Fubini's Theorem/Lemma

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

Let $\struct {X \times Y, \Sigma_1 \otimes \Sigma_2, \mu \times \nu}$ 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 $\Sigma_X$-measurable

and:


 * $J_f$ is $\Sigma_Y$-measurable.

Proof
We first show that $I_f$ is $\Sigma_X$-measurable.

We will first rewrite $I_f$ in a more amenable form.

Note that $f_x$ is $\nu$-integrable :


 * $\ds \int_Y \paren {f_x}^+ \rd \nu < \infty$

or:


 * $\ds \int_Y \paren {f_x}^- \rd \nu < \infty$

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


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

and from Negative Part of Vertical Section of Function is Vertical Section of Negative Part, we have:


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

So, $f_x$ is $\nu$-integrable :


 * $\ds \int_Y \paren {f^+}_x \rd \nu < \infty$

and:


 * $\ds \int_Y \paren {f^-}_x \rd \nu < \infty$

From Integral of Vertical Section of Measurable Function gives Measurable Function, we have:


 * $\ds x \mapsto \int_Y \paren {f^+}_x \rd \nu$ is $\Sigma_X$-measurable

and:


 * $\ds x \mapsto \int_Y \paren {f^-}_x \rd \nu$ is $\Sigma_X$-measurable.

From Set of Points for which Measurable Function is Real-Valued is Measurable, we therefore have:


 * $\ds \set {x \in X : \int_Y \paren {f^+}_x \rd \nu < \infty} \in \Sigma_X$

and:


 * $\ds \set {x \in X : \int_Y \paren {f^-}_x \rd \nu < \infty} \in \Sigma_X$

Set:


 * $\ds A = \set {x \in X : \int_Y \paren {f^+}_x \rd \nu < \infty}$

and:


 * $\ds B = \set {x \in X : \int_Y \paren {f^-}_x \rd \nu < \infty}$

Then:


 * $A \cap B \in \Sigma_X$

from Sigma-Algebra Closed under Countable Intersection.

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


 * $X \setminus \paren {A \cap B} \in \Sigma_X$

With that, we can write:

For brevity, write:


 * $C = A \cap B$

and:


 * $D = X \setminus \paren {A \cap B}$

Write $\Sigma_{X, C}$ for the trace $\sigma$-algebra of $C$ in $\Sigma_X$.

Likewise write $\Sigma_{X, D}$ for the trace $\sigma$-algebra of $D$ in $\Sigma_X$.

From Restriction of Measurable Function is Measurable on Trace Sigma-Algebra, we have:


 * $\ds x \mapsto \int_Y \paren {f^+}_x \rd \nu$ is $\Sigma_{X, C}$-measurable

and:


 * $\ds x \mapsto \int_Y \paren {f^-}_x \rd \nu$ is $\Sigma_{X, C}$-measurable.

From Pointwise Difference of Measurable Functions is Measurable, we have:


 * $\ds \int_Y \paren {f^+}_x \rd \nu - \int_Y \paren {f^-}_x \rd \nu$ is $\Sigma_{X, C}$-measurable.

From Constant Function is Measurable, we have:


 * $x \mapsto 0$ is $\Sigma_{X, D}$-measurable.

From Piecewise Combination of Measurable Mappings is Measurable: General Case, we have:


 * $I_f$ is $\Sigma_X$-measurable.

We will now show that $J_f$ is $\Sigma_Y$-measurable.

We will write $J_f$ in a similar form to $I_f$.

Note that $f^y$ is $\mu$-integrable :


 * $\ds \int_X \paren {f^y}^+ \rd \mu < \infty$

or:


 * $\ds \int_X \paren {f^y}^- \rd \mu < \infty$

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


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

and from Negative Part of Horizontal Section of Function is Horizontal Section of Negative Part, we have:


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

So, $f^y$ is $\mu$-integrable :


 * $\ds \int_X \paren {f^+}^y \rd \mu < \infty$

and:


 * $\ds \int_X \paren {f^-}^y \rd \mu < \infty$

From Integral of Horizontal Section of Measurable Function gives Measurable Function, we have:


 * $\ds y \mapsto \int_X \paren {f^+}^y \rd \mu$ is $\Sigma_Y$-measurable

and:


 * $\ds y \mapsto \int_X \paren {f^-}^y \rd \mu$ is $\Sigma_Y$-measurable.

From Set of Points for which Measurable Function is Real-Valued is Measurable, we therefore have:


 * $\ds \set {y \in Y : \int_X \paren {f^+}^y \rd \mu < \infty} \in \Sigma_Y$

and:


 * $\ds \set {y \in Y : \int_X \paren {f^-}^y \rd \mu < \infty} \in \Sigma_Y$

Set:


 * $\ds A = \set {y \in Y : \int_X \paren {f^+}^y \rd \mu < \infty}$

and:


 * $\ds B = \set {y \in Y : \int_X \paren {f^-}^y \rd \mu < \infty}$

Then:


 * $A \cap B \in \Sigma_Y$

from Sigma-Algebra Closed under Countable Intersection.

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


 * $Y \setminus \paren {A \cap B} \in \Sigma_Y$

With that, we can write:

For brevity, write:


 * $C = A \cap B$

and:


 * $D = Y \setminus \paren {A \cap B}$

Write $\Sigma_{Y, C}$ for the trace $\sigma$-algebra of $C$ in $\Sigma_Y$.

Likewise write $\Sigma_{Y, D}$ for the trace $\sigma$-algebra of $D$ in $\Sigma_Y$.

From Restriction of Measurable Function is Measurable on Trace Sigma-Algebra, we have:


 * $\ds y \mapsto \int_X \paren {f^+}^y \rd \mu$ is $\Sigma_{Y, C}$-measurable

and:


 * $\ds y \mapsto \int_X \paren {f^-}^y \rd \mu$ is $\Sigma_{Y, C}$-measurable.

From Pointwise Difference of Measurable Functions is Measurable, we have:


 * $\ds \int_X \paren {f^+}^y \rd \mu - \int_X \paren {f^-}^y \rd \nu$ is $\Sigma_{Y, C}$-measurable.

From Constant Function is Measurable, we have:


 * $y \mapsto 0$ is $\Sigma_{Y, D}$-measurable.

From Piecewise Combination of Measurable Mappings is Measurable: General Case, we have:


 * $J_f$ is $\Sigma_Y$-measurable.