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  "body": "\n## Proof\n\n1. Let $(\\Omega, \\mathcal{F}, P)$ be a [[probability space|0ZYBZ34RT2]] and let $A, B \\in \\mathcal{F}$ be [[events|00SQGJNRTB]] with $A$ a [[subset|0FE8Q5S42Y]] of $B$.\n2. The set $B \\setminus A = B \\cap A^c$ is in $\\mathcal{F}$, because a [[σ-algebra|04WSW8T3SY]] is closed under complements and finite intersections.\n3. By the [[complement rule|0PQEBTYDZM]] applied to the event $\\Omega$, $P(\\varnothing) = 1 - P(\\Omega) = 0$.\n4. Let $E_1 = A$, $E_2 = B \\setminus A$ and $E_n = \\varnothing$ for $n \\ge 3$. These sets are pairwise disjoint, because no point of $B \\setminus A$ lies in $A$.\n5. Their union is $A \\cup (B \\setminus A)$. This equals $B$, because $A \\subseteq B$.\n6. By the countable additivity of the [[probability measure|0VGMEBVW1M]] $P$ and step 3, $P(B) = P(A) + P(B \\setminus A)$.\n7. All three numbers are real. So $P(B \\setminus A) = P(B) - P(A)$.\n8. The measure $P$ takes values in $[0, 1]$, so $P(B \\setminus A) \\ge 0$. Hence $P(A) \\le P(B)$.\n\n## Idea\n\nWhen $A \\subseteq B$, the event $B$ splits into $A$ and the part of $B$ outside $A$. Additivity gives the probability of $B$ as the probability of $A$ plus a nonnegative remainder.\n",
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