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  "body": "\n## Proof\n\nWe bound the event of a large deviation at time $n$ by the event of a large deviation at some time $m \\ge n$.\n\n1. Let $C = \\set{\\omega \\in \\Omega : \\lim_{n \\to \\infty} X_n(\\omega) = X(\\omega)}$. By [[almost sure convergence|04P7D63W8P]], $P(C) = 1$.\n2. Fix $\\varepsilon \u003e 0$. For $n \\ge 1$, let $A_n = \\bigcup_{m \\ge n} \\set{\\abs{X_m - X} \u003e \\varepsilon}$. Each $A_n$ is an event, as a countable union of events.\n3. The sets decrease: $A_1 \\supseteq A_2 \\supseteq \\cdots$, because the union defining $A_{n+1}$ runs over fewer indices. Let $A = \\bigcap_{n \\ge 1} A_n$.\n4. Claim: $A \\cap C = \\varnothing$. Let $\\omega \\in C$. Since $X_m(\\omega) \\to X(\\omega)$, there is $N$ with $\\abs{X_m(\\omega) - X(\\omega)} \\le \\varepsilon$ for all $m \\ge N$.\n5. Then $\\omega \\notin A_N$, so $\\omega \\notin A$. This proves the claim.\n6. By step 4, $A \\subseteq \\Omega \\setminus C$. So $P(A) \\le 1 - P(C) = 0$.\n7. By the [[continuity of probability|0X8HBKCSYB]] from above, $P(A_n) \\to P(A) = 0$.\n8. For each $n$, $\\set{\\abs{X_n - X} \u003e \\varepsilon} \\subseteq A_n$, since $m = n$ appears in the union. So $0 \\le P(\\abs{X_n - X} \u003e \\varepsilon) \\le P(A_n)$.\n9. By steps 7 and 8, $P(\\abs{X_n - X} \u003e \\varepsilon) \\to 0$. Since $\\varepsilon \u003e 0$ was arbitrary, $X_n \\to X$ [[in probability|0EFDDXTVD6]].\n\n## Idea\n\nIf $X_n$ is far from $X$ at time $n$, then it is far from $X$ at some time after $n$. The events \"far at some time after $n$\" shrink to the event \"far infinitely often\", which has probability $0$ when the sequence converges almost surely.\n",
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