Chapter 6 Product Measure

Product Measures, Tonelli's Theorem, Fubini's Theorem.

Let $(\Omega_1,\mathcal{F}_1)$ and $(\Omega_2,\mathcal{F}_2)$ be two measurable spaces. Consider $\Omega:=\Omega_1\times\Omega_2$

Coordinate maps:

\[\begin{aligned} \rho_1((\omega_1,\omega_2))&=\omega_1\\ \rho_2((\omega_1,\omega_2))&=\omega_2 \end{aligned}\]

Define $\mathcal{F}:=\sigma(\rho_1,\rho_2)$, which is the “product $\sigma$-algebra”. Notation: $\mathcal{F}:=\mathcal{F_1}\times\mathcal{F}_2$.

\[\begin{aligned} \rho_1^{-1}(B_1)&=B_1\times\Omega_2,\qquad B_1\in\mathcal{F}_1\\ \rho_2^{-1}(B_2)&=B_2\times\Omega_1,\qquad B_2\in\mathcal{F}_2\\ \end{aligned}\] \[\rho_1^{-1}(B_1)\cap\rho_2^{-1}(B_2)=B_1\times B_2\]

Moreover,

\[\mathcal{I}:=\{B_1\times B_2:B_i\in\mathcal{F}_i\}\]

is a $\pi$-system generating $\mathcal{F}$.

Lemma 1

Let $\mathcal{H}$ be the class of functions $h:\Omega\to\mathbb{R}\text{ s.t. }h\in b\mathcal{F}$, which is bounded and measuable, and that

  1. $\forall\omega_1\in\Omega_1$, the map $\omega_2\to h(\omega_1,\omega_2)\in\mathcal{F}_2$
  2. $\forall\omega_2\in\Omega_2$, the map $\omega_1\to h(\omega_1,\omega_2)\in\mathcal{F}_1$

Then $\mathcal{H}=b\mathcal{F}$.

Proof of Lemma 1

\[A\in\mathcal{I}\implies\mathbb{1}_A\in\mathcal{H}\]
  1. Constant functions are in $\mathcal{H}$
  2. $h_n\geq0,h_n\leq h_{n+1}<K<\infty,h=\sup h_n,h_n\in\mathcal{H},h(\omega_1,\omega_2)=\sup h_n(\omega_1,\omega_2)\implies h\in\mathcal{H}$. $\mathcal{H}$ is a monotone class, therefore $\mathcal{H}$ contains all bounded $\sigma(\mathcal{I})$-measurable functions. But $\sigma(\mathcal{I})=\mathcal{F}$.

Lemma 2

Suppose $(\Omega_i,\mathcal{F}_i,\mu_i)$ are $\sigma$-finite. If $E\in\mathcal{F}$, then the functions

\[f(\omega_1):=\mu_2(\{(\omega_1,\omega_2):(\omega_1,\omega_2)\in E\})\\ f(\omega_2):=\mu_1(\{(\omega_1,\omega_2):(\omega_1,\omega_2)\in E\})\]

are measurable, and

\[\int f\mathrm{d}\mu_1=\int g\mathrm{d}\mu_2\]

Proof of Lemma 2

Suppose $\mu_i$’s are finite.

\[\mathcal{L}:=\{E\in\mathcal{F}:f\in\mathcal{F}_1\}\]

  1. $\mathcal{I}\subset\mathcal{F}$: If $E\in\mathcal{I},E=B_1\times B_2$, where $B_i\in\mathcal{F}_i$.

    \[f(\omega_1)=\mathbb{1}_{B_1}\mu_2(B_2)\in m\mathcal{F}_1\]

  2. $\mathcal{L}$ is a $\lambda$-system (Prove it!!). By Dynkin’s $\pi$-$\lambda$ Theorem, $\mathcal{L}\supset\sigma(\mathcal{I})=\mathcal{F}$. $f$ is $\mathcal{F}_1$-measurable $\forall E\in\mathcal{F}$, thus $g$ is $\mathcal{F}_1$-measurable $\forall E\in\mathcal{F}$.

  3. Define $\Pi_i$ on $(\Omega,\mathcal{F})$ by

    \[\Pi_1(E)=\int f\mathrm{d}\mu_1,\qquad\Pi_2(E)=\int g\mathrm{d}\mu_2\]

    $\Pi_1(\Omega)=\int_{\Omega_1}\mu_2(\Omega_2)\mathrm{d}\mu_1=\mu_2(\Omega_2)\mu_1(\Omega_1)$, and $\Pi_2(\Omega)=\mu_1(\Omega_1)\mu_2(\Omega_2)$, which are two set functions agreeing on $\Omega$.

  4. $\Pi_i$’s are countably additive set functions (Prove it!!).

  5. If $A\in\mathcal{I},\Pi_1(A)=\int\mathbb{1}_{B_1}\mu_2(B_2)\mathrm{d}\mu_1=\mu_1(B_1)\mu_2(B_2)$, and $\Pi_1(A)=\mu_1(B_1)\mu_2(B_2)$.

  6. $\Pi_1$ and $\Pi_2$ can be extended uniquely to $\mathcal{F}$ to define the product measure $\Pi$.

For the $\sigma$-finite case: $\Omega_1=\bigcup_{n\geq1}B_{1n},\mu_1(B_{1n})<\infty$, and $\Omega_2=\bigcup_{n\geq1}B_{2n},\mu_2(B_{2n})<\infty$. On each $\mathcal{F}_i$, define $\mu_{in}(E_i):=\mu_i(E_i\cap B_{in})$. Analogously define $\Pi_{1n}$ and $\Pi_{2n}$ and conclude by MCT and the result prove for the finite measure case.

We have in fact defined our product measure!

\[\Pi(E)=\int f\mathrm{d}\mu_1=\int g\mathrm{d}\mu_2.\]

Theorem 1 (Tonelli’s Theorem)

$(\Omega_i,\mathcal{F}_i,\mu_i)$ are $\sigma$-finite, $f\in (m\mathcal{F})^+$. Then,

\[\begin{aligned} g_1(\omega_1)&=\int f(\omega_1,\omega_2)\mu_2(\mathrm{d}\omega_2)\in m\mathcal{F}_1\\ g_2(\omega_2)&=\int f(\omega_1,\omega_2)\mu_1(\mathrm{d}\omega_1)\in m\mathcal{F}_2\\ \end{aligned}\] \[\int_{\Omega}f\mathrm{d}\Pi=\int_{\Omega_1}f_1\mathrm{d}\mu_1=\int_{\Omega_2}f_2\mathrm{d}\mu_2\]

Proof of Tonelli’s Theorem

If $f=\mathbb{1}_E,E\in\mathcal{F}$, the result follows from Lemma 2. By the linearity of integrals, it tolds for all non-negative simple measurable functions, MCT yields the claim for all non-negative measurable functions.

Theorem 2 (Fubini’s Theorem)

The conclusion of the above holds if $f\in m\mathcal{F}$, and $\int\lvert f\rvert\mathrm{d}\Pi<\infty$.

Warning: $\sigma$-finiteness is crucial. E.g.: $\Omega_i=[0,1],\mathcal{F}_i=\mathcal{B}[0,1],\mu_1=Leb,\mu_2$ is the counting measure. $E=\{(x,y):x=y\}\in\mathcal{F}$,

\[\begin{aligned} \int\int\mathbb{1}_{\{x=y\}}\mu_2(\mathrm{d}y)\mu_1(\mathrm{d}x)=\int\mu_1(\mathrm{d}x)=1\\ \int\int\mathbb{1}_{\{x=y\}}\mu_1(\mathrm{d}x)\mu_2(\mathrm{d}y)=\int0\mu_2(\mathrm{d}y)=0 \end{aligned}\]

Remark: Product measures are handy when it comes to joint laws of r.v.’s.