:


 * $\ds \int_X 1_{X \setminus N_1} \paren {\int_Y \paren {f_x}^- \rd \nu} \rd \mu = \int_{X \setminus N_1} \paren {\int_Y \paren {f_x}^- \rd \nu} \rd \mu$

We then have:

From Pointwise Multiplication preserves A.E. Equality, we have:


 * $I_f \times \chi_{X \setminus N_1} = I_f$ $\mu$-almost everywhere

so, by A.E. Equal Positive Measurable Functions have Equal Integrals, we have:


 * $\ds \int \paren {I_f \times \chi_{X \setminus N_1} } \rd \mu = \int I_f \rd \mu$

giving:


 * $\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu} - \int_{X \times Y} f^- \map \rd {\mu \times \nu} = \int_X I_f \rd \mu$

as desired.

We now show that:


 * $\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu} - \int_{X \times Y} f^- \map \rd {\mu \times \nu} = \int_Y J_f \rd \nu$

From Tonelli's Theorem, we have:


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

From Positive Part of Horizontal Section of Function is Horizontal Section of Positive Part, we have:


 * $\paren {f^+}^y = \paren {f^y}^+$

and so:


 * $\ds \int_Y \paren {\int_X \paren {f^+}^y \rd \mu} \rd \nu= \int_Y \paren {\int_X \paren {f^y}^+ \rd \mu} \rd \nu$

From Tonelli's Theorem, we also obtain:


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

From Negative Part of Horizontal Section of Function is Horizontal Section of Negative Part, we have:


 * $\paren {f^-}^y = \paren {f^y}^-$

and so:


 * $\ds \int_Y \paren {\int_X \paren {f^-}^y \rd \mu} \rd \nu = \int_Y \paren {\int_X \paren {f^y}^- \rd \mu} \rd \nu$

We therefore have:


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

From Almost All Horizontal Sections of Integrable Function are Integrable, we have:


 * $f^y$ is $\mu$-integrable for $\nu$-almost all $y \in Y$.

That is:


 * there exists a $\nu$-null set $N_2 \subseteq Y$ such that whenever $f^y$ is not $\nu$-integrable, $y \in N_2$.

To rewrite this in terms of $J_f$, we pass the $\nu$-integral to an $\nu$-integral over $Y \setminus N_2$.

Since $\sigma$-algebras are closed under relative complement, we have:


 * $Y \setminus N_2$ is $\Sigma_Y$-measurable.

From Characteristic Function Measurable iff Set Measurable, we have:


 * $1_{Y \setminus N_2}$ is $\Sigma_Y$-measurable.

so:


 * $\ds \paren {\int_X \paren {f^y}^+ \rd \mu} \times 1_{Y \setminus N_2}$ is $\Sigma_Y$-measurable

and:


 * $\ds \paren {\int_X \paren {f^y}^- \rd \mu} \times 1_{Y \setminus N_2}$ is $\Sigma_Y$-measurable.

From Characteristic Function of Null Set is A.E. Equal to Zero: Corollary, we have:


 * $1_{Y \setminus N_2} = 1$ $\nu$-almost everywhere.

So, from Pointwise Multiplication preserves A.E. Equality, we have:


 * $\ds \paren {\int_X \paren {f^y}^+ \rd \mu} \times 1_{Y \setminus N_2} = \int_X \paren {f^y}^+ \rd \mu$ $\nu$-almost everywhere

and:


 * $\ds \paren {\int_X \paren {f^y}^- \rd \mu} \times 1_{Y \setminus N_2} = \int_Y \paren {f^y}^- \rd \mu$ $\nu$-almost everywhere.

From A.E. Equal Positive Measurable Functions have Equal Integrals, we have:


 * $\ds \int_Y \paren {\int_X \paren {f^y}^+ \rd \mu} \rd \nu = \int_Y 1_{Y \setminus N_2} \paren {\int_X \paren {f^y}^+ \rd \mu} \rd \nu$

and:


 * $\ds \int_Y \paren {\int_X \paren {f^y}^- \rd \mu} \rd \nu = \int_Y 1_{Y \setminus N_2} \paren {\int_X \paren {f^y}^- \rd \mu} \rd \nu$

From the definition of the $\nu$-integral of an integrable function over a $\Sigma_Y$-measurable set, we have:


 * $\ds \int_Y 1_{Y \setminus N_2} \paren {\int_X \paren {f^y}^+ \rd \mu} \rd \nu = \int_{Y \setminus N_2} \paren {\int_X \paren {f^y}^+ \rd \mu} \rd \nu$

and:


 * $\ds \int_Y 1_{Y \setminus N_2} \paren {\int_X \paren {f^y}^- \rd \mu} \rd \nu = \int_{Y \setminus N_2} \paren {\int_X \paren {f^y}^- \rd \mu} \rd \nu$

We then have:

From Pointwise Multiplication preserves A.E. Equality, we have:


 * $J_f \times \chi_{Y \setminus N_2} = J_f$ $\nu$-almost everywhere

so, by A.E. Equal Positive Measurable Functions have Equal Integrals, we have:


 * $\ds \int \paren {J_f \times \chi_{Y \setminus N_2} } \rd \mu = \int J_f \rd \nu$

giving:


 * $\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu} - \int_{X \times Y} f^- \map \rd {\mu \times \nu} = \int_Y J_f \rd \nu$

We therefore obtain:


 * $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_{X \times Y} f^+ \map \rd {\mu \times \nu} - \int_{X \times Y} f^- \map \rd {\mu \times \nu}$

and hence:


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

Since:


 * $\ds \int_{X \times Y} f \map \rd {\mu \times \nu} < \infty$

we also have:


 * $I_f$ is $\mu$-integrable and $J_f$ is $\nu$-integrable

completing the proof.

Also see

 * Tonelli's Theorem