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  "body": "\n## Proof\n\nWe show that the complement of $\\limsup_n A_n$ has probability $0$.\n\n1. Let $L = \\limsup_n A_n = \\bigcap_{n \\ge 1} \\bigcup_{m \\ge n} A_m$. By [[De Morgan's laws|0VK0W4D5Y5]], $\\Omega \\setminus L = \\bigcup_{n \\ge 1} C_n$ with $C_n = \\bigcap_{m \\ge n} A_m^c$.\n2. Claim: let $J$ be a finite nonempty set of indices, and let $K \\subseteq J$. Put $E_j = A_j^c$ for $j \\in K$ and $E_j = A_j$ for $j \\in J \\setminus K$. Then $P\\left(\\bigcap_{j \\in J} E_j\\right) = \\prod_{j \\in J} P(E_j)$.\n3. We prove the claim by induction on the size of $K$. For $K = \\varnothing$, it is the definition of [[mutually independent|0GCVEV6BJA]] events.\n4. Let $K \\ne \\varnothing$, and assume the claim for all smaller sets $K$. Pick $l \\in K$, and let $D = \\bigcap_{j \\in J \\setminus \\set{l}} E_j$, with $D = \\Omega$ if $J = \\set{l}$.\n5. By the induction hypothesis, $P(D) = \\prod_{j \\ne l} P(E_j)$ and $P(A_l \\cap D) = P(A_l) P(D)$. So $P(A_l^c \\cap D) = P(D) - P(A_l \\cap D) = (1 - P(A_l)) P(D)$.\n6. By the [[probability of the complement|0PQEBTYDZM]], $1 - P(A_l) = P(A_l^c)$. So the product formula holds for $K$, which proves the claim.\n7. Fix $n \\ge 1$ and $N \\ge n$. By step 2, $P\\left(\\bigcap_{m = n}^{N} A_m^c\\right) = \\prod_{m=n}^{N} (1 - P(A_m))$.\n8. By the inequality [[$1 + x \\le e^x$|012C7GFF20]] with $x = -P(A_m)$, each factor satisfies $0 \\le 1 - P(A_m) \\le e^{-P(A_m)}$. So the product is at most $\\exp\\left(-\\sum_{m=n}^{N} P(A_m)\\right)$.\n9. Since $\\sum_{m \\ge 1} P(A_m) = \\infty$ and the finitely many terms with $m \u003c n$ are finite, $\\sum_{m=n}^{N} P(A_m) \\to \\infty$ as $N \\to \\infty$. So the bound in step 8 tends to $0$.\n10. The events $\\bigcap_{m=n}^{N} A_m^c$ decrease in $N$ to $C_n$. By the [[continuity of probability|0X8HBKCSYB]], $P(C_n) = \\lim_{N} P\\left(\\bigcap_{m=n}^{N} A_m^c\\right) = 0$.\n11. By step 1 and [[Boole's inequality|0EGSEE9J9R]], $P(\\Omega \\setminus L) \\le \\sum_{n \\ge 1} P(C_n) = 0$.\n12. So $P(L) = 1$. An outcome lies in $L$ exactly when, for each $n$, it lies in some $A_m$ with $m \\ge n$, that is, in infinitely many $A_m$. So almost surely infinitely many of the events occur.\n\n## Idea\n\nNot being in $\\limsup_n A_n$ means avoiding all $A_m$ from some index on. By independence, avoiding $A_n, \\dots, A_N$ has probability $\\prod (1 - P(A_m))$. A divergent sum $\\sum P(A_m)$ forces this product to $0$, by the bound $1 - x \\le e^{-x}$.\n",
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