[Measure Theory] Product Measure

In this post, we introduce a new type of measure called product measure, which allows us to deal with mutiple integrals of multiple indepedent variables. And the main goal is to prove the Fubini-Tonelli theorem, one of the most important and practical results in analysis.

Product $\sigma$-algebras

We start with constructing a $\sigma$-algebra for product measure.

$\color{blue}{\mathbf{Definition.}}$ Measurable rectangle
Let $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$ be two measure spaces. A measurable rectangle is a set of the form $A \times B$, where $A \in \mathcal{A}$ and $B \in \mathcal{B}$. If $E \subset X \times Y$, we define the $x$-section of $E$ by $$ \begin{aligned} E_x \equiv s_x (E) := \{y \in Y \; : \; (x, y) \in E \} \end{aligned} $$ and similarly define the $y$-section by $$ \begin{aligned} E^y \equiv t_y (E) := \{x \in X \; : \; (x, y) \in E \} \end{aligned} $$ Also, if $f$ is a function on $X \times Y$, we define the $x$-section $f_x$ of $f$ and the $y$-section $f^y$ of $f$ by $$ \begin{aligned} f_x (y) = f^y (x) = f(x, y). \end{aligned} $$

Clearly,

\[\begin{aligned} (A \times B) \cap (E \times F) = (A \cap E) \times (B \cap F), \end{aligned}\]

and

\[\begin{aligned} (A \times B)^c & = (A \times B^c) \cup (A^c \times B) \cup (A^c \times B^c) \\ & = (A \times B^c) \cup (A^c \times Y) \\ & = (X \times B^c) \cup (A^c \times B). \end{aligned}\]

And obviously, the sections of measurable sets and measurable functions again form measurable sets and measurable functions, respectively. We will prove this fact in the later lemma.

$\color{blue}{\mathbf{Definition.}}$ Product $\sigma$-algebra
Let $\mathcal{C}_0$ be the collection of finite unions of disjoint measurable rectangles: $$ \begin{aligned} \mathcal{C}_0 = \left\{ \bigcup_{i=1}^n (A_i \times B_i) \; \middle| \; A_i \in \mathcal{A}, B_i \in \mathcal{B}, n \in \mathbb{N} \text{ and } (A_i \times B_i) \cap (A_j \times B_j) = \emptyset \text{ for } i \neq j \right\} \end{aligned} $$ It is easy to check that $\mathcal{C}_0$ is an algebra of sets. Then we define the product $\sigma$-algebra $$ \begin{aligned} \mathcal{A} \times \mathcal{B} = \sigma(\mathcal{C}_0), \end{aligned} $$ the $\sigma$-algebra generated by $\mathcal{C}_0$.

$\color{#bf1100}{\mathbf{Proposition.}}$ Let $\mathcal{B}_2$ be the Borel $\sigma$-algebra on $\mathbb{R}^2$, that is, the smallest $\sigma$-algebra that contains all the open subsets of $\mathbb{R}^2$, and let $\mathcal{B}_1$ be the usual Borel $\sigma$-algebra on $\mathbb{R}$. Let $\mathcal{L}_2$ be the Lebesgue $\sigma$-algebra on $\mathbb{R}^2$, that is, the completion $\overline{\mathcal{B}_2}$ of $\mathcal{B}_2$.

1. $\mathcal{B}_2 = \mathcal{B}_1 \times \mathcal{B}_1$
2. $\mathcal{L}_2 = \overline{\mathcal{B}_1 \times \mathcal{B}_1}$

$\color{#bf1100}{\mathbf{Proof.}}$

Let

\[\begin{aligned} \mathcal{C} = \{ E \in \mathcal{A} \times \mathcal{B} \mid \lambda (E) = \mu \times \nu (E) \} \end{aligned}\]

It is obvious that $\mathcal{C}_0 \subset \mathcal{C}$. And we claim that $\mathcal{C}$ is a monotone class containing $\mathcal{C}_0$ if $\mu$ and $\nu$ are $\sigma$-finite.

(1) First we show $\mathcal{B}_2 \subseteq \mathcal{B}_1 \times \mathcal{B}_1$.

Note that

\[\begin{aligned} \{ (r_1 - \varepsilon_1, r_1 + \varepsilon_1) \times (r_2 - \varepsilon_2, r_2 + \varepsilon_2) : r_i. \varepsilon_i \in \mathbb{Q} \} \end{aligned}\]

is an open basis of $\mathbb{R}^2$. Since its elements are measurable rectangles and each open set can be written as union of elements of basis, we have $G_2 \subseteq \mathcal{B}_1 \times \mathcal{B}_1$ where $G_2$ is the collection of open sets in $\mathbb{R}^2$. Since $\mathcal{B}_1 \times \mathcal{B}_1$, we obtain $\mathcal{B}_2 = \sigma(G_2) \subseteq \mathcal{B}_1 \times \mathcal{B}_1$.

Finally, we show $\mathcal{B}_1 \times \mathcal{B}_1 \subseteq \mathcal{B}_2$. Note that since $\mathcal{B}_2$ is $\sigma$-algebra, it sufficies to show that any measurable rectangles are in $\mathcal{B}_2$. Fix any $B \in \mathcal{B}_1$. Define

\[\begin{aligned} \mathcal{C}(B) = \{ A \in \mathcal{B}_1 : A \times B \in \mathcal{B}_2 \}. \end{aligned}\]

We claim that $\mathcal{C}(B)$ is $\sigma$-algebra containing all open sets (thus $\mathcal{B}_1 \subseteq \mathcal{C}(B)$.)

1. product of open sets is open, hence $\emptyset \times B \in \mathcal{B}_2$.
2. Let $A \in \mathcal{C}(B)$. Then $A^c \in \mathcal{C}(B)$ as $A^c \times B = (\mathbb{R} \times B) \cap (A \times B)^c$.
3. Let $A_1, A_2, \dots \in \mathcal{C}(B)$. Then $\bigcap_{n=1}^\infty A_n \in \mathcal{C}(B)$ as $(\bigcap_{n=1}^\infty A_n) \times B = \bigcap_{n=1}^\infty (A_n \times B)$.

Define

\[\begin{aligned} \mathcal{C} = \{ B \in \mathcal{B}_1 : \mathcal{B}_1 \subseteq \sigma (B)\}. \end{aligned}\]

We claim that $\mathcal{C}$ is $\sigma$-algebra containing all open sets (thus $\mathcal{B}_1 \subseteq \mathcal{C}$.)

1. product of open sets is open, hence $\emptyset \in \mathcal{C}$.
2. Let $B \in \mathcal{C}$. Then $B^c \in \mathcal{C}$ as for any $A \in \mathcal{B}_1$, $A \times B^c = (A \times \mathbb{R}) \cap (A \times B)^c$.
3. Let $B_1, B_2, \dots \in \mathcal{C}$. Then for any $A \in \mathcal{B}_1$, $\bigcap_{n=1}^\infty B_n \in \mathcal{C}$ as $A \times (\bigcap_{n=1}^\infty B_n) = \bigcap_{n=1}^\infty (A \times B_n)$.

Thus, for any $A, B \in \mathcal{B}_1$, we showed that $A \times B \in \mathcal{B}_2$. Note that $\mathcal{B}_2$ is $\sigma$-algebra. Thus we have

\[\begin{aligned} \mathcal{B}_1 \times \mathcal{B}_1 = \sigma (\{ A \times B\}) \subseteq \mathcal{B}_2. \end{aligned}\]

(2) From (1), it is clear.

\[\tag*{$\blacksquare$}\]

Note that $\mathcal{L}_2 \neq \mathcal{L}_1 \times \mathcal{L}_1$. It can be simply verified as follows. Let $V$ be non-measurable set, and let \(N = V \times \{ \frac{1}{2}\}\). Since $N$ is $m_2$ null set and must be in $\mathcal{L}_2$, but not in $\mathcal{L}_1 \times \mathcal{L}_1$ as $V$ is non-measurable.

Product Measure

The next step is to obtain a measure on the product $\sigma$-algebra. Suppose $A \times B$ is a measurable rectangle that is a countable disjoint union of rectangles $A_j \times B_j$. Then,

\[\begin{aligned} \chi_A(x) \chi_B(y) = \chi_{A \times B} (x, y) = \sum_{j=1}^\infty \chi_{A_j \times B_j} (x, y) = \sum_{j=1}^\infty \chi_{A_j}(x) \chi_{B_j}(y) \end{aligned}\]

Integrate both terms with regard to $x$ and again with regard to $y$. Then we obtain

\[\begin{aligned} \mu (A) \nu (B) = \sum_{j=1}^\infty \mu(A_j) \nu(B_j) \end{aligned}\]

Hence, if $E \in \mathcal{C}_0$, i.e. the disjoint union of rectangles $A_1 \times B_1$, $\cdots$, $A_n \times B_n$ and we set

\[\begin{aligned} \mu \times \nu (E) = \sum_{j=1}^n \mu(A_j) \nu(B_j) \end{aligned}\]

then $\mu \times \nu$ is well-defined on $\mathcal{C}_0$ therefore a premeasure on algebra $\mathcal{C}_0$. And later, if $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$ are $\sigma$-finite, the extension of this premeasure to $\mathcal{A} \times \mathcal{B}$ will turn out to be a well-defined measure called product measure.

The following lemmas are useful facts to prove further step:

$\color{green}{\mathbf{Lemma.}}$

1. If $E \in \mathcal{A} \times \mathcal{B}$, then $E_x \in \mathcal{B}$ for each $x$ and $E^y \in \mathcal{A}$ for each $y$.
2. If $f$ is $\mathcal{A} \times \mathcal{B}$ measurable, then $f_x$ is $\mathcal{B}$-measurable and $f^y$ is $\mathcal{A}$-measurable for any $x \in X$, $y \in Y$.

$\color{green}{\mathbf{Proof.}}$

(1) Let

\[\begin{aligned} \mathcal{C} = \{ E \in \mathcal{A} \times \mathcal{B} \; | \; E_x \in \mathcal{A} \text{ for each } x \text{ and } E^y \in \mathcal{A} \text{ for each } y \} \subseteq \mathcal{A} \times \mathcal{B}. \end{aligned}\]

We will show that $\mathcal{C}$ is a $\sigma$-algebra containing the measurable rectangles, and hence is all of $\mathcal{A} \times \mathcal{B}$.

1. $\mathcal{C}$ contains all measurable rectangles
   

   If $E = A \times B$, $E_x = B$ of $x \in A$ and equal to $\emptyset$ if $x \notin A$. Thus $E^x \in \mathcal{B}$ for each $x$ when $E$ is a measurable rectangle.

2. $\mathcal{C}$ is $\sigma$-algebra
   

   Suppose $E \in \mathcal{C}$ and $x \in X$. It is easy to show that $$ \begin{aligned} (E^c)_x = (E_x)^c. \end{aligned} $$ Therefore $(E^c)_x$ is $B$ measurable, and $\mathcal{C}$ is closed under the operation of taking complements.
   Similarly, it is easy to see that $$ \begin{aligned} s_x \left( \bigcup_{i=1}^\infty E_i \right) = \bigcup_{i=1}^\infty s_x (E_i), \end{aligned} $$ therefore so $\mathcal{C}$ is closed under the operation of countable unions.

Hence, it implies that $\mathcal{A} \times \mathcal{B} \subset \mathcal{C}$:

\[\begin{aligned} E \in \mathcal{A} \times \mathcal{B} \Rightarrow E_x \in \mathcal{B} \text{ and } E^y \in \mathcal{A} \text{ for any } x \in X, y \in Y. \end{aligned}\]

(2) Let $B$ be any Borel set. Then $f^{-1} (B) \in \mathcal{A} \times \mathcal{B}$.

\[\begin{aligned} (f_x)^{-1} (B) & = \{ y \in Y \; | \; f_x (y) \in B \} \\ & = \{ y \in Y \; | \; f (x, y) \in B \} \\ & = \{ y \in Y \; | \; (x, y) \in f^{-1}(B) \} \\ & = (f^{-1}(B))_x \in \mathcal{B} \end{aligned}\]

Similarly, we can show that $(f^{-1}(B))^y \in \mathcal{A}$.

\[\tag*{$\blacksquare$}\]

Finally, we now come to the main results of this section. The following theorem relates integrals on $X \times Y$ to integrals on $X$ and $Y$:

$\color{red}{\mathbf{Theorem.}}$ Suppose $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$ are $\sigma$-finite measure spaces. If $E \in \mathcal{A} \times \mathcal{B}$, then

1. the functions $$ \begin{aligned} x \mapsto \nu (E_x) \text{ and } y \mapsto \mu (E^y) \end{aligned} $$ are measurable on $X$ and $Y$, respectively.
2. $$ \begin{aligned} \int \nu (E_x) \; d\mu(x) = \int \mu (E^y) \; d\nu(y). \end{aligned} $$

$\color{red}{\mathbf{Proof.}}$

(1) First, suppose $\mu$ and $\nu$ are finite.

Let

\[\begin{aligned} \mathcal{C} := \{ E \in \mathcal{A} \times \mathcal{B} \; | \; \text{ 1 and 2 hold. }\} \end{aligned}\]

We claim that $\mathcal{C}$ is a monotone class containing $\mathcal{C}_0$.

1. $\mathcal{C}$ contains $\mathcal{C}_0$
   

   If $E = A \times B$ for $A \in \mathcal{A}$ and $B \in \mathcal{B}$, then $$ \begin{aligned} \nu (E_x) & = \chi_A (x) \nu (B) \\ \mu (E^y) & = \chi_B (y) \mu (A) \end{aligned} $$ Since $A$ and $B$ are measurable in $\mathcal{A}$ and $\mathcal{B}$ respectively, $E \in \mathcal{C}$.

2. Closed under countable increasing union
   

   Let $\{ E_n \}$ be an increasing sequence in $\mathcal{C}$ and $E = \bigcup_{n=1}^\infty E_n$. Then for each $n \in \mathbb{N}$ $$ \begin{aligned} h_n (x) & := \nu ( (E_n)_x ) \\ k_n (y) & := \mu ( (E_n)^y ) \end{aligned} $$ are measurable, and increase pointwisely to $$ \begin{aligned} h (x) & := \nu ( E_x ) \\ k (y) & := \mu ( E^y ) \end{aligned} $$ As $E_n \in \mathcal{C}$ and monotone convergence theorem, we have $$ \begin{aligned} \int \mu (E^y) \; d \nu(y) & = \lim_{n \to \infty} \int \mu ((E_n)^y) \; d\nu(y) \\ & = \lim_{n \to \infty} \int \nu ((E_n)_x) \; d\mu(x) \\ & = \int \nu (E_x) \; d \mu(x) \end{aligned} $$ Hence $E \in \mathcal{C}$.

3. Closed under countable decreasing intersection
   

   Let $\{ E_n \}$ be a decreasing sequence in $\mathcal{C}$ and $E = \bigcap_{n=1}^\infty E_n$. Then for each $n \in \mathbb{N}$, $h_n$ and $k_n$ are measurable, and decrease pointwisely to $h$ and $k$. Since the functions $h_1 (x) = \nu ((E_1)_x)$ and $k_1 (y) = \mu ((E_1)^y)$ are integrable as $\mu$ and $\nu$ are finite, by the dominated convergence theorem, we have $$ \begin{aligned} \int \mu (E^y) \; d \nu(y) & = \lim_{n \to \infty} \int \mu ((E_n)^y) \; d\nu(y) \\ & = \lim_{n \to \infty} \int \nu ((E_n)_x) \; d\mu(x) \\ & = \int \nu (E_x) \; d \mu(x) \end{aligned} $$ Hence $E \in \mathcal{C}$.

As $\mathcal{A} \times \mathcal{B}$ is the smallest monotone class containing $\mathcal{C}_0$, $\mathcal{A} \times \mathcal{B} \subset \mathcal{C}$. Hence 1 and 2 hold for all $E \in \mathcal{A} \times \mathcal{B}$.

(2) Now let’s consider when $\mu$ and $\nu$ are $\sigma$-finite. We can write $X$ and $Y$ as the union of an increasing ${ F_i \times G_i }$ of measurable rectangles of finite measure.

Let $E \in \mathcal{A} \times \mathcal{B}$. We apply the preceding argument to $\mu \mid_{F_i}$ and $\nu \mid_{G_i}$ and $E \cap (F_i \times G_i)$. We have

\[\begin{aligned} (E \cap (F_i \times G_i))_x & = \{ y \in Y \; | \; (x, y) \in E \cap (F_i \times G_i) \} \\ & = E_x \cap \{ y \in Y \; | \; (x, y) \in F_i \times G_i \} \\ & = \begin{cases} E_x \cap G_i & \quad \text{ if } x \in F_i \\ \emptyset & \quad \text{ if } x \notin F_i \\ \end{cases} \end{aligned}\]

therefore

\[\begin{aligned} \int \nu ( (E \cap (F_i \times G_i))_x ) \; d\mu(x) = \int \chi_{F_i} (x) \nu (E_x \cap G_i) \; d\mu(x) \end{aligned}\]

and

\[\begin{aligned} \int \mu ( (E \cap (F_i \times G_i))^y ) \; d\nu(y) = \int \chi_{G_i} (y) \mu (E^y \cap G_i) \; d\nu(y). \end{aligned}\]

And note that

\[\begin{aligned} \int \nu ( (E \cap (F_i \times G_i))_x ) \; d\mu(x) = \int \mu ( (E \cap (F_i \times G_i))^y ) \; d\nu(y) \end{aligned}\]

By monotone convergence theorem, we have

\[\begin{aligned} \int \nu (E_x) \; d \mu(x) = \int \mu (E^y) \; d \nu(y) \end{aligned}\]

and

\[\begin{aligned} \nu ((E \cap (F_i \times G_i))_x ) & \uparrow \nu (E_x) \\ \mu ((E \cap (F_i \times G_i))^y ) & \uparrow \mu (E^y) \end{aligned}\]

hence $\nu (E_x)$ and $\mu (E^y)$ measurable.

\[\tag*{$\blacksquare$}\]

Based on the theorem, we now define a product measure $\mu \times \nu$ as follows:

$\color{blue}{\mathbf{Definition.}}$ Product measure
For measure spaces $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$, we define product measure $$ \begin{aligned} (\mu \times \nu) (E) & := \int \nu(E_x) \; d\mu(x) \\ & = \int \mu(E^y) \; d\nu(y) \end{aligned} $$ for $E \in \mathcal{A} \times \mathcal{B}$. It is easy to check $\mu \times \nu$ is a measure on $\mathcal{A} \times \mathcal{B}$.

$\color{blue}{\mathbf{Proof.}}$

1. Clearly we have $\mu \times \nu (\emptyset) = 0$.

2. If $E_1, \dots, E_n$ are disjoint, in $\mathcal{A} \times \mathcal{B}$, and $E = \bigcup_{j=1}^n E_j$ then $$ \begin{aligned} \nu (E_x) = \nu (\bigcup_{j=1}^n (E_j)_x ) = \sum_{j=1}^n \nu ((E_j)_x) \end{aligned} $$ Thus $$ \begin{aligned} \mu \times \nu (E) & = \int \nu (E_x) \; d\mu(x) = \sum_{j=1}^n \int \nu ((E_j)_x) \; d\mu(x) = \sum_{j=1}^n \mu \times \nu (E_j) \end{aligned} $$ Since $\bigcup_{j=1}^n E_j \uparrow \bigcup_{j=1}^\infty E_j$, done by the monotone convergence theorem.

\[\tag*{$\blacksquare$}\]

Note that if $E = A \times B$ is a measurable rectangle, then

\[\begin{aligned} \mu \times \nu (A \times B) = \mu (A) \nu (B), \end{aligned}\]

which is what it should be. Furthermore, a measure on $\mathcal{A} \times \mathcal{B}$ that satisfy this property is unique:

$\color{#bf1100}{\mathbf{Remark.}}$ Uniqueness of product measure
Let $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$ be measure spaces. Suppose $\mu$ and $\nu$ are $\sigma$-finite. Prove that if $\lambda$ is a measure on $\mathcal{A} \times \mathcal{B}$ such that $$ \begin{aligned} \lambda (A \times B) = \mu(A)\nu(B) \end{aligned} $$ whenever $A \in \mathcal{A}$ and $B \in \mathcal{B}$, then $\lambda = \mu \times \nu$ on $\mathcal{A} \times \mathcal{B}$.

$\color{#bf1100}{\mathbf{Proof.}}$

Let

\[\begin{aligned} \mathcal{C} = \{ E \in \mathcal{A} \times \mathcal{B} \mid \lambda (E) = \mu \times \nu (E) \} \end{aligned}\]

It is obvious that $\mathcal{C}_0 \subset \mathcal{C}$. And we claim that $\mathcal{C}$ is a monotone class containing $\mathcal{C}_0$ if $\mu$ and $\nu$ are $\sigma$-finite.

(1) $\mu$ and $\nu$ are finite.

Let $E_n \in \mathcal{C}$ and $E_n \uparrow E$.

\[\begin{aligned} \lambda (E) = \lim_{n \to \infty} \lambda (E_n) = \lim_{n \to \infty} \mu \times \nu (E_n) = \mu \times \nu (E) \end{aligned}\]

by continuity from below of measures. On the other hand, let $E_n \in \mathcal{C}$ and $E_n \downarrow E$.

\[\begin{aligned} \lambda (E) = \lim_{n \to \infty} \lambda (E_n) = \lim_{n \to \infty} \mu \times \nu (E_n) = \mu \times \nu (E) \end{aligned}\]

by continuity from above of measures and finiteness of $\mu$ and $\nu$. Hence $\mathcal{C}$ is a monotone class when $\mu$ and $\nu$ are finite.

(2) $\mu$ and $\nu$ are $\sigma$-finite

Let $X_n \uparrow X$ and $Y_n \uparrow Y$ with $\mu (X_n) < \infty$ and $\nu (Y_n) < \infty$ for each $n \in \mathbb{N}$. Denote $F_n = X_n \times Y_n$. Let any $E \in \mathcal{A} \times \mathcal{B}$ be given.

\[\begin{aligned} \mu \times \nu (E) = \lim_{n \to \infty} \underbrace{\mu \times \nu (E \cap F_n)}_{< \infty} = \lim_{n \infty} \lambda (E \cap F_n) = \lambda (E) \end{aligned}\]

by continuity from below.

\[\tag*{$\blacksquare$}\]

Completeness

Even when $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$ are complete, $(X \times Y, \mathcal{A} \times \mathcal{B}, \mu \times \nu)$ is almost never complete.

For example, let $(X, \mathcal{A}, \mu) = (Y, \mathcal{B}, \nu)$ be Lebesgue measure on $[0, 1]$ with the Lebesgue $\sigma$-algebra. Let $A$ be a non-measurable set in $[0, 1]$ and let \(E = A \times \{ 1/2 \}\). Then $E$ is not a measurable set with regard to $\mathcal{A} \times \mathcal{B}$. (If not, $E^{1/2}$ should be measurable on $\mathcal{A}$ by lemma.) However, \(E \subset [0, 1] \times \{ 1/2 \}\), which has zero measure with regard to $\mu \times \nu$, therefore $E$ is a null set.

But one can take the completion of $(X \times Y, \mathcal{A} \times \mathcal{B}, \mu \times \nu)$ without great difficulty. See [3] for details.

Fubini-Tonelli Theorem

The main application of product measures is the Fubini-Tonelli theorem, which allows one to interchange the order of integration.

$\color{red}{\mathbf{Theorem.}}$ Fubini-Tonelli Theorem
Suppose $f: X \times Y \to \mathbb{R}$ is measurable with respect to $\mathcal{A} \times \mathcal{B}$. Suppose $\mu$ and $\nu$ are $\sigma$-finite measures on $X$ and $Y$, respectively. If either

1. $f$ is non-negative or
2. $f$ is integrable with regard to $\mu \times \nu$,

then

1. for each $x \in X$, the function $y \mapsto f(x, y)$ is measurable with regard to $\mathcal{B}$;
2. for each $y \in Y$, the function $x \mapsto f(x, y)$ is measurable with regard to $\mathcal{A}$;
3. for each $x \in X$, the function $x \mapsto \int f(x, y) \; d\nu(y)$ is measurable with regard to $\mathcal{A}$;
4. for each $y \in Y$, the function $y \mapsto \int f(x, y) \; d\mu(x)$ is measurable with regard to $\mathcal{B}$;
5. $$ \begin{aligned} \int f(x, y) \; d(\mu \times \nu) & = \int \left( \int f(x, y) \; d\mu \right) d\nu \\ & = \int \left( \int f(x, y) \; d\nu \right) d\mu \\ \end{aligned} $$

$\color{red}{\mathbf{Proof.}}$

(1) Let $f = \chi_E$, for $E \in \mathcal{A} \times \mathcal{B}$. Note that $\int f(x, y) \; d\nu (y) = \nu (E_x)$ and $\int f(x, y) \; d\mu (x) = \mu (E^y)$. Then, by the previous theorem and lemma, it satisfies the Fubini-Tonelli theorem. So do any simple functions, as they are linear combinations of characteristic functions.

(2) Let $f$ be a non-negative function. Then $f$ is the increasing limit of simple functions. Using the monotone convergence theorem, it satisfies the Fubini-Tonelli theorem.

(3) Let $\int \vert f(x, y) \vert \; d(\mu \times \nu) < \infty$. Write $f = f^+ - f^-$ and apply the Fubini-Tonelli theorem to $f^+$ and $f^-$ respectively. Done.

\[\tag*{$\blacksquare$}\]

$\color{#bf1100}{\mathbf{Remark.}}$

1. Observe that if we know $$ \begin{aligned} \int \int | f(x, y) | \; d\mu (x) d\nu (y) < \infty, \end{aligned} $$ then since $\vert f(x, y) \vert$ is non-negative the Fubini theorem tells us that $$ \begin{aligned} \int | f(x, y) | \; d(\mu \times \nu) = \int \int | f(x, y) | \; d\mu (x) d\nu (y) < \infty \end{aligned} $$ We can then apply the Fubini theorem again to conclude $$ \begin{aligned} f (x, y) \; d ( \mu \times \nu ) = \int \int f(x, y) \; d\mu d\nu = \int \int f(x, y) \; d\nu d\mu. \end{aligned} $$ Thus in the hypotheses of the Fubini theorem, we could as well assume $\int \int \vert f(x, y) \vert \; d\mu \; d\nu < \infty$ or $\int \int \vert f(x, y) \vert \; d\nu \; d\mu < \infty$.
2. The hypothesis of $\sigma$-finiteness is necessary.
   $\color{#bf1100}{\mathbf{Example.}}$

   Let $X = Y = [0, 1]$, and $\mathcal{A} = \mathcal{B}$ be Borel $\sigma$-algebras on $[0, 1]$. And, let $\mu$ be a Lebesgue measure and $\nu$ be a counting measure.

   Set a diagonal \(D = \{ (x, x) \mid x \in [0, 1] \}\) on $X \times Y$ and let $f (x, y) = \chi_D$.

   
   

   (1) $f$ is measurable with respect to $\mathcal{A} \times \mathcal{B}$.

   It suffices to show that $D \in \mathcal{A} \times \mathcal{B}$. We can verify this simply, by defining a continuous function $h: X \times Y \to \mathbb{R}$ with $h(x, y) = x - y$, so that it is Borel measurable. And \(D = h^{-1} (\{0 \})\).

   Alternatively, fix $n \in \mathbb{N}$ and define a disjoint union of rectangles $E_n \in \mathcal{A} \times \mathcal{B}$:

   \[\begin{aligned} E_n = \bigcup_{i=0}^{2^{n-1}}\left[\frac{i}{2^{n-1}}, \frac{i+1}{2^{n-1}}\right] \times\left[\frac{i}{2^{n-1}}, \frac{i+1}{2^{n-1}}\right] . \end{aligned}\]

   The picture is to cover $D$ by a bunch of squares of side length $1/2^{n−1}$ evenly spaced along the diagonal. Then obviously

   \[\begin{aligned} D = \bigcap_{n=1}^\infty E_n. \end{aligned}\]

   (2) But it doesn’t satisfy the theorem.

   \[\begin{aligned} \int_{0}^1 \int_{0}^1 \chi_D (x, y) \; d\mu \; d\nu = 0 \end{aligned}\]

   but

   \[\begin{aligned} \int_{0}^1 \int_{0}^1 \chi_D (x, y) \; d\nu \; d\mu = \int_{0}^1 1 \; d\mu = 1. \end{aligned}\]

   Also, we claim that $\int \chi_D \; d(\mu \times \nu) = \infty$. By

   \[\tag*{$\blacksquare$}\]
3. The hypothesis of measurability of $f$ with regard to $\mathcal{A} \times \mathcal{B}$ is necessary.
   $\color{#bf1100}{\mathbf{Example.}}$

   Taking on faith a bit of set theory, there exists a set $X$ together with a partial order $``\leq”$ such that $X$ is uncountable but for any $y \in X$, the set \(\{ x \in X \; : \; x \leq y \}\) is countable. (For example, let $X$ be the set of countable ordinals.)

   The $\sigma$-algebra is the co-finite collection; the collection of subsets $A$ of $X$ such that either $A$ or $A^c$ is countable. And define $\mu$ on $X$ by

   \[\begin{aligned} \mu (A) = \begin{cases} 1 & \text{ if } A \text{ is countable } \\ 0 & \text{ if } A \text{ is uncountable } \end{cases} \end{aligned}\]

   Define \(f: X \times X \to \{0, 1\}\) by

   \[\begin{aligned} f(x, y) = \begin{cases} 1 & \text{ if } x \leq y \\ 0 & \text{ otherwise } \end{cases} \end{aligned}\]

   Then $\int \int f(x, y) \; d\mu(x) \; d\mu(y) = 1$ but $\int \int f(x, y) \; d\mu(y) \; d\mu(x) = 0$.

   However, it doesn’t contradict the theorem as $f$ is not measurable with respect to the product $\sigma$-algebra.

   \[\tag*{$\blacksquare$}\]
4. The hypothesis $f \in L^+ (X \times Y)$ or $f \in L^1 (\mu \times \nu)$ is necessary, in two respects.
   $\color{#bf1100}{\mathbf{Example\ 1.}}$

   Let $X = Y = [0, 1]$ with $\mu$ and $\nu$ both being Lebesgue measure. Let $g_i$ be continuous functions with support in $(1/(i + 1), 1/i)$ such that $\int_{0}^1 g_i (x) \; dx = 1$, $i = 1, 2, \dots$. Let

   \[\begin{aligned} f(x, y) = \sum_{i=1}^\infty [ g_i (x) - g_{i+1} (x)] g_i (y). \end{aligned}\]

   For each point $(x, y)$ at most two terms in the sum are non-zero, so note that the sum is actually a finite one. If we first integrate with respect to $y$, we get

   \[\begin{aligned} \int_{0}^1 f (x, y) \; dy = \sum_{i=1}^\infty [g_i (x) - g_{i+1} (x)] = g_1 (x). \end{aligned}\]

   Therefore

   \[\begin{aligned} \int_{0}^1 \int_{0}^1 f (x, y) \; dy \; dx = \int_{0}^1 g_1 (x) dx = 1. \end{aligned}\]

   On the other hand, integrating first with respect to $x$ gives $0$, so

   \[\begin{aligned} \int_{0}^1 \int_{0}^1 f(x, y) \; dx \; dy = 0. \end{aligned}\]

   However, it doesn’t contradict the theorem as

   \[\begin{aligned} \int_{0}^1 \int_{0}^1 |f(x, y)| \; dx \; dy = \infty. \end{aligned}\] \[\tag*{$\blacksquare$}\]
   $\color{#bf1100}{\mathbf{Example\ 2.}}$

   Let \(X = \{1, 2, \cdots \}\) and $\mu$ be a counting measure. Define $f: X \times X \to \mathbb{R}$:

   \[\begin{aligned} f(x, y) = \begin{cases} 1 & \quad x = y \\ -1 & \quad x = y + 1 \\ 0 & \quad \text{otherwise } \end{cases} \end{aligned}\]

   Then

   \[\begin{aligned} \int_X \int_X f(x, y) \; dx \; dy = \sum_{y=1}^{\infty} \sum_{x=1}^{\infty} f(x, y) = 0 \end{aligned}\]

   and

   \[\begin{aligned} \int_X \int_X f(x, y) \; dy \; dx = \sum_{x=1}^{\infty} \sum_{y=1}^{\infty} f(x, y) = 1 \end{aligned}\]

   However, it doesn’t contradict the theorem as

   \[\begin{aligned} \int_X \int_X |f(x, y)| \; dx \; dy = \infty. \end{aligned}\] \[\tag*{$\blacksquare$}\]
   $\color{#bf1100}{\mathbf{Example\ 3.}}$

   Let $X = Y = \mathbb{R}$ and $\mathcal{B}$ be a Borel $\sigma$-algebra. Define $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$:

   \[\begin{aligned} f(x, y) = \begin{cases} 1 & \quad x \geq 0 \text{ and } x \leq y \leq x + 1 \\ -1 & \quad x \geq 0 \text{ and } x + 1 < y \leq x + 2 \\ 0 & \quad \text{otherwise } \end{cases} \end{aligned}\]

   Then

   \[\begin{aligned} \int \int f(x, y) \; dx \; dy = 1 \end{aligned}\]

   and

   \[\begin{aligned} \int \int f(x, y) \; dy \; dx = 0 \end{aligned}\]

   However, it doesn’t contradict the theorem as

   \[\begin{aligned} \int \int |f(x, y)| \; dx \; dy = \infty. \end{aligned}\] \[\tag*{$\blacksquare$}\]
5. Let $X = Y$ be the positive integers and $\mu = \nu$ be counting measure. Write $c_{ij} = f(i, j)$. Then, the Fubini theorem implies that $$ \begin{aligned} \sum_{i=1}^\infty \sum_{j=1}^\infty c_{ij} = \sum_{j=1}^\infty \sum_{i=1}^\infty c_{ij} \end{aligned} $$ provided either that the $c_{ij}$ are non-negative or that $\sum_{i} \sum_{j} |c_{ij}| < \infty$.

There is no difficulty extending the Fubini theorem to the product of $n$ measures. If we have $\mu_1, \cdots , \mu_n$ all equal to $m$ i.e., Lebesgue measure on $\mathbb{R}$ with the Lebesgue $\sigma$-algebra $\mathcal{L}$, then the completion of $(\mathbb{R}^n, \mathcal{L} \times \dots \times \mathcal{L}, \mathcal{m} \times \dots \times m)$ is called $n$-dimensional Lebesgue measure.

Reference

[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Folland, Gerald B. Real analysis: modern techniques and their applications. Vol. 40. John Wiley & Sons, 1999.
[3] W. Rudin. Real and Complex Analysis, 3rd ed. McGraw-Hill, New York, 1987.