Independence, Fubini's Theorem and Product Spaces $(\Omega_1\times\Omega_2, \mu_1\otimes \mu_2)$ Following is an excerpt from Lecture 7 of these notes, about the independence of random variables. I have some very basic questions about what's going on. Please scroll down to read them.

Independence can always be viewed in a canonical way. Let $(\Omega,\mu)$ be a product space $(\Omega_1\times\Omega_2, \mu_1\otimes \mu_2)$ where $\mu_1$ and $\mu_2$. Suppose $X$ and $Y$ are random variables on $\Omega$ for which the value $X(\omega_1,\omega_2)$ depends only upon $\omega_1$, while $Y(\omega_1,\omega_2)$ depends only upon $\omega_2$. Then any integral (that converges appropriately) can be written as the product of integrals by Fubini's theorem.
$$\mathbb E[f(X)g(Y)] = \int f(X(s)) g(Y(t))\ \mathrm d\mu_1\otimes\mu_2 (s,t)$$
$$= \int f(X(s))\ \mathrm d \mu_1(s) \int g(Y(t))\ \mathrm d\mu_2(t)= \mathbb E[f(X)] \mathbb E[g(Y)]$$


*

*What do we mean by "canonical" here? I didn't get much on searching on Google.

*What is $\mu_1\otimes\mu_2$? I have never seen this notation before. It'd be helpful if someone could explain the product space (or point me to some references) to me.

*Where is Fubini's theorem being used here? From what I know, Fubini's theorem is a prescription for swapping integrals.

*Inside the integrals, I see that the author chose to write $\mathrm d\mu_1(s)$ instead of just $\mathrm d\mu_1$ (similarly in two other places). Is there a specific reason for this? I haven't seen such notation being used before. When integrating some function $f$ with respect to a measure $\mu$, we typically just write $\int_X f\ \mathrm d \mu$.

 A: *

*In mathematics, "canonical" is a word, that describes a construction or a representation of an object, which is in some sense obvious, very handy with respect to calculation and - in many cases - independent of a choice. Maybe have a look into this post.
Here, it refers to the construction of a product of measurable spaces. It obvious in the sense, that the product of the sets $\Omega_1$ and $\Omega_2$ is just the cartesian product $\Omega_1\times\Omega_2$. The niceness of the construction is the property, that the projections
$$\pi_i:\Omega_1\times\Omega_2\to \Omega_i,\quad \pi_i(\omega_1,\omega_2):=\omega_i \quad\text{ for }(\omega_1,\omega_2)\in\Omega_1\times\Omega_2$$
are measurable as functions on the measurable spaces
$$(\Omega_1\times\Omega_2,\mathcal{A}_1\otimes\mathcal{A}_2)\to (\Omega_i,\mathcal{A}_i).$$
Here, $\mathcal{A}_i$ denotes the $\sigma$-algebra on $\Omega_i$ and $\otimes$ refers to the product $\sigma$-algebra on $\Omega_1\times\Omega_2$.


*As Riquelme posted, $\mu_1\otimes\mu_2$ is the product measure of the two measures. It is defined on the product $\sigma$-algebra $\mathcal{A}_1\otimes\mathcal{A}_2$.


*Fubini's theorem is used in the second equation, where the product measure is split into the two measures. Since the functions $(s,t)\mapsto f(X(s))$ and $(s,t)\mapsto g(Y(t))$ only depend on one of the variables, one can integrate each function on its own. Note, that the projections $\pi_1(s,t):=s$ and $\pi_2(s,t):=t$ are themselves only dependent on one of the two variables, and that we could write $(s,t)\mapsto f(X(\pi(s,t)))$ for the first function (and the second respectively). In this sense, the projection functions are - by construction - the independent functions on a product space.


*This simply denotes the dependence of the integrant on the measure. In the simple case of
$$\int_\Omega f(\omega)\,d\mu(\omega)=\int_\Omega f\, d\mu,$$
one omits the dependence of $f$ on $\omega$ because of lazyness. If you want to omit the $(s,t)$ in the example above, you could again use the projections
$$\int_{\Omega_1\times\Omega_2}f(X(\pi_1))g(Y(\pi_2)) \:d(\mu_1\otimes\mu_2).$$
Whether you like this notion or not is up to personal preference, I guess.
