# 1.7. Product space


Let $X, Y$ be random variables on $(\Omega_1, \mathcal{F}_1, P_1)$ and $(\Omega_2, \mathcal{F}_2, P_2)$ respectively. In order to well-define sets such as $\{ X + Y \le 0 \}$, we should consider a random vector $(X,Y)$ on a product space $\mathbb{R} \times \mathbb{R}$ since $+: \mathbb{R}\times\mathbb{R} \to \mathbb{R}$ is defined this way. In addition, to measure the probability of such sets, we also need to define another product probability space $(\Omega, \mathcal{F}, P)$ where $\Omega = \Omega_1 \times \Omega_2$. Main question to answer in this subsection is: how can we define a proper $\sigma$-field $\mathcal{F}$ and a product measure $P$ on such $\Omega$?

## Product measure space

Let $(\Omega_1, \mathcal{F}_1, \mu_1)$ and $(\Omega_2, \mathcal{F}_2, \mu_2)$ be measure spaces with $\sigma$-finite measures $\mu_1, \mu_2$.
(i) $\Omega = \Omega_1 \times \Omega_2 := \{(\omega_1, \omega_2):~ \omega_1 \in \Omega_1, \omega_2 \in \Omega_2\}$ is the product space.
(ii) $\mathcal{S} := \{E_1 \times E_2:~ E_1 \in \mathcal{F}_1, E_2 \in \mathcal{F}_2\}$. $S \in \mathcal{S}$ is called a (measurable) rectangle.
(iii) $\mathcal{F} = \mathcal{F}_1 \times \mathcal{F}_2 := \sigma(\mathcal{S})$ is the $\sigma$-field of the product space.

Note that instead of directly product the two $\sigma$-fields, we define $\mathcal{F}$ with measurable rectangles. In fact, direct product of the $\sigma$-field does not yield a $\sigma$-field.

We now define a product measure on $(\Omega, \mathcal{F})$. Uniqueness of such measure is as important as existence of it. Even if we define a proper measure, it would be useless unless it is uniquely determined. Thankfully, Caratheodory’s theorem ensures uniqueness.

There uniquely exists a measure $\mu$ on $(\Omega, \mathcal{F})$ such that $\mu(E_1 \times E_2) = \mu_1(E_1) \mu_2(E_2),~ \forall E_1\in\mathcal{F}_1, E_2\in\mathcal{F}.$

$\mathcal{S}$ is a semi-algebra. Show (i) $\mu(A \times B) = \sum_{i=1}^\infty \mu(A_i \times B_i) = \sum_{i=1}^\infty \mu_1(A_i)\mu_2(B_i)$ for $A\times B = \cup_{i=1}^\infty (A_i \times B_i)$ and (ii) $\sigma$-finiteness of $\mu$. Use Caratheodory's theorem to get the result.

## Fubini’s theorem

Measure of a measurable rectangle is comprehensive. However it is cumbersome to compute and intuitively understand measurable sets that are not rectangles. We could understand it as the limit of measure of rectangles, but it is still a problem to calculate it - we do not know anything more than the existence of the measure. Fubini’s theorem is here for the rescue.

Let $(\Omega, \mathcal{F}, \mu)$ be the product space of $(\Omega_1, \mathcal{F}_1, \mu_1)$ and $(\Omega_2, \mathcal{F}_2, \mu_2)$. If $f \ge 0$ a.s. or $f \in L^1(\mu)$, then \begin{align} \int_{\Omega} f d\mu &= \int_{\Omega_2} \int_{\Omega_1} f(x,y) d\mu_1(x) d\mu_2(y)\\ &= \int_{\Omega_1} \int_{\Omega_2} f(x,y) d\mu_2(y) d\mu_1(x) \end{align}

Fubini’s theorem implies that for almost surely non-negative or integrable functions, we can calculate the integral with respect to the product measure as a double integral, and its order is negligible.

Finishing the section, I would like to remark some points.

1. We can even define $n$-dimensional product spaces to extend the notion even further.
2. If $(X,Y)$ is a random element on a metric space $S \times S$, then $X, Y$ are random elements on $S$. However the converse is not always the truth. In fact, it is known that the converse holds if $S$ is separable.1 Since $(\mathbb{R}, d)$ is a separable space, we can be sure that $(X,Y)$ is a random vector if and only if $X, Y$ are random variables.

Acknowledgement

This post series is based on the textbook Probability: Theory and Examples, 5th edition (Durrett, 2019) and the lecture at Seoul National University, Republic of Korea (instructor: Prof. Johan Lim).

1. Billingsley, 1999, Convergence of Probability Measures, 2nd edition.