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  "body": "\n## Proof\n\nWe show that each tail event is independent of itself. The tool is Dynkin's π-λ theorem.\n\n1. For each $i$, let $X_i$ take values in a measurable space $(E_i, \\mathcal{E}_i)$. For $k \\ge 0$, let $\\mathcal{F}_k = \\sigma(X_1, \\dots, X_k)$, with $\\mathcal{F}_0 = \\set{\\varnothing, \\Omega}$, and let $\\mathcal{G}_k = \\sigma(X_{k+1}, X_{k+2}, \\dots)$.\n2. Lemma: let $\\mathcal{P}, \\mathcal{Q} \\subseteq \\mathcal{F}$ be [[π-systems|AQN6ZN66NK]] with $P(A \\cap B) = P(A) P(B)$ for all $A \\in \\mathcal{P}$ and $B \\in \\mathcal{Q}$. Then the same holds for all $A \\in \\sigma(\\mathcal{P})$ and $B \\in \\sigma(\\mathcal{Q})$.\n3. Proof of the lemma: fix $B \\in \\mathcal{Q}$ and let $\\mathcal{D}_B = \\set{A \\in \\mathcal{F} : P(A \\cap B) = P(A) P(B)}$. It contains $\\Omega$, since $P(\\Omega \\cap B) = P(B)$.\n4. If $A \\subseteq A'$ are in $\\mathcal{D}_B$, then $(A' \\setminus A) \\cap B = (A' \\cap B) \\setminus (A \\cap B)$. So $P((A' \\setminus A) \\cap B) = (P(A') - P(A)) P(B) = P(A' \\setminus A) P(B)$.\n5. If $A_1 \\subseteq A_2 \\subseteq \\cdots$ are in $\\mathcal{D}_B$ with union $A$, then $A_j \\cap B$ increase to $A \\cap B$. By the [[continuity of probability|0X8HBKCSYB]], $P(A \\cap B) = \\lim_j P(A_j) P(B) = P(A) P(B)$.\n6. By steps 3 to 5, $\\mathcal{D}_B$ is a [[Dynkin system|69460Z6DMW]]. It contains $\\mathcal{P}$ by hypothesis. By [[Dynkin's π-λ theorem|BBHM1A36M7]], $\\sigma(\\mathcal{P}) \\subseteq \\mathcal{D}_B$.\n7. Now fix $A \\in \\sigma(\\mathcal{P})$ and let $\\mathcal{D}'_A = \\set{B \\in \\mathcal{F} : P(A \\cap B) = P(A) P(B)}$. By step 6, it contains $\\mathcal{Q}$. The argument of steps 3 to 5, with the roles exchanged, shows that it is a Dynkin system.\n8. By the π-λ theorem, $\\sigma(\\mathcal{Q}) \\subseteq \\mathcal{D}'_A$. This proves the lemma.\n9. Fix $k \\ge 0$. Let $\\mathcal{P}_k$ be the family of sets $\\bigcap_{i=1}^{k} \\set{X_i \\in B_i}$ with $B_i \\in \\mathcal{E}_i$, where the empty intersection is $\\Omega$.\n10. $\\mathcal{P}_k$ is a π-system, since $\\set{X_i \\in B} \\cap \\set{X_i \\in B'} = \\set{X_i \\in B \\cap B'}$. It lies in $\\mathcal{F}_k$, and it contains each set $\\set{X_i \\in B}$ with $i \\le k$, by taking the other $B_j = E_j$.\n11. So each $X_i$ with $i \\le k$ is measurable for $\\sigma(\\mathcal{P}_k)$. By the minimality of $\\mathcal{F}_k$, $\\mathcal{F}_k \\subseteq \\sigma(\\mathcal{P}_k) \\subseteq \\mathcal{F}_k$, so $\\sigma(\\mathcal{P}_k) = \\mathcal{F}_k$.\n12. Let $\\mathcal{Q}_k$ be the family of sets $\\bigcap_{i=k+1}^{k+j} \\set{X_i \\in B_i}$ with $j \\ge 1$ and $B_i \\in \\mathcal{E}_i$. Two such sets can be written with the same $j$, by adding factors with $B_i = E_i$.\n13. So, as in steps 10 and 11, $\\mathcal{Q}_k$ is a π-system with $\\sigma(\\mathcal{Q}_k) = \\mathcal{G}_k$.\n14. For $A = \\bigcap_{i \\le k} \\set{X_i \\in B_i} \\in \\mathcal{P}_k$ and $B = \\bigcap_{i = k+1}^{k+j} \\set{X_i \\in B_i} \\in \\mathcal{Q}_k$, the [[independence|0XCE070V8B]] of $X_1, \\dots, X_{k+j}$ gives $P(A \\cap B) = \\prod_{i=1}^{k+j} P(X_i \\in B_i) = P(A) P(B)$.\n15. By steps 14 and 2, $P(A \\cap B) = P(A) P(B)$ for all $A \\in \\mathcal{F}_k$ and $B \\in \\mathcal{G}_k$.\n16. Let $\\mathcal{T}$ be the tail σ-algebra. By definition, $\\mathcal{T} \\subseteq \\mathcal{G}_k$ for every $k$. So step 15 gives $P(A \\cap B) = P(A) P(B)$ for $A \\in \\bigcup_k \\mathcal{F}_k$ and $B \\in \\mathcal{T}$.\n17. Let $\\mathcal{U} = \\bigcup_{k \\ge 0} \\mathcal{F}_k$. The $\\mathcal{F}_k$ increase, so two sets of $\\mathcal{U}$ lie in a common $\\mathcal{F}_k$, and so does their intersection. Thus $\\mathcal{U}$ is a π-system.\n18. Each $\\set{X_m \\in B}$ lies in $\\mathcal{F}_m \\subseteq \\mathcal{U}$, and $\\mathcal{U} \\subseteq \\mathcal{G}_0 = \\sigma(X_1, X_2, \\dots)$. As in step 11, $\\sigma(\\mathcal{U}) = \\mathcal{G}_0$.\n19. $\\mathcal{T}$ is a [[σ-algebra|04WSW8T3SY]], as an intersection of σ-algebras. So it is a π-system with $\\sigma(\\mathcal{T}) = \\mathcal{T}$.\n20. By steps 16 to 19 and the lemma, $P(A \\cap B) = P(A) P(B)$ for all $A \\in \\mathcal{G}_0$ and $B \\in \\mathcal{T}$. So $\\mathcal{G}_0$ and $\\mathcal{T}$ are [[independent σ-algebras|09T8ZHST1V]].\n21. Let $A \\in \\mathcal{T}$. Since $\\mathcal{T} \\subseteq \\mathcal{G}_0$, step 20 applies with $B = A$, and gives $P(A) = P(A \\cap A) = P(A)^2$.\n22. The only real solutions of $p = p^2$ are $p = 0$ and $p = 1$. Hence $P(A) \\in \\set{0, 1}$.\n\n## Idea\n\nA tail event does not depend on any finite group $X_1, \\dots, X_k$, so it is independent of all of them. The π-λ theorem upgrades this to independence from the whole σ-algebra $\\sigma(X_1, X_2, \\dots)$. That σ-algebra contains the tail event itself, so the event is independent of itself, and $P(A) = P(A)^2$.\n",
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