[proofplan] We prove both directions directly from the definitions. In the forward direction, independence of the events gives the product formula not only for intersections of the $A_i$, but also for intersections in which any $A_i$ is replaced by $A_i^c$; this follows by finite additivity and inclusion-exclusion over the complemented indices. Since every Borel preimage of an [indicator random variable](/page/Indicator%20Random%20Variable) is one of $\varnothing$, $\Omega$, $A_i$, or $A_i^c$, this proves independence of the indicators. In the reverse direction, we test independence of the random variables against the Borel set $\{1\}$, whose preimage under $X_i$ is exactly $A_i$. [/proofplan]

[step:Declare the indicator maps and their Borel preimages] For each $i \in \{1,\ldots,n\}$, define \begin{align*} X_i:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)) \end{align*} \begin{align*} \omega&\mapsto \mathbb{1}_{A_i}(\omega). \end{align*} Since $A_i \in \mathcal F$, the map $X_i$ is measurable: for every Borel set $B \in \mathcal B(\mathbb R)$, \begin{align*} X_i^{-1}(B)= \begin{cases} \Omega, & 0 \in B \text{ and } 1 \in B, \end{align*} \begin{align*} A_i, & 1 \in B \text{ and } 0 \notin B, \end{align*} \begin{align*} A_i^c, & 0 \in B \text{ and } 1 \notin B, \end{align*} \begin{align*} \varnothing, & 0 \notin B \text{ and } 1 \notin B. \end{cases} \end{align*} Each of these four sets belongs to $\mathcal F$. [/step]

[step:Derive the product formula for arbitrary choices of events and complements]Assume that the events $A_1,\ldots,A_n$ are independent. Thus, for every nonempty set $J \subset \{1,\ldots,n\}$, \begin{align*} \mathbb P\left(\bigcap_{j\in J} A_j\right)=\prod_{j\in J}\mathbb P(A_j). \end{align*} We first prove that for every choice of sets $D_i \in \{A_i,A_i^c\}$, indexed by $i \in \{1,\ldots,n\}$, \begin{align*} \mathbb P\left(\bigcap_{i=1}^n D_i\right)=\prod_{i=1}^n \mathbb P(D_i). \end{align*} Let $T=\{i\in\{1,\ldots,n\}:D_i=A_i\}$ and $S=\{i\in\{1,\ldots,n\}:D_i=A_i^c\}$. Define \begin{align*} E_T:=\bigcap_{i\in T}A_i, \end{align*} with the convention that $E_\varnothing=\Omega$. By finite additivity applied to the disjoint decomposition determined by the events $(A_i)_{i\in S}$, equivalently by finite inclusion-exclusion, \begin{align*} \mathbb P\left(E_T\cap\bigcap_{i\in S}A_i^c\right) = \sum_{R\subset S}(-1)^{|R|}\mathbb P\left(E_T\cap\bigcap_{i\in R}A_i\right). \end{align*} For each $R\subset S$, independence gives \begin{align*} \mathbb P\left(E_T\cap\bigcap_{i\in R}A_i\right) = \prod_{i\in T\cup R}\mathbb P(A_i), \end{align*} where the empty product is $1$. Therefore \begin{align*} \mathbb P\left(E_T\cap\bigcap_{i\in S}A_i^c\right) = \sum_{R\subset S}(-1)^{|R|} \prod_{i\in T\cup R}\mathbb P(A_i). \end{align*} Factoring out the terms indexed by $T$ and expanding the finite product over $S$ gives \begin{align*} \mathbb P\left(E_T\cap\bigcap_{i\in S}A_i^c\right) = \left(\prod_{i\in T}\mathbb P(A_i)\right) \left(\prod_{i\in S}(1-\mathbb P(A_i))\right). \end{align*} Since $\mathbb P(A_i^c)=1-\mathbb P(A_i)$, this is exactly \begin{align*} \mathbb P\left(\bigcap_{i=1}^n D_i\right)=\prod_{i=1}^n \mathbb P(D_i). \end{align*}[/step]

[guided]Assume that the events $A_1,\ldots,A_n$ are independent. By definition, this means that every nonempty subfamily satisfies the product formula: \begin{align*} \mathbb P\left(\bigcap_{j\in J} A_j\right)=\prod_{j\in J}\mathbb P(A_j) \end{align*} for every nonempty set $J\subset\{1,\ldots,n\}$. Why do we need more than this? To prove independence of the indicator random variables, we must handle Borel preimages such as $X_i^{-1}(B)$. These preimages can be $A_i$, but they can also be $A_i^c$, $\Omega$, or $\varnothing$. Thus we must show that replacing any of the independent events by its complement preserves the same product formula. Fix a choice $D_i\in\{A_i,A_i^c\}$ for every $i\in\{1,\ldots,n\}$. Let \begin{align*} T:=\{i\in\{1,\ldots,n\}:D_i=A_i\} \end{align*} and \begin{align*} S:=\{i\in\{1,\ldots,n\}:D_i=A_i^c\}. \end{align*} Define the event \begin{align*} E_T:=\bigcap_{i\in T}A_i, \end{align*} with the convention $E_\varnothing=\Omega$. The event whose probability we need is \begin{align*} E_T\cap\bigcap_{i\in S}A_i^c. \end{align*} We now remove the complements by finite inclusion-exclusion. Since $S$ is finite, finite additivity of the [probability measure](/page/Probability%20Measure) gives \begin{align*} \mathbb P\left(E_T\cap\bigcap_{i\in S}A_i^c\right) = \sum_{R\subset S}(-1)^{|R|} \mathbb P\left(E_T\cap\bigcap_{i\in R}A_i\right). \end{align*} Each event inside the probability on the right is an intersection only of events of the form $A_i$, so the assumed independence applies. Hence, for every $R\subset S$, \begin{align*} \mathbb P\left(E_T\cap\bigcap_{i\in R}A_i\right) = \prod_{i\in T\cup R}\mathbb P(A_i). \end{align*} Substituting this into the inclusion-exclusion formula gives \begin{align*} \mathbb P\left(E_T\cap\bigcap_{i\in S}A_i^c\right) = \sum_{R\subset S}(-1)^{|R|} \prod_{i\in T\cup R}\mathbb P(A_i). \end{align*} The factors with indices in $T$ do not depend on $R$, so they factor out: \begin{align*} \mathbb P\left(E_T\cap\bigcap_{i\in S}A_i^c\right) = \left(\prod_{i\in T}\mathbb P(A_i)\right) \sum_{R\subset S}(-1)^{|R|} \prod_{i\in R}\mathbb P(A_i). \end{align*} The remaining finite sum is the expansion of the product \begin{align*} \prod_{i\in S}(1-\mathbb P(A_i)). \end{align*} Therefore \begin{align*} \mathbb P\left(E_T\cap\bigcap_{i\in S}A_i^c\right) = \left(\prod_{i\in T}\mathbb P(A_i)\right) \left(\prod_{i\in S}(1-\mathbb P(A_i))\right). \end{align*} Because $\mathbb P(A_i^c)=1-\mathbb P(A_i)$ for each $i$, this becomes \begin{align*} \mathbb P\left(\bigcap_{i=1}^nD_i\right) = \prod_{i=1}^n\mathbb P(D_i). \end{align*} This proves the product formula for all choices of the events and their complements.[/guided]

[step:Use the preimage classification to prove independence of the indicators] Let $B_1,\ldots,B_n \in \mathcal B(\mathbb R)$ be arbitrary Borel sets. For each $i\in\{1,\ldots,n\}$, define \begin{align*} C_i:=X_i^{-1}(B_i). \end{align*} By the preimage classification above, each $C_i$ is one of $\varnothing$, $\Omega$, $A_i$, or $A_i^c$. If some $C_i=\varnothing$, then \begin{align*} \mathbb P\left(\bigcap_{i=1}^n C_i\right)=0=\prod_{i=1}^n\mathbb P(C_i). \end{align*} If no $C_i$ is empty, then each $C_i$ is one of $\Omega$, $A_i$, or $A_i^c$. The indices with $C_i=\Omega$ do not affect either the intersection or the product, because intersecting with $\Omega$ changes no set and $\mathbb P(\Omega)=1$. Applying the product formula from the previous step to the remaining indices gives \begin{align*} \mathbb P\left(\bigcap_{i=1}^n X_i^{-1}(B_i)\right) = \prod_{i=1}^n\mathbb P\left(X_i^{-1}(B_i)\right). \end{align*} Since the Borel sets $B_1,\ldots,B_n$ were arbitrary, the random variables $X_1,\ldots,X_n$ are independent. [/step]

[step:Recover event independence by testing the indicators at the value $1$] Conversely, assume that the random variables $X_1,\ldots,X_n$ are independent. Let $J\subset\{1,\ldots,n\}$ be nonempty. For each $i\in\{1,\ldots,n\}$, define the Borel set \begin{align*} B_i:= \begin{cases} \{1\}, & i\in J, \end{align*} \begin{align*} \mathbb R, & i\notin J. \end{cases} \end{align*} Then \begin{align*} X_i^{-1}(B_i)= \begin{cases} A_i, & i\in J, \end{align*} \begin{align*} \Omega, & i\notin J. \end{cases} \end{align*} Using independence of the random variables with these Borel sets gives \begin{align*} \mathbb P\left(\bigcap_{i=1}^n X_i^{-1}(B_i)\right) = \prod_{i=1}^n\mathbb P\left(X_i^{-1}(B_i)\right). \end{align*} Substituting the identified preimages, \begin{align*} \mathbb P\left(\bigcap_{j\in J}A_j\right) = \left(\prod_{j\in J}\mathbb P(A_j)\right) \left(\prod_{i\notin J}\mathbb P(\Omega)\right). \end{align*} Since $\mathbb P(\Omega)=1$, this becomes \begin{align*} \mathbb P\left(\bigcap_{j\in J}A_j\right) = \prod_{j\in J}\mathbb P(A_j). \end{align*} Because $J$ was an arbitrary nonempty subset of $\{1,\ldots,n\}$, the events $A_1,\ldots,A_n$ are independent. [/step]