Let $X=\bigcup_{i=1}^{\infty}X_{i}$ be a union of mutually independent Poisson point processes on $S\subseteq\mathbb{R}$, and let $B\in\mathscr{B}(S)$ be bounded. Then: $$\tag{1} \mathbb{P}(X\cap B=\emptyset)=\prod_{i=1}^{\infty}\mathbb{P}(X_{i}\cap B=\emptyset). $$
My question: Why does (1) above hold true?
My thoughts: A sequence of random variables $X_{1},\dots,X_{n}$ are said to be independent if their induced sigma algebras $\sigma(X_{1}),\dots,\sigma(X_{n})$ are independent. That is: $$\tag{2} \mathbb{P}\left(\bigcap_{i=1}^{n}A_{i}\right)=\prod_{i=1}^{n}\mathbb{P}(A_{i}),\quad \forall A_{i}\in\sigma(A_{i}). $$ I've been trying to relate (2) to (1), but I can't seem to make the connection. What am I missing??
Suppose $\{X_i\}_{i=1}^{\infty}$ are mutually independent Poisson Point Processes (PPPs) on the real number line with intensities $\{\lambda_i\}_{i=1}^{\infty}$. For any region $B \in \mathcal{B}(\mathbb{R})$ with finite length $|B|$ you can define $N_i(B)$ as the number of points from process $i$ in the region $B$. Then $\{N_i(B)\}_{i=1}^{\infty}$ are mutually independent random variables and $$N_i(B) \sim Poisson(\lambda_i \cdot |B|) \quad \forall i \in \{1, 2, 3, ...\}$$ and the probability that no points from any of the PPPs fall in the region $B$ is: \begin{align} P\left[\cap_{i=1}^{\infty}\{N_i(B)=0\}\right] &= \prod_{i=1}^{\infty}P[N_i(B)=0]\\ &=\prod_{i=1}^{\infty} e^{-\lambda_i \cdot Length(B)}\\ &=e^{-(\sum_{i=1}^{\infty}\lambda_i)\cdot |B|} \end{align}
Here we are using the fact that if $\{X_i\}_{i=1}^{\infty}$ are mutually independent random variables then for any Borel sets $\{C_i\}_{i=1}^{\infty}$ of the real number line and for any positive integer $n$ we have $$ P[\cap_{i=1}^n \{X_i \in C_i\}] = \prod_{i=1}^n P[X_i \in C_i] $$ You can take $n\rightarrow \infty$ (and formally use continuity of probability) to conclude $$ P[\cap_{i=1}^{\infty} \{X_i \in C_i\}] = \prod_{i=1}^{\infty} P[X_i \in C_i] $$ Defining $C_i=\{0\}$ for all $i$ and observing that the event $\{X_i \in \{0\}\}$ is equivalent to the event $\{X_i=0\}$, we get $$ P[\cap_{i=1}^{\infty} \{X_i=0\}] = \prod_{i=1}^{\infty} P[X_i=0]$$
It can be shown that if we define $\lambda = \sum_{i=1}^{\infty} \lambda_i$ and we assume $\lambda<\infty$, and if we define $$ N(B) = \sum_{i=1}^{\infty} N_i(B)$$ then $N(B) \sim Poisson(\lambda)$. In general, the superposition of mutually independent PPPs is again a PPP with the sum rate.