is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$”

Question

Sometimes it seems to me, that when we are considering independent variables on a sample space of the form $\Omega^n$, we use the assumed independence of variables to show that $P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$, where $P'$ is a yet unknown measure on $\Omega^n$.
Whereas I would find it natural (supposing we know of a probability distribution $P$ on $\Omega$) to instead define a distribution on $\Omega^n$, derived from $P$ (as the multiplication of individual probabilities), and then show that our $X_i$'s are indeed independent.
So why is the first way of doing it more commonly used and hot do I show using the independence that $P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$ ? (I managed to do it the other way around and show independence using the above product measure)

To end it with Russell: "My mathematical lecturers never showed me any reason to suppose probability theory was anything but a tissue of fallacies".

Answer 1

As to why it's more commonly used, I can't say...for example, when I was learning this stuff I saw these in the opposite order.

To be clear, for what follows, we are in the case that $\Omega$ is finite (or countably infinite), so that you can recover what $P$ does to arbitrary subsets of $\Omega$ from what it does to singletons (by $P(S) = \sum_{s \in S} P(s)$). (otherwise the implication $P'(\omega_1, \ldots, \omega_n) = P(\omega_1) \ldots P(\omega_n) \Rightarrow X_i$ are independent doesn't hold: if say $n = 2, \Omega = \mathbb{R}$ with $P$ given by integrating a cdf, the probability of any point is 0, so all we know about $P'$ is that it gives no weight to individual points)

For the desired implication, what does it mean that the "variables" are independent? Well, the $i^{th}$ variable $X_i: \Omega^n \rightarrow \Omega$ is simply projection onto the $i^{th}$ coordinate.

Now then, to see what $P'$ of $(\omega_1, \ldots, \omega_n)$ is, we note that this point is exactly the subset of $\Omega^n$ where $X_1 = \omega_1 \ldots X_n = \omega_n$. So by independence $$P'(X_1 = \omega_1 \cap \ldots \cap X_n = \omega_n) = \prod_i P'(X_i = \omega_i)$$

So it looks we want not just the independence of the $X_i$, but also that $P'(X_i = \omega) = P(\omega)$ for all $\omega$ and $i$ to get the desired result.