Functions, Random Variables and Integration#

Cool facts about Measurable Functions#

1. Inverse images of set functions preserve set operations. i.e. for $f:\mathbb{X}\rightarrow\mathbb{Y}$ and $A, A_i\subset\mathbb{Y}, i\in \mathcal{I}$,

   $$f^{-1}\left(\displaystyle\bigcup_{i\in \mathcal{I}} A_i\right) = \displaystyle\bigcup_{i\in \mathcal{I}} f^{-1}(A_i) ~~ \text{and} ~~ f^{-1}(\mathbb{Y}\setminus A) = \mathbb{X} \setminus f^{-1}(A)$$

   For a measurable set function $f$, this implies that $\{f^{-1}(B) : B \in \mathcal{Y}\}$ is a $\sigma$-Field and is contained in $\mathcal{X}$ . Hence, we want $\mathcal{Y}$ to be no larger than $\mathcal{X}$ to have measurable functions. Furthermore, this can be used to show that the measurability of $f$ can be established by looking only at a collection of sets $\mathcal{A} \subset \mathcal{Y}$ that generates $\mathcal{Y}$.

   Example

   let $\mathcal{A}$ be the set of all half-lines $A_t = (−\infty, t]$ for $t \in \mathbb{R}$ will generate $\mathcal{B}(\mathbb{R})$. Thus, $f$ is measurable as long as the sets $\{x : f(x) \leq t\}$ are measurable.

2. For any $A \in \mathcal{X}$, the indicator functions $f(x) = \mathbf{1}[x \in A]$ are measurable. The $\sigma$-Field generated by $f^{−1}$ is simply $\{\varnothing, A, A^c, \mathbb{X}\}\subset\mathcal{X}$.

3. For measurable functions $f, g : \mathbb{X} \rightarrow \mathbb{R}$, the functions $f + g$ and $fg$ are measurable.

4. For measurable functions, $\{f_i\}_{i=1}^\infty$ from $\mathbb{X}$ to $\mathbb{R}$, the following are also measurable: $\sup_i f_i$, $\inf_i f_i$, $\displaystyle\limsup_i f_i$, $\displaystyle\liminf_i f_i$ and $\displaystyle\lim_i f_i$ if it exists.

   Proof

   In set notation, $\{x : \sup_i f_i(x)\leq t\}=\displaystyle\bigcap_{i=1}^\infty\{x : f_i(x)\leq t\}$ where the righthand side is a countable intersection of measurable sets and hence measurable. Similarly, $\{x : \inf_i f_i(x)\leq t\}=\displaystyle\bigcup_{i=1}^\infty\{x : f_i(x)\leq t\}$ and $\displaystyle\limsup_i f_i = \inf_i \sup_{j\geq i} f_i$ and $\displaystyle\liminf_i f_i = \sup_i\inf_{j\geq i} f_i$. If $\displaystyle\lim_i f_i$ exists then $\displaystyle\limsup f_i=\displaystyle\lim_i f_i=\displaystyle\liminf_i f_i$. $\square$

5. Let $f : \mathbb{X}\rightarrow\mathbb{R}$ be a continuous function, then it is measurable.

   Proof

   If $U$ is an open set in $\mathbb{R}$, then $f^{−1}(U)$ is open in $\mathbb{X}$ (definition of a continous function between two topological spaces). Thus, the set $f^{−1}(U)$ is measurable. Since the open sets of $\mathbb{R}$ generate $\mathcal{B}(\mathbb{R})$, the function $f$ is measurable. $\square$

6. Given a collection of functions $f_i: \mathbb{X}\rightarrow\mathbb{Y}, i\in \mathcal{I}$, we can make them measurable by constructing the measurable space $(\mathbb{X},\mathcal{X})$ where $\sigma(\{f_i\}_{i\in \mathcal{I}})\subseteq\mathcal{X}$ is the $\sigma$-field generated by the sets $f^{-1}_i(B)$ for all $i$ and $B\in\mathcal{Y}$.

Almost Surely / Almost Everywhere

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space. For two functions $f, g : \Omega\rightarrow\mathbb{R}$, we say that $f = g$ a.e. (almost everywhere) when the set $N = \{\omega : f(\omega) \neq g(\omega)\}$ has measure $\mu(N) = 0$.

In probability theory, “almost everywhere” is replaced with “almost surely” abbreviated a.s. and it is equivalently is written “with probability 1” or w.p.1.

Example

Let $([0, 1], \mathcal{B}, \lambda)$ be the standard measure space of Borel sets on the unit interval with Lebesgue measure. Let $f(t)=0$ for all $t\in(0,1]$ and $g(t) = 0$ on $(0, 1]\setminus\mathbb{Q}$ and $g(t)=1$ on $(0, 1]\displaystyle\cap\mathbb{Q}$.

Then we have $f=g$ a.e. that is $\lambda((0, 1)\displaystyle\cap\mathbb{Q}) = 0$. To prove that $\lambda(\mathbb{Q})=0$, we enumerate each rational number $\{q_m\}_{m=1}^\infty$ and surround each $q_m$ with an interval $(q_m-\frac{\epsilon}{2^m}, q_m + \frac{\epsilon}{2^m})$. For $\epsilon > 0$, $\displaystyle\sum_{m=1}^\infty \lambda((q_m-\frac{\epsilon}{2^m}, q_m + \frac{\epsilon}{2^m}))=\displaystyle\sum_{m=1}^\infty\frac{\epsilon}{2^{m-1}}=2\epsilon$. Since $\epsilon$ is arbitrary, we have $\lambda(\mathbb{Q})=0$.

Monotone Convergence Theorem#

Monotone Convergence Theorem

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $\{f_i\}_{i=1}^\infty$ be measurable functions from $\Omega$ to $[-\infty, \infty]$ such that $f_i \uparrow f$ $\mu$-a.e. and $\displaystyle\int f_1\mathrm{d}\mu > -\infty$. Then $\displaystyle\int f_i\mathrm{d}\mu \uparrow \displaystyle\int f\mathrm{d}\mu$. (This means that we can swap the limit and the integral)

Proof

First, we need to check that $f$ is measureable. For $c\in\mathbb{R}$, we consider the sets $(c, \infty]$, which generate the Borel $\sigma$-Field. Since $f_i\uparrow f$, $f^{-1}((c, \infty]) = \displaystyle\bigcup_{i=1}^\infty f_i^{-1}((c, \infty])$ and $f_i^{-1}((c, \infty]) \in \mathcal{F}$, thus $f$ is measureable. (by the first cool fact about measurable functions)

Next, assume that $f_1 \geq 0$, and for each $f_i$ we take simple functions $g_{ij}$ such that $g_{ij} \uparrow f_i$ as $j\rightarrow\infty$. Thus, by MCT for Simple Functions, $\displaystyle\int g_{ij}\mathrm{d}\mu\uparrow\displaystyle\int f_i\mathrm{d}\mu$. Furthermore, let $g_i^* = \max\{g_{1i}, \cdots , g_{ii}\}$. These $g_i^*$ are simple functions and $g_i^*\uparrow f$. Once again, MCT for Simple Functions implies that $\displaystyle\int g_i^*\mathrm{d}\mu\uparrow\displaystyle\int f\mathrm{d}\mu$. But since $g_i^* \leq f_i$ by construction, $\displaystyle\int g_i^*\mathrm{d}\mu \leq \displaystyle\int f_i\mathrm{d}\mu \leq \displaystyle\int f\mathrm{d}\mu$. Thus, $\displaystyle\int f_i\mathrm{d}\mu\uparrow\displaystyle\int f\mathrm{d}\mu$. This means we are done for non-negative fucntions ($f_1\geq 0$).

Now, we assume that $f \leq 0$. In this case $f_i \uparrow f$ implies that $−f_i \downarrow −f$. Let $h=-f$ and $h_i=-f_i$, we have $0\leq \displaystyle\int h\mathrm{d}\mu\leq \displaystyle\int h_i\mathrm{d}\mu$. Next, note that $0\leq h_1-h_i\uparrow h_1-h$. Applying the above result gives that $\displaystyle\int (h_1-h_i)\mathrm{d}\mu\uparrow \displaystyle\int (h_1-h)\mathrm{d}\mu$. Since all of the $h$ have finite integrals, we are allowed to subtract to get that $\displaystyle\int h_i\mathrm{d}\mu\downarrow\displaystyle\int h\mathrm{d}\mu$ and thus $\displaystyle\int f_i\mathrm{d}\mu\uparrow\displaystyle\int f\mathrm{d}\mu$.

For a general function $f = f^+ - f^-$, we have $f_i^+\uparrow f^+$ and $f_i^-\downarrow f^-$ and $\displaystyle\int f^-\mathrm{d}\mu < \infty$. So by the above special cases, $\displaystyle\int f_i^+\mathrm{d}\mu\uparrow\displaystyle\int f^+\mathrm{d}\mu$ and $\displaystyle\int f_i^-\mathrm{d}\mu\downarrow\displaystyle\int f^-\mathrm{d}\mu$. Finally, $\displaystyle\int f_i\mathrm{d}\mu \uparrow \displaystyle\int f\mathrm{d}\mu$. $\square$

We only require $f_i \uparrow f$ to hold almost everywhere to establish the result. Hence convergence can fail on a set of measure (probability) zero and we still have convergence of the integrals.

Secondly, we can redo the above proof for $f_i \downarrow f$ with $\displaystyle\int f_1\mathrm{d}\mu < \infty$ to get a similar result for decreasing sequence.

Fatou’s Lemma #

Fatous’ Lemma

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $\{f_i\}_{i=1}^\infty$ be non-negative measurable functions from $\Omega$ to $[-\infty, \infty]$. Then, $\displaystyle\int \displaystyle\liminf f_i\mathrm{d}\mu \leq \displaystyle\liminf \displaystyle\int f_i\mathrm{d}\mu$

Proof

Recall that $\displaystyle\liminf_{i\rightarrow\infty} = \sup_j \inf_{i>j} f_i$. Hence, let $g_j=\inf_{i\geq j} f_i$. Then, $g_j\uparrow \displaystyle\liminf_{i\rightarrow\infty} f_i$ and $f_1\geq 0$ by assumption. So MCT for Simple Functions says that $\displaystyle\int g_j\mathrm{d}\mu\uparrow\displaystyle\int\displaystyle\liminf_{i\rightarrow\infty}f_i\mathrm{d}\mu$. By construction, $g_j\leq f_i$ for any $i\geq j$, thus $\displaystyle\int g_j\mathrm{d}\mu \leq \displaystyle\int f_i\mathrm{d}\mu$ for any $i\geq j$ and subsequently, $\displaystyle\int g_j\mathrm{d}\mu\leq \inf_{i\geq j}\displaystyle\int f_i\mathrm{d}\mu$. Taking $j\rightarrow\infty$ gives $\displaystyle\lim_{j\rightarrow\infty}\displaystyle\int g_j\mathrm{d}\mu = \displaystyle\int\displaystyle\liminf f_i\mathrm{d}\mu\leq\displaystyle\liminf\displaystyle\int f_i\mathrm{d}\mu$. $\square$

Dominated Convergence Theorem#

Dominated Convergence Theorem

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $\{f_i\}_{i=1}^\infty$ and $g$ be measurable functions and absolutely integrable. If $|f_i| \leq g$ for all $i$ and $f_i(\omega) \rightarrow f(\omega)$ for each $\omega \in \Omega$ (i.e. pointwise convergence), then $f$ is absolutely integrable and $\displaystyle\int f_i\mathrm{d}\mu \rightarrow \displaystyle\int f\mathrm{d}\mu$.

Proof

Let $f_i^\wedge = \inf_{j\geq i} f_j$ and $f_i^\vee = \sup_{j\geq i} f_j$. Then $f_i^\wedge \leq f_i \leq f_i^\vee$. We have that $f_i^\wedge\uparrow f$ and $\displaystyle\int f_1^\wedge\mathrm{d}\mu\geq -\displaystyle\int g\mathrm{d}\mu > -\infty$. So MCT for Simple Functions implies that $\displaystyle\int f_i^\wedge\mathrm{d}\mu\uparrow \displaystyle\int f\mathrm{d}\mu$.

Do the same for $f_i^\vee$, we have that $f_i^\vee\downarrow f$ and hence that $\displaystyle\int f_i^\vee\mathrm{d}\mu\downarrow\displaystyle\int f\mathrm{d}\mu$. Since $\displaystyle\int f_i^\wedge\mathrm{d}\mu\leq \displaystyle\int f_i\mathrm{d}\mu\leq \displaystyle\int f_i^\vee\mathrm{d}\mu$, we have the desired result that $\displaystyle\int f_i\mathrm{d}\mu\rightarrow \displaystyle\int f\mathrm{d}\mu$. $\square$

Lebesgue-Stieltjes Measure#

Let $(\mathbb{X}, \mathcal{X})$ and $(\mathbb{Y}, \mathcal{Y})$ be two measurable spaces. Let $\psi : \mathbb{X}\rightarrow\mathbb{Y}$ be a measurable function and $\mu$ be a measure on $\mathcal{X}$. Then we can define $\nu = \mu \circ \psi^{-1}$ to be a measure of $\mathcal{Y}$. This allows us to turn Lebesgue measure into Lebesgue-Stieltjes measures.

Theorem

Let $F : \mathbb{R} \rightarrow \mathbb{R}$ be non-constant, right-continuous, and non-decreasing. Then, there exists a unique measure $\mathrm{d}F$ on $\mathbb{R}$ such that for all $a, b \in \mathbb{R}$ with $a < b$,

$$\mathrm{d}F((a, b]) = F(b) - F(a)$$

Proof

Let $F(\infty) = \displaystyle\lim_{x\rightarrow\infty} F(x)$ and $F(-\infty) = \displaystyle\lim_{x\rightarrow-\infty}F(x)$. We define an open interval $I = (F(-\infty), F(\infty))$ and define $g(y) = \inf\{x\in\mathbb{R} : y\leq F(x)\}$. We want to define $\mathrm{d}F$ to be $\lambda \circ g^{−1}$ where $\lambda$ is Lebesgue Measure on $\mathbb{R}$, so we need to show that this makes sense.

We first show that $g$ is left-continuous and non-decreasing and for $y \in I$ and $x \in \mathbb{R}, g(y) \leq x$ if and only if $y \leq F(x)$. To show this, fix a $y\in I$ and let $J_y = \{x\in\mathbb{R} : y\leq F(x)\}$. As $F$ is non-decreasing, if $x\in J_y$ and $x'\geq x$ then $x'\in J_y$. As $F$ is right-continuous, if $x_n\in J_y$ and $x_n\downarrow x$ then $x\in J_y$. Therefore, $J_y = [g(y), \infty)$ and $g(y)\leq x$ if and only if $y\leq F(x)$. Secondly, for $y\leq y'$, we have that $J_{y'}\subseteq J_y$ and thus $g(y)\leq g(y')$. So if $y_n\uparrow y$, then $J_y = \displaystyle\bigcap_{n=1}^\infty J_{y_n}$ and thus $g(y_n)\rightarrow g(y)$, which implies that $g$ is left continuous and non-decreasing.

From the above, $g$ is Borel Measurable and thus defining $\mathrm{d}F = \lambda \circ g^{-1}$ gives us that

$$\mathrm{d}F((a, b]) = \lambda(\{y : g(y) > a, g(y) \leq b\}) = \lambda((F(a), F(b)]) = F(b) - F(a)$$

Furthermore, this measure, $\mathrm{d}F$, is unique by using the same arguments as before for Lebesgue Measure. $\square$

In the case that $F : \mathbb{R} \rightarrow [0, 1]$ such that the interval $I = [0, 1]$, we have a cumulative distribution function, which induces a measure on the real line. This allows us to do things like integrate with respect to such measures——i.e. take an expectation.

Radon Measure#

Definition

Let $(\Omega, \mathcal{B}, \mu)$ be a measure space where $\mathcal{B}$ is the Borel $\sigma$-field. The measure $\mu$ is said to be a Radon Measure if $\mu(K) < \infty$ for all compact $K \in \mathcal{B}$.

• $\mathrm{d}F$ is a Radon Measure.

• Every non-zero Radon Measure on $\mathcal{B}(\mathbb{R})$ can be written as $\mathrm{d}F = \lambda \circ g^{−1}$ for some $F$.

  If $\mu$ is a Radon Measure on $\mathbb{R}$, then we can define $F$ as

  $$F(x) = \begin{cases} \mu((0, x]) & \text{if } x > 0 \\ -\mu((x, 0]) & \text{if } x < 0 \\ \end{cases}$$

  Thus, $F(b) − F(a) = \mu((a, b])$ for $a < b$ and hence $\mu = \mathrm{d}F$ by uniqueness.

Monotone Class#

Definition

A collection of subsets $\mathcal{M}$ of $\Omega$ is said to be monotone if

1. for $\{A_i\}_{i=1}^\infty$ s.t. $A_i \in \mathcal{M}$ and $A_i \uparrow A = \displaystyle\bigcup_{i=1}^\infty Ai$, then $A \in \mathcal{M}$.
2. for $\{A_i\}_{i=1}^\infty$ s.t. $A_i \in \mathcal{M}$ and $A_i \downarrow A = \displaystyle\bigcup_{i=1}^\infty Ai$, then $A \in \mathcal{M}$.

Note that if a field $\mathcal{A}$ is also monotone, then it is a $\sigma$-field (by definition of $\sigma$-field.)

Monotone Class Theorem

Let $\mathcal{A}$ be a field and $\mathcal{M}$ be monotone such that $\mathcal{A}\subset\mathcal{M}$. Then, $\sigma(\mathcal{A}) \subseteq \mathcal{M}$

Proof

This proof is similar to the one of Dynkin $\pi$-$\lambda$ Theorem.

Existence and Uniqueness of Product Measure#

Existence and Uniqueness Theorem of Product Measure

Let $(\mathbb{X}, \mathcal{X} , \mu)$ and $(\mathbb{Y}, \mathcal{Y}, \nu)$ be $\sigma$-finite measure spaces. We denote the product $\sigma$-Field to be $\mathcal{X} \times \mathcal{Y}$. (the cartesian product of two $\sigma$-Fields may not be a $\sigma$-Field) Let $π$ be a set function on $\mathcal{X}\times\mathcal{Y}$ such that for $A \in \mathcal{X}$ and $B \in \mathcal{Y}$, $π(A \times B) = \mu(A)\nu(B)$. Then, $π$ extends uniquely to a measure on $(\mathbb{X} \times \mathbb{Y}, \mathcal{X}\times\mathcal{Y})$ such that for any $E \in \mathcal{X} \times \mathcal{Y}$,

$$\pi(E) = \displaystyle\int\displaystyle\int \mathbf{1}_E(x, y)\mathrm{d}\mu(x)\mathrm{d}\nu(y) = \displaystyle\int\displaystyle\int \mathbf{1}_E(x, y)\mathrm{d}\nu(y)\mathrm{d}\mu(x)$$

Lemma

Let $(\mathbb{X}, \mathcal{X} , \mu)$ and $(\mathbb{Y}, \mathcal{Y}, \nu)$ be finite measure spaces, and let

$$\mathcal{F} = \left\{E \subset \mathbb{X}\times\mathbb{Y} : \displaystyle\int\displaystyle\int \mathbf{1}_E(x, y)\mathrm{d}\mu(x)\mathrm{d}\nu(y) = \displaystyle\int\displaystyle\int \mathbf{1}_E(x, y)\mathrm{d}\nu(y)\mathrm{d}\mu(x)\right\}$$

Then $\mathcal{X}\times\mathcal{Y} \subseteq \mathcal{F}$. ( This means that $(\mathbb{X}\times\mathbb{Y}, \mathcal{X}\times\mathcal{Y}) \equiv (\mathbb{Y}\times\mathbb{X}, \mathcal{Y}\times\mathcal{X})~$)

Proof

Let $E = A \times B$ for $A \in \mathcal{X}$ and $B \in \mathcal{Y}$, i.e. $E \in \mathcal{R}$. Then,

$$\begin{aligned} \displaystyle\int\displaystyle\int \mathbf{1}_E(x, y)\mathrm{d}\mu(x)\mathrm{d}\nu(y) &=\mu(A) \displaystyle\int \mathbf{1}_B(y)\mathrm{d}\nu(y) \mu(A)\nu(B) =\nu(B)\mu(A)\\ &=\nu(B) \displaystyle\int \mathbf{1}_A(x)\mathrm{d}\mu(x) =\displaystyle\int\displaystyle\int \mathbf{1}_E(x, y)\mathrm{d}\nu(y)\mathrm{d}\mu(x) \end{aligned}$$

Therefore, $\mathcal{R} \subset \mathcal{F}$.

Also, for disjoint $R_1, R_2 \in \mathcal{R}$, $\mathbf{1}_{R_1\displaystyle\cup R_2} = \mathbf{1}_{R_1} + \mathbf{1}_{R_2}$. Hence, $\mathcal{F}$ contains finite disjoint unions of $\mathcal{R}$. This implies that the field generated by the set of rectangles $\mathcal{A} \subset \mathcal{F}$. (Dudley 3.2.3)

Next, consider $\{E_i\}_{i=1}^\infty, E_i\in\mathcal{F}$. Then if $E_i\uparrow E$, then MCT says

$$\displaystyle\int\displaystyle\int \mathbf{1}_{E_i}(x, y)\mathrm{d}\mu(x)\mathrm{d}\nu(y) ~\big\uparrow \displaystyle\int\displaystyle\int \mathbf{1}_{E}(x, y)\mathrm{d}\mu(x)\mathrm{d}\nu(y)$$

and

$$\displaystyle\int\displaystyle\int \mathbf{1}_{E_i}(x, y)\mathrm{d}\nu(y)\mathrm{d}\mu(x) ~\big\uparrow \displaystyle\int\displaystyle\int \mathbf{1}_{E}(x, y)\mathrm{d}\nu(y)\mathrm{d}\mu(x)$$

Thus, $E \in \mathcal{F}$ and the same holds if $E_i \downarrow E$. Therefore, $\mathcal{F}$ is a monotone class. Finally, applying the Monotone Class Theorem shows that $\mathcal{X} \times \mathcal{Y} = \sigma(\mathcal{A}) \subset \mathcal{F}$. $\square$

Proof : Existence and Uniqueness of Product Measure

First, we consider the case that $\mu$ and $\nu$ are finite measures and $\pi(A\times B) = \nu(A)\nu(B)$. Then we extend to a set function

$$\pi(E)\vcentcolon=\displaystyle\int\displaystyle\int \mathbf{1}_E(x, y)\mathrm{d}\mu(x)\mathrm{d}(y)$$

for any $E\in \mathcal{X}\times\mathcal{Y}$. The above lemma says that the definiton makes sense and the order of integration can be reversed for any $E \in \mathcal{X} \times \mathcal{Y}$. Linearity of integral implies that $\pi$ is finitely additive. Apply MCT shows that $\pi$ is countably additive.. Thus, $\pi$ is a measure on $\mathcal{X} \times \mathcal{Y}$.

To show that $\pi$ is unique, let $\rho$ be some other set function such that $\rho(A \times B) = \mu(A)\nu(B)$ for $A \times B \in \mathcal{R}$. Let $\mathcal{M} = \{E \subset \mathbb{X} \times \mathbb{Y} : \pi(E) = \rho(E)\}$. Then, $\mathcal{M}$ is a monotone class because for $E_i \uparrow E =\displaystyle\bigcup_{i=1}^\infty E_i$, we can rewrite $E = \displaystyle\bigcup_{i=1}^\infty D_i$ where $D_1 = E_1$ and $D_i = E_i \setminus E_{i−1}$ for $i \geq 2$ are disjoint. So by countable additivity, $\pi(E)=\rho(E)$ and we can do the same for $E_i \downarrow E$. Thus, by Monotone Class Theorem $\mathcal{X} \times \mathcal{Y} \subseteq \mathcal{M}$. Therefore, $\pi$ is unique on $\mathcal{X} \times \mathcal{Y}$ for finite measures $\mu$ and $\nu$.

Now let $\mu$ and $\nu$ be $\sigma$-finite measures. Let $\{A_i\}_{i=1}^\infty$ and $\{B_i\}_{i=1}^\infty$ be disjoint partitions of $\mathbb{X}$ and $\mathbb{Y}$, respectively, such that $\mu(A_i)<\infty$ and $\nu(B_i)<\infty$. Then, for any $E \in \mathcal{X} \times \mathcal{Y}$, we define $E_{ij} = E\displaystyle\cap (A_i\times B_j)$. From the above finite measure case,

$$\displaystyle\int\displaystyle\int \mathbf{1}_{E_{ij}}(x, y)\mathrm{d}\mu(x)\mathrm{d}\nu(y) = \displaystyle\int\displaystyle\int \mathbf{1}_{E_{ij}}(x, y)\mathrm{d}\nu(y)\mathrm{d}\mu(x)$$

Sum over all $i$ and $j$ and apply MCT again to get

$$\pi(E) = \displaystyle\int\displaystyle\int \mathbf{1}_{E}(x, y)\mathrm{d}\mu(x)\mathrm{d}\nu(y) = \displaystyle\int\displaystyle\int \mathbf{1}_{E}(x, y)\mathrm{d}\nu(y)\mathrm{d}\mu(x)$$

Futhermore, Monotone convergence implies that $\pi$ is countably additive and hence a measure on $\mathcal{X}\times\mathcal{Y}$. For any other measure $\rho$ such that $\rho(A \times B) = \mu(A)\nu(B)$, countably additivity and uniqueness for finite measures implies that

$$\pi(E) = \displaystyle\sum_{i,j}\pi(E_{i,j}) = \displaystyle\sum_{i,j}\rho(E_{i,j}) = \rho(E)$$

Hence, the extension of $\pi$ to $\mathcal{X} \times\mathcal{Y}$ is unique. $\square$

Fubini-Toneli#

Fubini-Toneli Theorem

Let $(\mathbb{X}, \mathcal{X} , \mu)$ and $(\mathbb{Y}, \mathcal{Y}, \nu)$ be $\sigma$-finite measure spaces, and let $f : \mathbb{X} \times \mathbb{Y} \rightarrow \mathbb{R}$ be measurable with respect to $\mathcal{X} \times \mathcal{Y}$ such that either $f\geq 0$ (non-negative) or $\displaystyle\int\displaystyle\int |f|\mathrm{d}(\mu\times\nu)<\infty$ (absolutely integrable). Then,

$$\displaystyle\int\displaystyle\int f\mathrm{d}(\mu\times\nu) = \displaystyle\int\displaystyle\int f(x,y)\mathrm{d}\mu(x)\mathrm{d}\nu(y) = \displaystyle\int\displaystyle\int f(x,y)\mathrm{d}\nu(y)\mathrm{d}\mu(x)$$

Also, $\displaystyle\int f(x, y)\mathrm{d}\mu(x)$ is $\mathcal{Y}$-measurable and $\displaystyle\int f(x, y)\mathrm{d}\nu(y)$ is $\mathcal{X}$-measurable.