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  "body": "\n## Proof\n\nWe prove continuity from below with disjoint pieces, then continuity from above with differences.\n\n1. Let $(\\Omega, \\mathcal{F}, P)$ be a [[probability space|0ZYBZ34RT2]]. Let $A_1 \\subseteq A_2 \\subseteq \\cdots$ be events and let $A = \\bigcup_{n=1}^{\\infty} A_n$.\n2. Let $F_1 = A_1$ and $F_n = A_n \\setminus A_{n-1}$ for $n \\ge 2$. Each $F_n$ is in $\\mathcal{F}$, because $\\mathcal{F}$ is a [[σ-algebra|04WSW8T3SY]].\n3. The sets $F_n$ are pairwise disjoint. Let $m \u003c n$. Then $F_m \\subseteq A_m \\subseteq A_{n-1}$, because the sequence increases, and $F_n$ has no point in $A_{n-1}$.\n4. We show $P(A_n) = \\sum_{i=1}^{n} P(F_i)$ by induction on $n$. For $n = 1$ this is $P(A_1) = P(F_1)$.\n5. For the inductive step, $A_{n-1} \\subseteq A_n$. By [[monotonicity of probability|0464Z0JP2E]], $P(A_n) = P(A_{n-1}) + P(A_n \\setminus A_{n-1}) = P(A_{n-1}) + P(F_n)$. The induction hypothesis then gives the claim for $n$.\n6. We show $\\bigcup_n F_n = A$. Each $F_n \\subseteq A_n \\subseteq A$.\n7. Conversely, let $\\omega \\in A$ and let $n$ be the least index with $\\omega \\in A_n$. If $n = 1$, then $\\omega \\in F_1$. If $n \\ge 2$, then $\\omega \\notin A_{n-1}$, so $\\omega \\in F_n$.\n8. By the countable additivity of the [[probability measure|0VGMEBVW1M]] $P$, $P(A) = \\sum_{i=1}^{\\infty} P(F_i)$.\n9. The sum of a series is the limit of its partial sums. By step 4, these partial sums are $P(A_n)$. So $P(A) = \\lim_{n \\to \\infty} P(A_n)$. This is the first claim.\n10. Now let $B_1 \\supseteq B_2 \\supseteq \\cdots$ be events and let $B = \\bigcap_{n=1}^{\\infty} B_n$. Let $D_n = B_1 \\setminus B_n$, an event.\n11. The sets $D_n$ increase, because $B_{n+1} \\subseteq B_n$ gives $B_1 \\setminus B_n \\subseteq B_1 \\setminus B_{n+1}$.\n12. We claim $\\bigcup_n D_n = B_1 \\setminus B$. A point $\\omega \\in B_1$ lies outside $B$ exactly when $\\omega \\notin B_n$ for some $n$, that is, when $\\omega \\in D_n$ for some $n$.\n13. By steps 11 and 12 and the first claim, $P(B_1 \\setminus B) = \\lim_{n \\to \\infty} P(B_1 \\setminus B_n)$.\n14. We have $B \\subseteq B_1$ and $B_n \\subseteq B_1$. By monotonicity of probability, $P(B_1 \\setminus B) = P(B_1) - P(B)$ and $P(B_1 \\setminus B_n) = P(B_1) - P(B_n)$.\n15. So $P(B_1) - P(B_n) \\to P(B_1) - P(B)$. Subtracting both sides from the constant $P(B_1)$ gives $P(B_n) \\to P(B)$. This is the second claim.\n\n## Idea\n\nAn increasing union is a disjoint union of the \"new parts\" $A_n \\setminus A_{n-1}$. Countable additivity makes $P(A)$ the sum of the new parts, and the partial sums are exactly $P(A_n)$. For a decreasing sequence, the differences $B_1 \\setminus B_n$ increase, so the first part applies to them. All the numbers are finite, so one can subtract.\n",
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