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This is literally the theorem from the textbook:

In order for [random variables] $X_1, \ldots, X_n$ to be independent, it is sufficient that for all $x_1, \ldots, x_n \in (-\infty, \infty]$:

\begin{align*} P(X_1 \le x_1, \ldots, X_n \le x_n) = \prod\limits_{i=1}^n P(X_i \le x_i) \\ \end{align*}

Proof: Let $\mathcal{A}_i =$ the sets of the form $\{ X_i \le x_i \}$. Since

\begin{align*} \{X_i \le x\} \cap \{X_i \le y\} = \{X_i \le x \land y \} \end{align*} where $(x \land y)_i = x_i \land y_i = \min\{x_i, y_i\}$. $\mathcal{A}_i$ is a $\pi$-system. Since we have allowed $x_i = \infty$, $\Omega \in \mathcal{A}_i$. Exercise 1.3.1. implies $\sigma(\mathcal{A}_i) = \sigma(X_i)$, so the result follows from Theorem 2.1.7.

Is $\mathcal{A}_i = \{X_i \le x_i\} = X_i^{-1}((-\infty, x_i))$? That seems like a single set rather than a collection of sets. Also, this interpretation doesn't seem to be closed under intersection.

I understand $\{X_i \le x\} \cap \{X_i \le y\} = \{X_i \le x \land y \}$, but how about $\{X_1 \le x_1\} \cap \{X_2 \le x_2\}$. It does't seem that intersection would equal some $\{X_i \le x_i\}$ for some $1 \le i \le n$.

It seems my reading of $\mathcal{A}_i$ is not correct, but I'm not sure how else to read that.

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The definition of $\mathcal A_i$ is $$\mathcal A_i:= \{ \{X_i \leq x_i\}: x_i \in \mathbb R \cup\{\infty\} \}.$$ It isn't just one set, and the proof goes on without intersecting sets of the form $\{X_i \leq x_i\} \cap \{X_j \leq x_j\}$, they just fix $i$ and prove that $\sigma(A_i) = \sigma(X_i)$.

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  • $\begingroup$ The expression $\{X_i \le x_i\}$ can be $\{X_1 \le x_1\}$ for $i=1$ or $\{X_2 \le x_2\}$ for $i=2$. $\{X_1 \le x_2\}$ would not be allowed as the indices are different. Next there are exactly $n$ random variables $X_1, \ldots, X_n$, and $n$ constant numbers $x_1, \ldots, x_n$. Is this right? $\endgroup$
    – clay
    Oct 7, 2021 at 14:08
  • $\begingroup$ For $i=2$, your expression would read $\mathcal{A}_2 := \{\{ X_2 \le x_2 \} : x_2 \in \mathbb{R} \cup \{ \infty \} \}$. That looks like a set of one and only one set. $\endgroup$
    – clay
    Oct 7, 2021 at 14:13
  • $\begingroup$ If I define $\{a : a \in \mathbb R\}$, the set isn't a singleton, in fact, is $\mathbb R$. $x_2$ is a variable as $a$ in the set i mentioned, it isn't a fixed value. If you like, you are intersecting sets of the form $\{X_i \leq a \}$ with $a \in \mathbb R \cup \{\infty\}$. $\endgroup$
    – thewatcher
    Oct 7, 2021 at 19:52

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